Physics for second secondary

Transcription

1 Physics for second secondary

2 Particle Physics Science concerned with the study of particles and waves ~ 1 ~ Wave - Has mass - Energy with no mass or volume Volume - Waves reflect, refract, diffract & interfere. Motion Mechanical Electromagnetic Charge - Need a medium to - Need no media to travel through. travel through. - Solid, Liquid, Gas - travel through space. Ex : Electron - Sound waves - Light waves. Fundamental Physical quantities Derived - Mass ( M ) - Can be expressed in terms of - Length ( L ) fundamental ones. - Time ( T ) Ex: Volume V = S x S x S = L x L x L = L 3. Measured in (m 3 ). Measuring systems of units Gausial Metric S.I. - Mass g kg kg - Length cm m m - Time s s s s... Velocity V... t... V... Acceleration a t... Force Work F W Density m a F m Vol S

3 Conversions : Big unit x 10 + prefix Small unit Small unit x 10 - prefix Big unit Ex : Kg x 10 3 g g x 10-3 kg - Prefixes for multiples of the S.I. of units Multiple Prefix Symbol Multiple Prefix Symbol factor factor tera T 10-2 centi c 10 9 giga G 10-3 milli m 10 6 mega M 10-6 micro 10 3 kilo K 10-9 nano n 10 2 hecto h pico p 10 1 deka da femto f 10-1 deci d atto a N.B: Angstrom A o = m Forces & weights are measured in Newton = kgm/s 2 ( corresponding to dyne ) Work & energy are measured in Joule = kgm 2 /s 2 ( corresponding to Erg ) Only similar physical quantities can be added or subtracted. Ex: 10 kg + 20 Kg = 30 kg. 10 m + 20 cm ( cannot be added ). - Calculate the force acting on a body of 5 kg. to accelerate it by 10 cm/s 2. ~ 2 ~

4 Conversion to S. I. of units - cm x 10-2 m L ( length) - cm 2 x 10-4 m 2 A ( area ) - cm 3 x 10-6 m 3 Vol ( Volume ) - Liter x 10-3 m 3 Vol ( Volume ) - gm x 10-3 kg M ( Mass ) - hr x 60 2 sec t ( Time) - km/h x 5/18 m/s v ( velocity) - cm/s 2 x 10-2 m/s 2 g or a ( acceleration ) - Complete: km = 1000 x 10 3 = 10 6 m cm =... =... m kg =...=... g gm/cm 3 =...=... kg/m cm 2 =.=... m cm 3 = = m Liter =..= m nm =....=... m. - Mass : is the quantity of material forming a body. - Measured in kg. ( kg.w.) - Weight : is the gravitational force acting on a body. - Measured in N ( as it is a force ) - F g = m x g ( g is the acceleration due to gravity = 9.8 m/s 2 ). Ideal devices : Are those which conserve energy.(respect the conservation of energy) Input energy = Output energy Non-ideal devices: Are those which lose energy while working. Efficiency of a device = 100 % output energy input energy ~ 3 ~

5 Geometrical shapes: Areas: Volumes: - Square L x b = side 2 - Cube L x b x h = side 3 - Rectangle L x b - Cuboid L x b x h - Circle r 2 - Cylinder r 2 h - Sphere 4 r 2 - Sphere 4 3 r 3 - Triangle 1 2 B x h - Prism 1 2 B x h x L The Greek alphabet Alpha Beta Gamma Delta Epsilon Zeta Eta Theta Iota Kappa Lambda Mu Nu Xi Omicron Pi Rho Sigma Tau Upsilon Phi Chi Psi Omega Graph: - is a relation between two variables where one is expressed on the ( X ) axis (abscissa ) while the other on the ( Y ) axis ( ordinate ). - Once the co-ordinates are located & connected we get a slope. Y Y Y tan X X X The value of the Slope 2 1. Y - If the slope is a straight line then its value is constant, Δ Y while if it is not then it represents a variable quantity. ΔX θ X - The value of the slope increases when ( ) increases. - The scale of the X axis is independent of that of the Y axis. ~ 4 ~

6 - How to make a scale? A scale is a value chosen to be represented on the squares of the graph paper & corresponds to a specific value for each variable, keeping the value of each square constant. Ex: ( ) ( Scale of 1 : 10 ) - If the numbers representing the variables in the table contain decimals as ( ), then 1. If the difference between each 2 successive values is constant including ( 0 ), this constant can be used as a scale. 2. The numbers may be all multiplied or divided by the same power 10 in order to remove the point & make the numbers simpler. ( X 10 ) 3. This may also make some decimals useless & turn the values easier to be drawn ( x 10 ) this makes the third number after the point useless if a scale of ( 1 ) is used for each square. 4. To choose the right scale the maximum number needed according to the table must be divide by the number of squares in the paper thus to be sure that the paper will suit the graph. Ex: A maximum of 400 : a scale of 20 = 20 squares. If the paper doesn t carry them then we have to increase the scale. ~ 5 ~

7 ~ 6 ~

8 UNIT ( 1 ) CHAPTER ( 1 ) WAVE MOTION * The wave: Is a disturbance that propagates through a medium transferring energy in the direction of its propagation. - May travel in straight lines in all directions forming concentric spheres around its source. Ex: When a stone is dropped in calm water energy is transferred through the water particles in all directions around the stone forming concentric circles ( wave ) Types Mechanical Electromagnetic - Is a disturbance that propagates - Is a wave which needs no media, but through a medium. can propagate through space. - May be transverse or longitudinal - Are all transverse. - Water waves - Light waves - Sound waves - Radio waves, x ray and gamma ray. * Mechanical waves: - Disturbance that propagate through a medium transferring energy in the direction of its propagation. -Conditions to produce mechanical waves: 1- Source of disturbance or vibrator as a simple pendulum, tuning fork, stretched wire or a bob held in a spiral spring. 2- Kind of disturbance able to be transferred to the medium. 3- Kind of medium that carries the disturbance. 2A A A ~ 7 ~

9 Physical quantities describing a wave: 1- Displacement (y-x-s-d) S(cm) - Is the distance between the rest position of a vibrating C λ body and its position at any instant. - Distance covered in a certain direction. T 2. Amplitude (A) d - Maximum displacement for a vibrating body. A T t(sec) - Distance between two points where the velocity is maximum at one and 0 at the other. 3. Complete vibration (wave form) (λ) - Motion performed by a vibrating body to pass through a fixed point two successive times in the same direction. N.B: During a complete vibration the body covers 4 A. EX: During a complete vibration, a body covers 20 cm then it s A = 4. Frequency ()(f) 20 5cm 4 - Is the number of complete vibrations covered by a vibrating body per second. ( ) 1.. Frequency ( ) number of complete vibrations oscillations ( Hertz vibr cycles oscill ) t ( sec ond) T sec sec sec 5. Periodic time (T) - Time taken to perform one complete vibration. time(sec) 1 T seconds number of oscillatio ns 6. Distance (d): - Is the distance covered by the wave in a certain direction. d number of waves. 7. Velocity ( V ): - Is the distance covered by the wave per unit time ( Sec.) d V m( cm) t s ~ 8 ~

10 - A mechanical wave is a disturbance that propagates through a medium, this takes place by transferring the disturbance of the source to the medium particle which vibrate similarly and transfer the motion to the next particles in contact and so on. - This results in that the disturbance (vibration) travels in the medium as wave motion in all directions as concentric spheres ( circles for water waves). N.B: - A wave always begins with a crest, the first crest emitted (in the series) is called the wave front and constitutes the outer sphere. * Remember: 1.What is meant by : 1. Mechanical wave. - Is a... that propagate through a medium transferring... in the direction of its propagation. 2. The amplitude of vibration is 2 cm. - 2 cm is the... for a vibrating. - 2 cm is the distance between two successive points where the velocity of the vibrating body at one is... and at the other The distance covered by a vibrating body in a certain direction is 3 cm. - 3 cm is the... of the vibrating body. 4. The distance between 2 successive point along the path of a vibrating body where its velocity at one is zero & at the other is maximum is 5 cm. - 5 cm is the... of vibration. 5. The frequency of vibration of a body is 50 Hz ( Vibr/s ). -50 (vibrations) is the number of... per sec. 6. The periodic time is 0.1 sec sec is time taken to perform... complete vibration. ~ 9 ~

11 7. S Crest 10 A t trough 4 cm d - From the data of the previous figure find: - The wavelength = - Amplitude = - Periodic time = - Wave velocity = - Frequency = * Types of mechanical waves: 1- Transverse wave: (Demonstration) - Fix a rope from one end. - Catch the other end and move your hand up & down, you note that a wave travels along the rope. Or Let a mass (M) attached to a vertical spring & a horizontal rope move vertically in a harmonic motion & observe the wave flowing through the rope Crest: Is the point having the maximum displacement in the positive direction. Trough: Is the point having the maximum displacement in the negative direction. Transverse wave: Is the wave where the medium particles vibrate perpendicular to the direction of the wave propagation. Consists of successive crests and troughs. ~ 10 ~

12 2- Longitudinal wave:(demonstration) - Catch one end of a spring suspended horizontally. - Vibrate your hand, as it moves to the right, the turns of the spring are brought close. This is called Compression. - As it moves left the turns of the spring move further apart, this is called Rarefaction. Or use a mass (M) moving on a smooth surface Freely between two springs as in the fig. *Compression: is the position where the particles of the medium get closer from each other than their normal positions. * Rarefaction : is the position where the particles of the medium get far apart from each other than their normal positions. * Longitudinal wave : Is the wave in which the medium particles vibrate along the direction (parallel to) the wave propagation. - Consists of successive compression and rarefaction. N.B:- Sound waves are transferred in air as longitudinal waves. ~ 11 ~

13 Transverse wave Longitudinal wave Crest Trough -Is the wave where the particles of the - Is the wave where the particles of the medium vibrate perpendicular to the wave medium vibrate along ( parallel to) the propagation direction. wave propagation direction. -Consist of crests and troughs -Consist of compressions and rarefactions -λ: Is the distance between two successive -λ: Is the distance between the centres of crests or troughs. two successive compressions or rarefactions. Ex: Water waves. Ex: Sound waves. N.B:- Transverse waves are more common in liquids and solids than in gases since they need a strong intermolecular attraction force between the molecules to occur while that of gases is weaker than the minimum force needed. - In solids both longitudinal and transverse waves may travel but longitudinal waves are faster because of the intermolecular attraction forces which are strong and spaces are nearly absent and longitudinal waves vibrate the particles along the wave propagation direction. Relation between Frequency, wave length and wave velocity: Wave length ():- Is the distance between 2 successive Crests or troughs. - Is the distance between the centers of 2 successive compressions or rarefactions. - Both A and C or B and D are moving in the same direction with the same velocity at the same time. * They are said to be in Phase. ~ 12 ~

14 Wave length: Is the distance between any 2 successive points having the same phase. Frequency: Is the number of complete vibrations per second. Wave velocity: Is the distance covered by the wave per unit time. -Relation between the time (t) or the periodic time (T) and the wavelength of a wave. * Wave propagation law: -If (x) is the distance covered by the wave in a time(t). Therefore and If Therefore X V t X V X t t T where T 1 T Therefore V ( Wave propagation law) Wave velocity = Frequency x wave length. N.B : - As the same kind of wave keeps its velocity constant in the same medium 1 then for two waves of the same kind travelling in the same medium. where υ υ V λ 1 ~ 13 ~

15 - As a wave keeps its frequency constant while passing from a medium to another while the velocity changes according to the medium density then V Where V V V υ Guide to answer the question (What is meant by?): λ N.B:1. Always answer with...cm is the wavelength of a... wave. 2. Find the wavelength as d. num. of waves 3.If the two points are similar the number of waves = the difference between their numbers. ( ex: the distance between the 1 st and the 5 th crests, means 5-1=4 waves) 4. If the distance is between the beginning of the 2 nd wave and the end of the 4 th, modify the last to be the beginning of the 5 th and follow the previous rule. 5. If the distance is between the 1 st crest (compression) and the 4 th trough (rarefaction), the number of waves is found as: ((4-1) + ½ ), and if it is between The 2 nd trough and the 4 th crest, the number of waves is found as: ((4-2) ½ ) * Remember: 1.What is meant by : 1. The wavelength of a transverse wave is 1 cm. - 1 cm is the distance between 2 successive The distance between the first and the third crests of a wave is 4 cm is the wavelength of... wave. 3. The distance between a crest and the next trough in a wave is 1 cm is the wavelength of... wave. 4. The distance between the beginning of the 1 st wave and the end of the 4 th wave is 5 cm is the wavelength of the wave. ~ 14 ~

16 5. The wavelength of a sound wave is 2 cm. - 2 cm is the distance between the centers of 2 successive... or The wavelength of a water wave is 4 cm. - 4 cm is the distance between 2 successive... or The wavelength of a wave is 3 cm. - 3 cm is the distance between 2 successive The region where the particles of the medium get far apart than their normal positions. - Rarefaction in a longitudinal wave. 9. The region where the particles of the medium get closer to each other than normal. - Compression in a longitudinal wave. 10. The distance between the first crest and the third trough of a wave is 2.5 cm is the wavelength of transverse wave. 2.Give reasons:- 1. Sound requires a material medium to be transmitted. Bec. sound waves are mechanical longitudinal waves which are transmitted by vibrating the particles of the medium. 2. A light wave doesn t require a material medium to travel. Bec. light waves are electromagnetic waves which can travel through space. 3. Sound wave is a longitudinal wave. Bec. the vibrations of the medium particles are parallel to the direction of the wave propagation. 4. Astronauts use radio transmitters to communicate. Bec. sound waves are longitudinal mechanical waves which require a material medium to travel through & can't travel through vacuum. ~ 15 ~

17 5. Transverse waves are more common in liquids & solids than in gases. Bec. to be transmitted the transverse waves require a minimum intermolecular attractive force between the molecules, higher than the one found in gases. 6. Longitudinal waves are faster in solids than liquids and gases. Bec. The intermolecular spaces in solids are narrower than liquids and gases. * Solved examples: 1. The number of water waves passing a certain point per sec. is 12, each of wavelength 0.1 m.find the wave velocity of water v = v = 0.1 x 12 = 1.2 m/s 2. The light wave vel. in space is km/s, given that the wavelength of light is 5000 A 0. Find frequency of light. v = c = 3 x 10 8 m/s = 5000 x = 5 x 10-7 m v = therefore 8 V Hz 3. Two waves propagate through a medium with a similar speed. Knowing that the frequency of the first is 256 Hz and the ratio between their wavelengths is 1 : 2 respectively, find the frequency of the second. V V Hz ~ 16 ~

18 * Waves: -Frequency:- T T Periodic time:- noof vibr. 1 t T Rules time 1 T number of oscillations - Velocity:- V d t - Wave propagation law:- V - - V V at constant frequency at constant velocity Distance covered by wave:- d no of vibr. ~ 17 ~

19 Homework 1-Choose the correct answer: 1. Sound is an example of. ( Electromagnetic wave longitudinal wave transverse wave ) 2. What kind of these waves can transfer in space..... ( Light waves Sound waves Water waves ) 3.The distance travelled by a vibrating body during one complete period equal to. the amplitude of vibration. ( 2 times 4 times equal to) 4. A pendulum is pulled aside & released to oscillate freely so it takes 5 s. to move from X to Y, the frequency of oscillation of the pendulum is ( 50 Hz 10 Hz 5 Hz Hz Hz ). X Y 5. A pendulum oscillates in a simple harmonic motion as represented by the figure below. The frequency of the pendulum is... ( 0.5 Hz 2 Hz 3 Hz ). displacement time (sec) 6. The time taken by the vibrating body to move from its rest position to maximum displacement position is... the periodic time. ( half equal to quarter ). 7. The figure below represents a wave propagating along a string with speed of 330 m s X (cm) +1 cm -1 cm distance (cm) ~ 18 ~

20 a) The frequency of the wave is.. (1280 Hz 640 Hz 320 Hz 8250 Hz 400 Hz). b) The amplitude of the wave is (1 cm 2 cm 4 cm 8 cm 16 cm). 8. The figure represents a longitudinal wave moving through water in a glass tank of length 9 m. the frequency of the wave is 500 Hz. a) This wave could be which of the following types of waves.. (visible light radio waves micro wave sound wave x-ray). b) The wavelength of the longitudinal wave is.. ( 1.5 m 3 m 4.5 m 9 m 18 m). c) The speed of the wave is most nearly. (500 m/s 750 m/s 1500 m/s 3000 m/s 4500 m/s). 9. which point in the wave shown in the below diagram is in phase with point A... C F A B D E (A B C D E). 10. A girl on the beach watching water waves, sees 4 waves passes in 2 seconds, each with a wavelength of 0.5 m. the speed of the wave is.. (0.25 m/s 0.5 m/s 1 m/s 0.2 m/s 4 m/s). 11. The wave on a lake causes a buoy to oscillate up and down 90 times per minute. The frequency of the waves in hertz is... (90 Hz 60 Hz 1.5 Hz 0.6 Hz). 12.When the frequency of the same wave decreases in the same medium the... (wavelength increases wavelength decreases speed increases speed decreases - wavelength decreases & speed increases). ~ 19 ~

21 13. Dolphins communicate using ultrasonic waves of Hz. If the sound speed in water is 1500m/s. then the wavelength of these waves is... (10 m 1 m 0.1 m m m). 14. When a wave travels from a medium to another, the frequency of the wave (remains constant increases decreases). 2- State the scientific term:- 1- The distance between two points where the velocity is maximum at one and zero at the other. (.) 2- Disturbance that propagate through a medium transferring energy in the direction of its propagation. (....) 3- The maximum displacement of a vibrating body. (....) 4- The number of complete vibrations covered by a vibrating body per second. ( ) 5- The distance covered by the wave per unit time. (....) 6 - Motion performed by a vibrating body to pass through a fixed point two successive times in the same direction. (...) 7- The distance between the rest position of a vibrating body and its position at any instant. (.. ) 8- Time taken to perform one complete vibration. (..) 9- The wave in which the medium particles vibrate along the direction the wave propagation. (......) 10- The point having the maximum displacement in the negative direction. (.. ) 11- The wave where the medium particles vibrate perpendicular to the direction of the wave propagation consists of successive crests and troughs. (.....) 12- The distance between any 2 successive points having the same phase. (..... ) 13- The position where the particles of the medium get far apart from each other than their normal positions. (....) ~ 20 ~

22 3-What is meant by: 1. The amplitude of a wave is 2 cm The frequency of a vibrating body is 100 Hz. & calculate its periodic time The periodic time of a vibrating body=0.01 s Complete vibration A body is doing 1200 complete vibration in one minute The region where the particles of the medium get far apart than their normal positions The region where the particles of the medium get closer to each other than their normal positions The distance between two successive crests in a wave is 2 cm.... ~ 21 ~

23 9.The distance between a compression and next rarefaction = 2 m The distance between the beginning of the first and that of the third wave is 4 cm The distance between the beginning of the first wave and the end of the fifth is 5 cm The wave length of a water wave is 3 cm The speed of a wave = 3 m/s The wavelength of a sound waves=2 m Complete: 1. Mechanical wave is...such as Electromagnetic wave is a disturbance which propagates in as well as in a Sound waves are transferred in as.. waves. 4. The frequency is measured in Motion performed by a vibrating body to pass through a fixed point two successive times in the same direction is called The product of frequency x periodic time = A... wave requires a material medium to be transferred. 8.. is the maximum displacement for a vibrating. ~ 22 ~

24 9. For the shown simple harmonic motion describe the complete path of one complete vibration:- a. Starting from a b b. Starting from d a waves are longitudinal, while... waves are transverse. 11. The transverse waves consists of and. 12. The wave in which the medium particles vibrate parallel to the direction of the wave propagation consists of successive..and The crest is. 14. The distance between two successive crests can be calculated from the relation The distance between two points having the same phase is called The.....is the product of the frequency x wave length. 17. The wavelength of a wave periodic time = 18. The frequency of a wave in a certain medium is doubled then its wavelength is... 5-Give reason for: 1. Astronauts use radio transmitters to communicate A light wave doesn t require a material medium to travel Sound propagates in gases in the form of longitudinal waves only.... c d a b ~ 23 ~

25 4. Transverse waves are more common in liquids & solids than in gases Compare between: 1.Mechanical and electromagnetic waves. 2.Transverse and longitudinal waves. 3. Crest and trough. 4. Compression and rarefaction. ~ 24 ~

26 7-Essay questions: 1. A stretched string is attached at one end to a vibrating tuning fork. Illustrate with drawing: a) The propagation of a pulse crest. b) The propagation of a pulse trough. c) The propagation of one complete transverse wave. 2. Draw a diagram of a longitudinal wave and a transverse wave having the same frequency and wavelength. 3. In the following figures when the tuning fork oscillates: A a) What type of mechanical wave passes in:- - String (A) - Spiral spring (B) B ~ 25 ~

27 4. Deduce the relation between the frequency, wave length and wave velocity. * Problems: 1. A generator produces 16 vibrations in 4 s. Calculate its periodic time & its frequency. ( 4 Hz 0.25 sec ) 2. S C 20 A t T 10 cm d - From the data of the previous figure find: The number of waves, the wavelength, the frequency, the wave velocity, the amplitude, the periodic time, the distance covered by 5 waves, the displacement after 5 seconds, then state what does each of the symbols ( A, C & T ) represent. ~ 26 ~

28 3. An oscillating source of frequency 100 Hz. Find the time between the passage of the first crest and the 20 th one through a point in the direction of the propagation of the wave. ( 0.19 sec.) 4. A transverse wave of 50 Hz. is moving along a thread. Knowing that the distance between a crest & a trough is 0.5 m. Calculate the period of the wave & its wavelength. ( 0.02 sec 1 m. ) 5. If the time taken to accomplish 100 vibrations is 20 sec. & the wave produced has a velocity of 20 cm/sec. Calculate the period, frequency, & wavelength of the wave. ( 0.2 sec 5 Hz 0.04 m. ) ~ 27 ~

29 6. If 6 waves pass a certain point in 60 sec. and the distance between the beginning of the 1 st wave and the end of 5 th waves was 75 m. Calculate periodic time, frequency, Wavelength and velocity of propagation. ( 10 sec 0. 1Hz 15 m 1.5 m/s ) 7. A train at rest produces a whistle of 300 Hz. Knowing that the waves reach a man standing 0.99 km. apart in 3 sec. Calculate the wavelength in meter. ( 1.1 m ) 8. A radio transmitter sends waves of 1 MegaHz. Calculate the wavelength if you know that the sound speed is 300 thousands m/s. ( 0.3 m. ) ~ 28 ~

30 9.Calculate the number of waves produced by a tuning fork of 165 Hz. Before its sound reaches a man 100 m away from it, knowing that the sound speed in air is 330 m/s. ( 50 vibr. ) 10. A Wireless radar station emits waves of velocity m sectowards a satellite, after 0.03 second the same station receives back the waves. Calculate the distance between the station and the satellite. ( m.) 11.If the light velocity in space is 3x10 5 km/s & the wavelength of its wave is 5000 A o.find the frequency of light. ( 6X10 14 Hz. ) 12. A thunderstorm takes place 405 km away from an observer. Find the time interval 8 between seeing the lightning & hearing the thunder ( 330 m s, c 310 m s) sound ( sec.) ~ 29 ~

31 13. Two sound waves propagate through air. Knowing that the frequency of the first is 256 Hz. & the ratio between their wavelengths is 1:2 respectively. Find the frequency of the second ( 128 Hz. ) 14. A school bell sends waves of 102 vibr. /sec. Knowing that it sends 12 vibrations before hearing its sound at the nearest class, & that the sound speed in air is 340 m/s. Calculate the distance between the bell & the nearest class. ( 40 m ) 15. Using the data on the graphs below find: v λ λ 1/υ - The slope of each of them and the mathematical relation relating the variables. 16. The opposite graphs show the variation of frequency ( ) υ A with ( 1 ) for a wave in two different media (A) and (B). B In which medium the velocity of the wave is faster? 1 ~ 30 ~

32 17. The relation between the frequency of some sound waves in ( Hz ) and their wavelengths in ( m ) is given in the following table. λ (m ) x 20 ν (Hz) Y Draw the graphical relation between the frequency on the Y axis and the reciprocal of the wavelength on the X axis, then from the graph find : 1. The value of each of ( X ) and ( Y ). 2. The sound velocity in air. ~ 31 ~

33 ~ 32 ~

34 - Light is a form of energy essential for life. CHAPTER ( 2 ) LIGHT AS A WAVE MOTION - Light travels as waves in all directions in straight lines. - Light has the general properties of waves, travels at a speed c = 3 x 10 8 m/s in space and constitutes the visible part of the spectrum of electromagnetic waves. - An electromagnetic wave consists of an electric field perpendicular to a magnetic field and both are perpendicular to the wave propagation direction. Light reflection & refraction: - Let a light ray fall on a glass block. - A part of the light is reflected & apart is refracted depending on the difference in the optical densities of the two media (ability to allow light to pass) ( as optical density increases less light penetrates) - Measure both angles of incidence and reflection. - The angle of incidence (φ) is equal to the angle of reflection (φ). ( 1 st law of reflection). - The incident ray, reflected ray and the normal are all on the same plane perpendicular to the reflecting surface. ( 2 nd law of reflection). - Measure both, the angle of incidence & that of refraction. - Repeat the experiment and re-measure the new angles. Sin Φ - Find the ratio sin V1 const. 1n2 sin V 1n 2 2-1n 2 is a constant called the refractive index between the 2 media. - Incident ray, refracted ray and the normal are all on the same plane perpendicular to the interface separating the 2 media. Ф Reflected ray Refracted ray sin θ ~ 33 ~

35 Where sin V sin V n ( refractive index from medium (1) to medium (2) 1 2 and light refracts passing different media as it changes its velocity due to the change of the medium density. i.e: The refractive index between 2 media ( 1 n 2 ): is equal to the ratio between the light velocity in the first medium to its velocity in the second medium. Or: The ratio between the sin of the angle of incidence to the sin of the angle of refraction of light. N.B: The light velocity in a denser medium is less than in a less dense one as V 1 -Less dense to denser VV 1 2 Denser to less dense V V refract closer to the norm. 2 refracts away from the norm. - The velocity of light in space is constant, equal to 3x10 8 m/s and is denoted by C. and is approximately equal to its velocity in air and is greater than its velocity in any other medium. - If the velocity of light in any other medium is denoted by ( V ). c Therefore the refractive index of light passing from space to another medium is 1n2 V This is called the absolute refractive index of the medium (n). c light velocity in space ( air ) n V light velocity in a medium c N. B: The absolute refractive index of air = 1, as c v and nair 1 V air ~ 34 ~

36 Absolute refractive index: - Is the ratio between the velocity of light in space to its velocity in the medium. - Is the ratio between the sin of the angle of incidence in space and the sin of the angle of refraction in the medium. - Consider 2 media where light velocities are v 1 and v 2 respectively. - Their absolute refractive indices will be : n 1 c V 1 and n2 V1 V1 Where and 1n2 n V V 1 2 n 1 2 n n This is the relative refractive index from the first medium to the second and is equal: the ratio between the absolute refractive index of the second medium to that of the first one. 2 1 Or: The ratio between the sin of the angle of incidence to the sin of the angle of refraction of light. Or: The ratio between the light velocity in the first medium to its velocity in the second medium. Snell s law. sin n 2 1n2 and 1n2 n1 n sin n1 n 2 2 sin sin or n sin n sin ( Snell ' s Law) c V 2 N.B : n 1 2 V1 sin n2 1 2 V sin n ~ 35 ~

37 What is meant by: 1. Optical density of a medium. Ability of the medium to 2. The refractive index between water and glass is is the ratio between the light velocity in. to its velocity in 1.2 is the ratio between the sin of the angle of... in... to the sin of the angle of... in The absolute refractive index of glass is is the ratio between the light velocity in... to its velocity in.. 4. The relative refractive index between water and glass is is the ratio between the absolute refractive index of. to that of 1.2 is the ration between the light velocity in.. to its velocity in.. 8 Solved examples: ( C Vair 3 10 m ) s 1. In each of the following figures where the ray falls parallel to mirror (Y): The angle of incidence of the ray reflected from mirror (X) on mirror (Y) =.. The angle of incidence of the ray reflected from mirror (Y) on mirror (X) =.. X 45 X Y Y 2. If the light velocity in glass is 2x10 8 m/s. Calculate the refractive index of glass. 30 n glass 8 C V 210 glass 3. If light falls on a glass block whose refractive index is 1.5, at an angle of 30 o. Find the angle between the reflected and the refracted rays. angle of reflection angle of incidence 30 n glass n air 0 sin sin1 sin 30, sin, 20 sin n 1.5 glass anglebetween reflected and refracted rays o o o ~ 36 ~

38 4. If the absolute refractive index of water is 4 3 and that of glass is 3 2, Calculate: The relative refractive index from water to glass. The relative refractive index from glass to water. w g n n g w n 3 g n 4 w nw n 3 g A light wave whose wavelength is 600 nm. falls on a liquid forming an angle of 30 o with its surface, knowing that the angle of refraction of that wave is 45 o find its wavelength in the liquid and the refractive index of that liquid. 1 sin 600 sin 60,, nm sin sin 45 n n 2 2 liq air sin nliq sin 60,, nliquid 1.22 sin 1 sin 45 ~ 37 ~

39 Light interference: Double slits experiment: Thomas young: - A source of monochromatic light sends cylindrical waves through a slit, sending the waves to a screen with 2 narrow slits S 1 and S 2. - These slits lying on the same wave front act as 2 sources in phase (emitting waves of the same frequency and amplitude.) - These sources are coherent which send similar cylindrical waves on a third screen. - An interference of wave patterns is observed as illuminated and dark regions called interference fringes since waves are superposed. - If : is the wavelength of monochromatic light R : is the distance between the slits and the third screen d : is the distance between the 2 slits - y distance between 2 successive similar fringes is calculated as: N.B: The experiment can be used to: - Prove the interference of light. R Y d - Find the frequency of a monochromatic light by knowing the wavelength. N.B: - Constructive interference - Destructive interference Where Path difference m where 1 Path difference ( m ) 2 and ( m is an integer = 0,1,2, ) - The waves aren t in phase or the - The waves are in phase. Phase difference = 180 o. ~ 38 ~

40 Constructive interference Destructive interference Takes place when compression meets Takes place when a compression meets a compression, rarefaction & rarefaction, a rarefaction or a crest meets a trough. crest meets crest or trough & trough. Both waves reinforce each other Both waves weaken each other The path difference = m λ The path difference = ( m + ½ ) λ N.B: for a path difference = 10 m when λ = 0.5 m ( interference is constructive as the number of waves = 10/0.5 = 20 λ.) ( integer). Light diffraction: - When light passes through a circular opening and is obtained on a screen, a circular spot is obtained. - This spot is found to have bright and dark fringes encircling it forming (the Airy s disc). This is due to that light diffracts when passing through the circular opening where each point acts as a light source which sends secondary waves which interfere. N.B: diffraction increases as the opening becomes narrower and is obvious when the wavelength is very long compared with the size of the hole and as light wavelengths are very short, this makes it difficult to observe light diffraction in our daily life. - From all the previous, light has the general properties of the waves, so light is a wave motion. ~ 39 ~

41 Refraction is the bending of a ray while passing from a medium to another one different in density Takes place in two different media The wave changes its velocity & wavelength. Diffraction is the change in the path of wave while passing through a narrow hole or a sharp edge Takes place in the same medium The wave doesn't change neither its velocity nor its wavelength. * Remember: 1.What is meant by: 1. The ability of a medium to refract light. 2. The bouncing of an incident ray back from a reflecting surface. 3. The bending of ray while passing from a medium to another one different in density. 4. The change of the path (spreading) of a wave while passing through a narrow hole or a sharp edge. 5. The superposition ( union ) of two waves having the same frequency, amplitude & wavelength coming along the same direction. 6. The sound produced due to the superposition of two waves having the same amplitude and slightly different frequencies. 7. Interference resulting when the path difference between two waves is = m λ. 8. Interference resulting when the path difference between two waves is = ( m + ½ ) λ. 2.Choose the correct answer: 1. When two positive waves each of amplitude ( A ) coming in opposite directions superpose, the resultant wave has an amplitude equal to... ( A - 2 A - Zero ). 2. When a positive wave of amplitude ( A ) superposes with another negative wave of amplitude ( A),the resultant wave has an amplitude equal to..... ( A - 2 A - Zero ). ~ 40 ~

42 3. As the medium density increases the light velocity... as... ( increases decreases remains constant ) ( V 1 - V - 1 V ). 3. What is meant by: 1. The refractive index from water to air is ½ is the ratio between the light velocity in the water to the air. 2. The absolute refractive index of water is is the ratio between the light velocity in space to its velocity in the water. 3. The relative refractive index between air and water is is the ratio between the absolute refractive index of water to that of air. 4.Give reasons for: 1. The absolute refractive index of a medium is always greater than 1. Bec. it is the ratio between the light velocity in space to its velocity in the medium & light travels through space with its highest velocity. 2. The relative refractive index may have a value less or more than one. Bec. Relative refractive index n 1 2 n n V V where n so if the wave comes from a less dense medium to a denser one V1 V2 then n1 n2 & 1n2 1 & vice versa. 3. The interference becomes clearer as the distance between the two slits decreases in Young s experiment. Bec. the distance between two successive fringes Δ Y is inversely proportional to the distance between the two slits ( d ) as more clear. R Y so as d decrease interference becomes d 4. The intermediate fringe in Young's experiment is always a bright one. Bec. the paths of the waves emitted by both slits are equal in that case & the path difference between them = 0, so interference is always constructive. ~ 41 ~

43 * Solved examples:- 1. In young s double slit experiment the separating distance between the two slits was 0.2 mm and the distance between the slits and the screen on which the fringes are formed was 120 cm. If the distance between two successive illuminated fringes is 3 mm. Calculate the wavelength of the monochromatic light in Angstrom. R d 3 y, m 5000 A o Two sound sources are fed from same amplifier of frequency 680 Hz and the velocity of sound in air 340 m s, what type of interference takes place at: a) A point at distance 18 m from first source and 28 m from second source. b) A point at distance 6 m from first source and m from second source. c) A point at distance from first source and from the second source m 680 a) Path difference m 10 Path differencein terms of 20 Constructive interference. 0.5 b) Path difference m 4.25 Path differencein terms of 8.5. Destructive interference. 0.5 c) Path difference m 6 Path difference in terms of 12. Constructive interference. 0.5 ~ 42 ~

44 * Critical angle and total internal reflection: - Light passing from a more dense medium to a less dense one refracts away from the normal. - As the angle of incidence increases, the angle of refraction also increases. - When the angle of refraction = 90 and the refracted ray moves parallel ( grazing ) to the interface, the angle of incidence is known as Critical angle. Critical angle: Is the angle of incidence in the denser medium corresponding to an angle of refraction of 90 in the less dense medium. According to Snell s law: n 1 sin = n 2 sin n 1 sin c = n 2 sin 90 ( as θ = 90 ) n 1 sin c = n 2 n 1 is the refractive index of the denser medium. n 2 is the refractive index of the less dense medium. n n 2 1 sinc sin 90 o sin If the less dense medium is air Then n 2 = 1 and n sin c = 1 Where critical angle] c 1 n [Helps to find the refractive index of a medium of known sin C N.B: - If the angle of incidence () in the denser medium exceeds the critical angle between two media ( c ), light suffers a total internal reflection at the interface and doesn t pass to the less dense medium. ~ 43 ~

45 Conc.: 1. A ray falling from a less dense to a denser medium refracts always towards the normal. 2. If the ray is coming from the denser medium to the less dense medium. - The Φ C between the two media must be calculated to be able to identify the path of the ray. ( will reflect or refract). - If Φ < Φ C the ray refracts & follows the law of refraction. n 1 2 V1 sin n2 1 2 V sin n If Φ > Φ C the ray reflects and follows the laws of reflection. ( angle of incidence = angle of reflection ) * Conditions to get total internal reflection: 1. Light falls from the denser to the less dense medium. 2. Angle of incidence > critical angle between the two media.(φ > Φ C) Applications on total internal reflection: 1- Optical fibers: - Are thin hollow transparent fibrous tubes in which light when penetrating suffers successive reflections since > c till emerging from the other end. - Bundles of these tubes are used in medical treatments and examinations using endoscopes and in communications as light can carry electrical signals in fiber-optic cables. ~ 44 ~

46 2- Totally reflecting prism: - Is an isosceles glass prism of angles 45, 45 and > c since c between air and glass is about Is used in some equipments to change the light direction by 90 or even by 180 as in the periscope used in submarines or in the binoculars. - It is preferable than plane mirrors because: 1- The reflection efficiency is 100% while the mirror may absorb some rays. 2- Mirror may lose its metallic luster making reflection weaker while in case of the prism this doesn t happen, this is granted by covering the surface on which light has to be reflected by a coat of a substance whose refractive index is less than glass such as Cryolite (AlF 3 ) or (MgF 2 ) to grant the reflection of all rays even if some escape from the surface of the prism. 3- Mirage: - When the weather is hot at noon, the smooth asphalt of highways appear wet as well as in desert water may be observed. Explanation: - Air in contact with the ground is heated and becomes less dense (with a smaller refractive index) than the upper cold air layers. - Due to the gradual increase of density with height as well as for refractive indices. - Light reflected from the top of an object (Palm tree) suffers several refractions away from the normal gradually until its angle of incidence() exceeds the critical angle ( c ) where ~ 45 ~

47 it is reflected at the ground surface to the eye of the observer who sees the image of the object at the extension of the reflected ray. - The observer sees the palm tree as well as its inverted image. * Remember: 1. What is meant by: 1. The critical angle between glass and air is 42 o o is the angle of incidence is glass ( denser) corresponding to an angle of refraction of 90 o in air ( less dense ). 2. Give reason for:- 1. A ray coming from a less dense to a denser medium (air & water) refracts towards the normal. Bec. Relative refractive index sin V 1 2 1n 2 sin V 2 where V 1 1 so if the wave comes from a less dense medium to a denser one V1 V2 then sin sin & vice versa. 2. Reflecting prisms are preferred than mirrors. Bec. the prism doesn t reduce the light intensity & doesn t lose its lustre. 3. Mirages are observed during hot summer at noon. Bec. the solar radiations heat the ground & so the air in contact gets hot, less dense than the upper layer, so light waves suffer several refractions away from the normal until φ exceeds φ c where they are reflected at the ground surface allowing the observer to see another image at their extensions. 4. Optical fibres are used in medicine. Bec. light penetrating through them suffers several reflections as it falls at φ > φ c due to their narrowness so it is able to penetrate through the other end and helps to examine internal body organs as stomach & intestines. ~ 46 ~

48 5. A ray falling from a denser to a less dense medium may emerge tangent to the interface. Bec. the angle of incidence is equal to the critical angle between the two media. 6. A wave falling perpendicular on the interface between 2 media is not refracted. sin 0 Bec. the angle of incidence is equal to zero as 1 n 2 and sin then 0. sin 1n2 3. State the scientific term: 1. Thin hollow transparent fibrous tubes in which light suffers successive reflections. 2. Prism used to change the path of the rays by 90 o or 180 o. 3. Phenomenon occurring due to successive light refractions in air followed by a total internal reflection. 4. Reflection of light at the interface separating a denser and the less dense transparent media. 4. Choose the correct answer: 1. A light ray falling perpendicular to a reflecting surface is reflected.. ( to the normal to the incident ray to the reflecting surface. ) 2. A light ray falling perpendicular to a reflecting surface is reflected.. ( back on itself perpendicular to itself Isn t reflected ) 3. A light ray falling perpendicular to a reflecting surface is reflected. ( at angle of 90 o at an angle of 0 o at an angle of 180 o ) 4. A light ray falling perpendicular to the interface separating two media. ( reflects at an angle of incidence equal to the angle of reflection refracts at an angle of refraction equal to the angle of incidence - crosses the interface undeviated ) 5. When light passes from a medium with a high refractive index to a medium of low refractive index it refracts.. the normal to the interface. ( towards - away from - parallel to ) ~ 47 ~

49 * Solved example: -The angle of incidence of the ray reflected from mirror Y on mirror X = X 45 o Y -Follow the path of the ray falling to the prism until it emerges out: n=1.5 n = n = ~ 48 ~

50 * Deviation through a triangular prism: - A ray AB falling on xy emerges in the direction CD - Light refracts twice at B and C deviating by an angle. - Angle of deviation : - Is the angle between the extension of the incident e ray and the emergent one. let 1 be the at xy., 1 be the ref. at xy. 2 be the at xz ( 2 nd ( inner ) angle of incidence ) φ 2 2 be the emergence at xz A A the refractive angle or angle of the prism. 1( at φ 2 = θ 1 ) Since e = 180 o and ( A + e ) = 180 o A θ 1 A Then 1 2 and = ( 1-1 ) + ( 2-2 ) Therefore 12 A i.e : The angle of deviation depends on the refractive angle (A) of the prism and the angle of incidence ( 1 ). - It could be practically demonstrated that decreases as 1 increases until reaching a minimum value after which a gradual re-increase is observed. - At minimum deviation it was found that: (Conditions of minimum deviation) 1 = 2 let them = o and 1 = 2 let them = o and since A = A = 2 o o A 2 Ф o 1 ~ 49 ~

51 and since o = A Therefore o = 2 o - A o o 2 and since n 2 1n2 n1 sin sin Therefore n ( o ) sin 2 ( ) sin 2 for a prism in air or in general 1n2 n n prism medium ( o ) sin 2 ( ) sin 2 N.B: Triangular prism is used to study light deviation and dispersion. *Experiment to determine the path of the ray through the glass prism or to show the relation between the angles practically or to find the refractive index of the material of the prism: 1. Place an equilateral glass prism vertically on a cardboard. 2. Fix 2 pins ( a ) & ( b ) 10 cm apart & one of them is very close to one of the prism s face. The line joining them constitutes the incident ray. 3. From the other face observe the two pins & fix 2 other pins ( c ) & ( d ) along the same line with the previous pins. The line joining ( c ) & ( d ) constitutes the emergent ray. 4. Trace the limits of the prism & remove it, then trace the lines joining ( ab ) & ( cd ) and extend them until they meet to draw the angle of deviation ( α ) 5. Trace the normal to each face of the prism & measure the angles of incidence (φ 1) refraction ( θ 1 ) the second angle of incidence ( φ 2 ) & the angle of emergence (θ 2 ). 6. We find that: A 1 2 and 1 2 A ~ 50 ~

52 - To find the refractive index of the material of the prism: 7. Repeat the experiment several times while changing the angle of incidence & tabulate the results. Angle of the prism A Angle of incidence φ 1 Angle of refraction θ 1 Inner angle of incidence φ 2 Angle of emergence θ 2 Angle of deviation α 8. Find the minimum angle of deviation ( α o ) where ( 1 = 2 and 1 = 2 )and the corresponding angles (φ o ) & ( θ o ) then find the refractive index from the relation Glass block:- n sin ( 1 2 sin o ) 2 ( ) 2 - When a ray falls on one face of glass block it refracts towards the normal as it passes to the denser medium. 1 This ray falls on the other face of the block at a 1 angle of incidence ( 2 ), equal to the angle of refraction 2 ( 1 ) (alternate angles) and emerges at an angle of 2 emergence ( 2 )equal to the angle of incidence ( 1 ) as the refractive indices of air and the block are constants. - The glass block is considered to be made of two prisms made of the same material and having equal refractive angles so the deviation and dispersion of the first are counteracted by those of the second and the emergent ray is parallel to the incident ray as the angle of incidence equal angle of emergence and angle of refraction equal second angle of incidence. ~ 51 ~

53 - Light dispersion through a triangular prism: - When light falls on a prism at the position of minimum deviation, the emergent light is observed to be split into the 7 colors of the spectrum arranged from upside down as: red, orange, yellow, green, blue, indigo, violet. Explanation: - Each color has a specific and so a specific n. - and since ( o ) sin n 2 ( ) sin 2 so n depends on A and o and A is constant for a certain prism. therefore the angle of minimum deviation o changes by changing the refractive index (n) which by its turn changes according to the wave length. Therefore it can be observed that the violet color with the shortest wavelength has the greatest deviation angle or maximum refractive index as well as the smallest critical angle and the red with the longest wavelength has the least angle of deviation and the minimum refractive index and the largest critical angle. * The thin prism: - A thin prism has a very small refractive angle A ( less than 10 o ) and is considered to be always in a state of minimum deviation where: o sin n 2 ( ) sin 2 o A A and since ( ) and ( ) in radian. 2 2 are very small angles and tend to zero when measured ~ 52 ~

54 oa oa A A Therefore we consider that sin ( ) and sin ( ) Therefore o ( ) n 2 ( ) 2 o 2 o n ( ) ( ) 2 An = o + A o = An - A α o A α o A (n-1) n A( n 1 ) n2 for a thin prism in air, or ( 1 ) if it is in any medium. o For a thin prism the angle of minimum deviation is represented as: o = A (n -1) o A n 1 - Chromatic dispersion: - Light falling on a thin prism is dispersed into the colors of spectrum. The deviation angles of red and blue rays will be: o r = A ( n r - 1)... (1) where A refractive angle of prism o b = A (n b - 1)... (2) n r refractive index for red n b refractive index for blue. By subtracting (1) from (2): o b - o r = A (n b - n r )... (3) ( o b - o r ) is known as ( the angular size of the spectrum for yellow rays) or ( the angular dispersion for the yellow spectrum) -Angular size of yellow spectrum: is the difference between the angle of deviation of the blue color and that of the red color. ~ 53 ~

55 - Yellow rays are those found between red and blue colors and their angle of deviation is: o y = A ( n y - 1)... (4) Where n y is the refractive index of yellow rays considered to be - By dividing (3) by (4): Where ( o b o y o r ) n A( n b A ( n n y n r 1) ob or nb n b r r o y ny 1 nb nr {( ) 1 } 2 ( ) ob oy or independent of the angle of the prism ( A ). ) n y n b n 2 indicates the dispersion power denoted by () and is r Dispersion power: is the ratio between the angular size of the yellow spectrum & the angle of deviation of the yellow color. * Remember: 1. Give reasons:- 1. The triangular prism disperses the white light. Bec. white light consists of 7 colours, each of a specific wavelength & refractive index ( n ) so each is refracted in a different manner as n o A sin 2 A sin 2 is constant so α o depends on n & each colour deviates differently. & for the same prism A ~ 54 ~

56 2. A ray passing through a glass block is not dispersed. Bec. the glass block consists of 2 prisms made of the same material and are placed opposite to each other so one disperses the light & the other converges the rays as they both form an achromatic group of prisms. 3. Deviation of a prism doesn t only depend on the angle of incidence. Bec. although white light falls at the same angle of incidence, each colour emerges at a different position as deviation depends on λ which affects ( n ) of the prism material for each colour changing α as o A sin n 2. A sin 2 4. The violet colour has a deviation angle greater than that of red one. Bec. deviation depends on the refractive angle of the prism ( constant for the same prism ) & the refractive index of its material for each colour & n violet > n red since its λ is shorter and ( n α 1/λ ). 5. The thin prism is considered always at minimum deviation. Bec. all of its angles are very small, especially the angle of the prism and so the angle of deviation. 2. What is meant by:- 1. The deviation angle of a prism is 40 o o is the angle between the extension of the incident ray and the emergent one. 2. The angular dispersion of a prism is 8 o. - 8 o is the difference between the angle of deviation of the blue colour and that of the red colour. 3. The dispersive power of a prism is is the ratio between the angular size of the yellow spectrum & the angle of deviation of the yellow colour. ~ 55 ~

57 * Solved examples:- 1. A triangular prism of refractive angle of 60 o, is adjusted to the minimum deviation position, when the angle of minimum deviation is 37.2 o. Find its refractive index. ( o ) sin sin n 1.5 ( ) 60 sin sin A thin prism of refractive angle 6 o and refractive index 1.5. Calculate its angle of deviation. An ( 1) 6(1.5 1) 3 o 0 3. Calculate the angular size to a thin prism its refractive index to the blue light 1.66 and its refractive index to the red light 1.44 and the angle of the prism is A( n n ) 10( ) 2.2 b r b r 0 ~ 56 ~

58 *Light: 1-Reflection & Refraction:- Rules Reflection: φ = θ Refraction: 1 n 2 sin V sin V 1 2 n n n2 sin -Snell:- sin c n1 sin 2-Interference:- -Young:- R y d 3-Prism:- -Angle of deviation:- 2 1 A A -Refractive angle OR Angle of the prism:- 2 -Minimum deviation: -Refractive index:- -Angle of refraction:- 1 n 2 o n n 2 1 A 2 A o -Angle of incidence:- o 2 -Thin prism:- A sin 0 2 A sin 2 -Angle of min. deviation:- A( n 1) 0 -Chromatic dispersion:- A n n ) b r ( b r 1 b r -Dispersion power:- y n b n n y r 1 *Refractive index of yellow rays:- n y n b n 2 r ~ 57 ~

59 1- Complete: Homework 1. The luminous (light) density of a medium is Snell s law states that n n Fiber-optics are used in medical &..using Mirage is caused due to the Angle of deviation in a triangular glass prism is.. 6. At the minimum deviation of a triangular prism, the angle of.= the angle of The red color has a wave length... than that of the violet color, while the last has a... refractive index. 8. The angle of deviation of a thin prism is calculated as The angular dispersion for the yellow spectrum =.....= The dispersion power What is meant by: 1. The refractive index from water to air is The absolute refractive index of a medium is The relative refractive between two media is ~ 58 ~

60 4. The critical angle between water and air is 48 o The deviation angle of a prism is 35 o A prism has an equal angle of deviation for two angles of incidences 40 o & 60 o The angular dispersion of a prism is 3 o The dispersive power of a prism is Give reasons for:- 1. The absolute refractive index of a medium is always greater than The relative refractive index may have a value less or more than one The central fringe in young s experiment is a bright one Interference becomes clearer as the distance between the two slits decreases in young s experiment.... ~ 59 ~

61 5. A ray falling from a denser to a less dense medium may emerge tangent to the interface A wave falling perpendicular on the interface between 2 media is not refracted optical fibers are used in medicine The reflecting prism is preferable over the plane mirror or any metallic reflecting surface Mirages are observed during hot summer at noon The triangular prism disperses the white light A ray passing through a glass block is not dispersed Deviation of the prism doesn t only depend on the angle of incidence The red colour has the least angle of deviation while the violet has the biggest one.... ~ 60 ~

62 14. The thin prism is considered always at minimum deviation Choose: 1- In vacuum all electromagnetic waves have the same.. (amplitude frequency wavelength intensity speed). 2- A light ray falls on the interface between two medium at an angle of incidence of 60 o refracts by an angle 45 o. the relative refractive index between the first medium and the second medium is. ( ). 3- If the refractive index ( 1 n 2 ) between 2 media =0.6 then all the followings are correct o except ( n 2 > n 1 n 1 > n 2 v 1 < v 2 - c ). 4- A beam of light incident on a piece of glass and most of the beam is refracted through the glass. Which of the following properties of light does not change as it is refracted ( speed frequency wavelength intensity ). 5- If the wavelength for the light passing through the slits is doubled, the distance from the central bright fringe to the first bright fringe is.. ( halved doubled tripled remains the same). 6- Light passes through two parallel slits and falls on a screen. The pattern produced is due to the interference and ( reflection refraction diffraction absorption ). 7- When a ray of light falls on the mirror A in a direction parallel A 30 to mirror B the reflected ray falls on the mirror B with an angle 30 of incidence equal to.. B ( 90 o 60 o 30 o 0 o ). 8- In the previous figure the reflected beam on the mirror B fall once more on the mirror A with an angle of incidence equals to. ( 60 o 45 o 30 o 0 o ). ~ 61 ~

63 9- When a ray of light falls with an angle from a medium with small refractive index to a medium of high refractive index, it refracts ( closer to the normal away from the normal perpendicular to the interface parallel to the interface). 10- A light ray falling perpendicular to a reflecting surface is reflected... ( to the normal to the incident ray to the reflecting surface. ) 11- A light ray falling perpendicular to a reflecting surface is reflected... ( back on itself perpendicular to itself Isn t reflected ) 12- A light ray falling perpendicular to a reflecting surface is reflected... ( at angle of 90 o at an angle of 0 o at an angle of 180 o ) 13- A light ray falling perpendicular to the interface separating two media... ( reflects at an angle of incidence equal to the angle of reflection refracts at an angle of refraction equal to the angle of incidence- crosses the interface undeviated ). 14- The internal reflection occurs when ( light passes from air into water light refracts as it exists glass into air light passing through glass is refracted inside the glass the angle of incidence is less than the critical angle ). 15- The maximum angle of refraction in the less dense medium is If the ray of the light falls normally on one face of a triangular prism of refractive index = 1.5. It emerges from the prism with an emergence angle equals to ( C ). ( 90 o 60 o 0 o 30 o ). 17- An equilateral prism has two angles of incidence 40 o & 60 o having the same angle of deviation, then its angle of minimum deviation is... ( ). 18- A thin prism of refractive index 1.5 and angle 4 o, the angle of deviation equals. ( 2 o 3 o 4 o 1 o ). ~ 62 ~

64 5- What is the use of each of: 1. The double slits in the young s experiment Optical fibers Reflecting prism Triangular glass prism Thin prism The scientific base: 1. The optical fibers The reflecting prism State the scientific term: 1. Ratio between speed of light in space and in glycerine. ( ) 2. R d ~ 63 ~ (... ) 3. Illuminated and dark regions observed on the superposition of two light waves. (....) 4. Angle of incidence in the denser medium corresponding to an angle of refraction of 90 o in the less dense one. (.....) 5. Thin hollow transparent fibrous tubes in which light suffers successive reflections. (....) 6. Prism used to change the path of the rays by 90 o or 180 o. (....) 7. Phenomenon occurring due to successive light refractions in air followed by a total internal reflection. (....)

65 8. Reflection of light at the interface separating a denser and the less dense transparent media. ( ) 9. Optical pieces used to deviate & disperse light. ( ) 10. The position of prism when the angle of incidence equal to the angle of emergence of the light beam. ( ) 8- Show with drawing: 1. An experiment to prove the interference of light waves. 2. The use of a (45 o - 45 o - 90 o ) prism to change the direction of a light beam by 90 o. 3. The occurrence of the total reflection without using any optical devices. ~ 64 ~

66 9- Essay questions: 1. Deduce the Snell s law. 2. Define the critical angle, then deduce the mathematical relation between it & the refractive index of the medium. 3. State the conditions necessary to obtain a total internal reflection. ~ 65 ~

67 4. Deduce the relation between the refractive index of the prism, its angle & the angle of minimum deviation. 5. Deduce the relation used to calculate the angle of deviation in a thin prism. 6. Show that the dispersion power of a prism depends on the type of its material. ~ 66 ~

68 7- Compare between: 1.Reflection & refraction. 2. Constructive & destructive interference. 3. Refraction & diffraction. * Problems: 1. Calculate the refractive index of light passing from air to glass, knowing that the velocity of light in air is 3 x10 8 m/s & in glass 2 x10 8 m/s. ( 1.5 ) ~ 67 ~

69 2. A light ray falls on the interface between two media whose refractive index is 3.If the angle of incidence is 60 o. Calculate the angle of refraction. ( 30 o ) 3. If the refractive index of water is 1.32 & that of glass is 1.5& the light velocity in air is 3x10 8 m/s. Calculate: a) The refractive index of glass to water. b) The refractive index of water to glass. c) The light velocity in glass ( 1.32/ /1.32-2x10 8 m/s ) 4. If light falls on the interface between air & glass at an angle of 30 o & its velocity in air is 3x10 8 m/s & in glass 2x10 8 m/s. Calculate the absolute refractive index of glass & the angle of refraction. ( o ) ~ 68 ~

70 5. A light ray falls on water surface at an angle of 60 o. Determine the direction of both reflected & refracted rays given that the refractive index of water is ( 60 o - 40 o ) 6. If the refractive index of transparent plastic is Calculate the light velocity in it. (c= m/s) ( 2.027x10 8 ) 7. If the refractive index of carbon tetrachloride is Calculate its critical angle. ( 43 o ) ~ 69 ~

71 8. A wave of 500 Hz travels in two media where its wavelengths are 16 & 12 cm. respectively. Calculate its velocity in both media. If the angle of incidence is 30 o. find the angle of refraction. ( m/s - 22 o 1 27 ) 9. A light ray falls from air on the surface of rectangular glass block so that it makes an angle of 30 o with the surface of the block, find the refractive index of glass, given that the ray is deviated from its original path by 18 o. (1.29) 10. A light beam is incident from air on a liquid surface making an angle of 30 o with its surface, the ray is deviated from its path by 20 o.find the refractive index of the liquid. (1.34) ~ 70 ~

72 11. A yellow light passes from a medium ( A ) to another medium ( B ). Knowing that the wavelength in ( A ) is 6590A o & its angle of incidence is 45 o. Calculate its refractive index as well as its wavelength at ( B ), knowing that the angle of refraction is 60 o. ( x10-7 m. ) 12. A light ray falls on a glass block forming an angle of 30 o with its surface. Knowing that the reflected & the refracted rays are both perpendicular, find the refractive index of glass. ( ) 13.Calculate the radius of a floating circular screen required to prevent the light coming from a source found, 6 cm below water surface to penetrate, given that the refractive index of water is 2. (6 cm.) ~ 71 ~

73 14. In Thomas Young experiment the distance between the two slits was 1.2x10-4 m. & the distance between the two slits & the screen was 4x10-3 m. Calculate the wavelength of monochromatic light given that the distance between two successive fringes is 0.1 m. ( 3x10-3 m. ) 15. In young s experiment, if the distance separating the two slits is m and the distance between the double slits and the screen is 0.75 m and the distance between two successive illuminated fringes is m. Calculate the frequency of the used light.( the illuminated velocity of light = m/s.) ( Hz.) ~ 72 ~

74 16. In young s double slit experiment, the fringe separation observed using green light was found to be mm giving the green lamp a wavelength of 550 nm is replaced by a purple on giving wavelength 400 nm in the violet colour and 600 nm in the red colour the remainder of the apparatus is undistributed. Calculate:- a) The distance between the fringes formed by the violet light. b) The distance between the fringes formed by the red light. ( m m.) 17. Follow the path of the ray until it emerges out of the prism whose refractive index is n = ~ 73 ~

75 18. Follow the path of the ray falling normally on the prism, until it emerges out of the prism whose refractive index is 1.5. and placed in kerosene whose refractive index is n = n = Follow the path of each ray until it emerges out of the prism whose refractive index is 1.5. Knowing that each ray falls perpendicular to the face on which it falls n= 1.5 n = ~ 74 ~

76 20. Follow the path of the ray falling on the prism n = Follow the path of the same ray if the prism is placed in kerosene whose refractive index is Calculate the refractive index of an equilateral prism, If light falls perpendicular to one of its sides and emerges tangential to the other side. (1.15) ~ 75 ~

77 22. Calculate the angle of incidence at which a beam of light is incident on the face (ab) of a triangular prism (abc) made of glass and of equal sides so that the beam emerges tangential to the face (ac) knowing that it s refractive index is 1.5. ( 27 o ) 23. A light ray fell on a prism at an angle of 45 o & emerged perpendicular to its surface. Calculate the refractive index of the prism & the deviation angle if the refractive angle is 30 o. ( o ) 24. A triangular prism has a refractive angle of 60 o & a refractive index of Calculate its angle of minimum deviation. ( 40 o 4 52 ) ~ 76 ~

78 25. Calculate the angle of minimum deviation for a prism whose refractive index is 1.6 placed in kerosene of refractive index 1.28.If the angle of the prism is 60 o, calculate the angles of incidence, emergence and refraction of the ray. (17 o o o o ) 26. Light falls on an equilateral prism whose angle of minimum deviation is 30 o. Calculate the refractive index of the prism. ( 1.41 ) 27. A triangular prism has a refractive angle of 75 o and made of a material whose refractive index is 2.Find the minimum angle of incidence on one side, allowing a ray to emerge out through the opposite side. (45 o ) ~ 77 ~

79 28.The graph in front of you represents the relationship between the incidence angles(φ 1 )and the deviation angles (α) of light ray falling on one face of a triangular prism. (α) 37 o A 48.5 o (φ 1 ) Using the values shown in the figure Calculate:- 1-The emergence angle of the ray. 2-The refractive angle of the prism. 3-The refractive index of the prism material. (48.5 o 60 o 1.49) ~ 78 ~

80 29.A triangular prism has a refractive angle of 75 o and is made of a material whose refractive index is 2. If a ray falls on one side of that prism and emerges tangential to the other side. Find the angle of incidence of that ray. (45 o ) 30. A thin prism of refractive angle 4 o & refractive index 1.6. Calculate its angle of deviation. ( 2.4 o ) ~ 79 ~

81 31. A thin prism whose refractive index is 1.8 deviate the rays falling on it by 2 o when immersed in a liquid. Knowing that the refractive index of the liquid is calculate the refractive angle of the prism. (6 o ) 32. Two thin prisms made of a similar material having their refractive angles 10 o & 8 o respectively, the refractive index of their material is 1.5 are placed next to each other. Find the final deviation angle if: a) They are placed upright next each other. b) One is placed upside down beside the other. ( 9 o 1 o ) ~ 80 ~

82 33. A prism has a refractive angle of 10 o, a refractive index of for red color and for blue. Calculate: a) The angle of deviation of both red & blue colors. b) The angular size of the spectrum. c) The dispersion power of the prism. ( 5 o o o ) ~ 81 ~

83 34. The following table shows the relation between the sin of the angle of incidence and the sin of the angle of refraction between two media. Sin Φ X Sin θ Y - Draw the relation between sin Φ on the Y axis and Sin θ on the X axis then find: - The values of both X and Y. - The relative refractive index between the two media. - The light velocity in the second medium if its velocity in the first is 3 x 10 8 m/s. - The absolute refractive index if the second medium if that of the first is 1. ~ 82 ~

84 Test ( 1 ) 1. Compare between: a) Transverse & longitudinal waves. 2. Complete: 1. The period is... & the frequency is... & the relation between both is The wave velocity is inversely proportional to the... of the medium. 3. During the interference of waves having the same... and...,... fringes are observed. 4.An equilateral triangular prism has two angles of incidence 40 o and 60 o having the same angle of deviation, then the angle of minimum deviation of the prism is. a) Complete: the paths of the incident rays falling on the isosceles right angled glass prism. b) Show by drawing only how can a ray pass through a triangular prism without deviation. 4. Give reasons for: 1. Reflecting prisms are preferable than mirrors..... ~ 83 ~

85 2. The glass prism disperses the white light. 3. The intermediate fringe in Young s experiment is always bright Problems: 1. a) If the critical angle of glass is 42 o, 45 o draw the path of the ray in the fig. Y 45 o 2. A triangular prism of refractive index 1.5 placed in benzene whose refractive index is 1.2. If the prism angle is 60 o. Calculate the angle of minimum deviation & the angles of incidence, refraction & emergence. 3. A swimming pool of depth 2 m is completely filled with water. If a light source is placed 4 m apart from the border, at a height of 3 m. Calculate the length of the dark part of the base of the pool, given that the refractive index of water is 4/3. ~ 84 ~

86 4. Calculate the ratio between the frequencies of two sound waves whose wavelengths are respectively 10 and 20 A o. 5. In the fig. - A is... - is... A - is... ~ 85 ~

87 School book revision questions ~ 86 ~

88 ~ 87 ~

89 ~ 88 ~

90 ~ 89 ~

91 ~ 90 ~

92 UNIT ( 2 ) CHAPTER ( 3 ) CHARACTERISTICS OF FLUIDS AT REST. (STATIC FLUIDS ) 1. Density:( ρ ) Is the mass of the material per unit volume. Density ( ) mass ( m) kg Volume ( Vol ) m 3 M L 3 ex: Al 2700 kg / m 3 Ag kg / m 3 Fe 7900 kg / m 3 Pb kg / m 3 Cu 8900 kg / m 3 Au kg / m 3 2. Specific gravity: ( Relative density ) ( Specific weight ) -Is the ratio between the density of the material & that of water at the same t o. -Is the ratio between a certain mass of the substance & the mass of an equal volume of water at the same temperature. m F subs tan ce subs tan ce g subs tan ce 0 0 rel at the same temperature. unitless M L of water m of anequalvol. of water F g of anequalvol. of water - It has no unit since it is a ratio. N.B : - Density is a characteristic for the material. - Density is inversely proportional to the temperature. - In mixtures and alloys : m t = m 1 + m 2 (ρ t vol t = ρ 1 vol 1 + ρ 2 vol 2 ). Applications on Density 1. Measuring the density of the electrolyte in the accumulator can be used to identify its state. As density decreases on discharging due to the conversion of H 2 SO 4 into PbSO 4 and re-increases on recharging. 2. The measurement of the blood s density may be used to identify some diseases such as Anemia where the blood density decreases due to the lack of RBCs. ( Normal density kg/m 3 ).Also some diseases may raise the concentration of urine density ( 1020 kg/m 3 ) due to the increase of the amount of salts. ~ 91 ~

93 3. Pressure: (P) Is the force acting perpendicular per unit area. F N P Pascal kg m s M L T A m ( 1 2 ) If the force is inclined by an angle (θ) to the perpendicular F F cos θ F cos θ P F sin θ A N.B : Pressure of an elephant s foot is less than that of a pointed healed shoe. Camel has broad pads. The edge of a knife is sharp. The needle is pointed. Pressure inside a fluid & how it is measured ( Hydraulic pressure ) The pressure acting on a point lying on the plane ( X ) of area ( A ) found at a depth ( h ) from the liquid surface is expressed as: weight of theliquid column above( X ) P Area ( A) P m g A vol g A Ah g g h A N J P g h Pascal kg m s M L T m m ( ) 2 3 For closed vessels. N.B:- The relation expresses the pressure exerted by the liquid only ( in a closed container ), but if the surface is exposed to the atmosphere it is then affected by the atmospheric pressure ( P a ) and the relation becomes: Pt Pa g h ( for opened vessels ) and is called absolute pressure. - The total pressure exerted on a point inside a static liquid is proportional to its depth ( h ) and the density (ρ) of the liquid regardless to the area ( A ) or the shape of the container. ~ 92 ~

94 - Pressure acts in all directions at a point inside a liquid. - Pressure is the same at all points lying on the same horizontal level. - The absolute pressure in case of many immiscible liquids placed 1 h 1 together in a container is calculated as : P t = P a + P 1 + P 2 + P 3 2 h 2 = P a + 1 g h g h g h 3 3 h 3 - The base of the dam s wall must be thicker than its top to be able to withstand the higher pressure to which it is exposed as P α h (depth). * Graphs used to show the relation between P & h in a liquid: 1. In closed vessels:p liq. P maximum P liq. ρg 0 h 0 h surface bottom bottom surface 2. In opened vessels:p T P maximum P T P a P P h h P 2 1 P g a h h 0 h surface bottom bottom surface ~ 93 ~

95 *Solved example: 1. A solid parallelogram of dimensions 5, 10 & 20 cm & density 5000 kg/m 3 is placed on a flat surface. Calculate its maximum & minimum pressures.( g = 10 m/s 2 ). Max. P. = F A N m or P max = ρgh max.= = 10 4 N m 2 Min.P. = F A or P min = ρgh min.= = 2500 * Balance of liquids in a U shaped tube: N m 2 N m - Pour some H 2 O in the U tube, the level of H 2 O in both branches P a P a will be similar since pressure on both sides is equal to the P a. - Pour in one branch oil which is immiscible with water & consider two points (D) at the interface and (A) at the same level. - Oil reaches a height ( h oil ) above the point ( D ) while water reaches a height ( h w ) above the point ( A ). - P of the column AB = P of the column CD P a + w g h w = P a + oil g h oil w h w = oil h oil o hw. h w o rel oil - By measuring h w & h oil, the specific gravity of oil can be calculated. - The device can be used also to prove that for different liquids to obtain an equal pressure 1 h 2 ~ 94 ~

96 *Solved example:- 1- The cross sectional area of the narrow side of a U tube is 1 cm 2, while that of its wide side is 2 cm 2. The tube is partially filled with water of density 1000 kg/m 3, then a quantity of oil of density 800 kg/m 3 was poured in the narrow side until its height reached 5 cm. Calculate the height of water upon the separating level between oil & water. g h g h h h h h cm. 2-A U-shaped tube of cross sectional area 2 3 2cm has an amount of water. 9cm of kerosene has been poured in one side, so the height difference of water in the two sides is 3.6cm. Find the volume of benzene poured in the other side till the level of water becomes the 3 3 same in the two sides. H 1000 kgm, kgm 2O benzene 900 v= 9cm 3 K 3.6 H 2 O K b Vk 9 hk 4.5cm A 2 k h H 2O k 3.6,, h H 2O k hb, h b k k h b 800kg m 800, hb 4cm Vol A h cm 3 b k. 3 ~ 95 ~

97 Atmospheric pressure: Is due to the weight of an air column of unit cross-sectional area and height equal to that of the atmosphere, can be measured using the Torricellian ( Hg )barometer. - Mercury barometer ( Torricelli ) - A glass tube, one m. long filled with Hg and inverted in a basin filled with mercury. - The mercury in the tube starts to fall down leaving h a space called ( Torricellian space where P = 0 ). - The atmospheric pressure is equal to the pressure of the mercury column whose height is equal to the difference between the Hg levels in and out of the tube. - Consider two points ( A ) & ( B ) at the same level,one inside and the other outside the tube. - Since P A = P B = P a - Therefore P a = g h ( Hg ) + 0 ( vacuum ) - Since h = 0.76 m & Hg = kg /m 3. - Therefore P a = 0.76 x x 9.8 = x 10 5 N/m 2. or Pascal. N.B: 1 bar = 10 5 Pascal = 10 5 N/m 2. N/m 2 ( Pascal ) (P a )= 1Atm. 76 cm.hg bar mm.hg (Torr) P a = x 10 5 Pascal = x 10 5 N/m 2 = bar = 760 torr ( mm.hg ) = 76 cm.hg = 0.76 m.hg = 1 atm.(atmosphere). - If the tube is inclined the volume of the Torricellian space decreases while the height of the Hg column isn t changed as P α h only regardless to the shape and area of the tube. Standard atmospheric pressure: is equivalent to the pressure exerted by a Hg column of 1 m 2 in section and 0.76 m height at 0 o C at sea level. N.B :When dealing with Hg,pressure can be calculated as: P t = P a (in cm.hg) + h of a Hg column. ~ 96 ~

98 Ex: Absolute pressure on the base P t = P a + h = = 86 cm.hg. Hg *Solved example:- 1. Find the total pressure & the total force acting upon the base of a tank containing salt water whose density is 1030 kg/m 3, if the cross sectional area of the tank is 1000 cm 2, the water height is 1m and its surface is exposed to air, g = 10 m/s 2 & P a = x 10 5 N/m 2. P = P a + g h = x x 10 x 1 = x 10 5 N/m 2 F = P x A = x 10 5 x 1000 x 10-4 = x 10 4 N. 2. In the fig. the cross section of the opened tube (a) of length 2.5m is 100 cm 2. a Find the absolute force acting on each side of the cubic container whose side is 2m long, if water fills it to the top of the tube. ( P a = 10 5 N /m 2, water density is 1000 kg/m 3, g = 10m/s 2. ) F PA Base F P g h Side N ( a 1) ( ) Side F P g h Side N ( a 2) ( ) Top F P g h Side a N ( a 3) ( ) ( )( )) A submarine is located horizontally in sea water, the interior of the submarine is maintained at sea level atmospheric pressure. Find the force acting on one of the submarine s windows of circular shape of radius 21 cm and whose center is at depth of 50 m from sea level if the density of sea water is F P A T P g h P A a g h r (2110 ) N a Kg m.( 2 g 10 m s ) ~ 97 ~

99 4. The table represents the values of pressures at different depths inside a liquid:- Pr.( Bar) X Depth(m) Draw a graphical representation to show ( P ) on the ordinate & ( h ) on the abscissa. Then find:- 1. The density of water. 2. Pressure at ( X ). (g = 9.8 m/s 2 ) H2O 5 P ( ) 10 h g (1 0) 9.8 P P g h Kg m. 5 X a ( ) 15 P N m bar 5 2 X P (h) depth ~ 98 ~

100 * Determination of the pressure of an enclosed gas: - Manometers: ( pressure gauge): - Devices used to measure the pressure difference ( P ) between that of an enclosed gas and P a. or to calculate the absolute pressure of an enclosed gas, when knowing P a. - The liquid used to separate both the gas and air in the U tube is Hg, in case of big pressure differences & H 2O of low density, in case of small differences. 1 As for the same P, h - The cross section of the tube doesn t affect the pressure. - Consider two points ( A ) & ( B ) at the same level. = P g = P a P g P a P g P a P g = P a P g = P a + g Δh P g = P a - g Δh P g P a = 0 P g P a = ρgδh p g P a = - ρgδh Pg P N a g h 2 ( Pascal ) m - Note: ΔP can be calculated in cm Hg as ΔP= P g P a = Δh in cm Hg, only if Hg is used. - Controlling the quantity of air in the Eustachian canal, controls the pressure on the ear drum. * This is done by swallowing or chewing a gum. ~ 99 ~

101 Applications on pressure: 1. Blood flows through arteries and veins in a steady flow which may change into turbulent flow producing noise detected by the stethoscope. When the cardiac muscle contracts blood pressure is 120 mm Hg & is called Systolic pressure and when it is relaxed pressure becomes 80 mm Hg and is called Diastolic pressure. 2. As a tire is well inflated ( high pressure) the area in contact with ground decreases, while it increases when it is under inflated, this leads to the increase of friction & overheating of the tire. ( Measured by the pressure gauge ) -N.B: Pressure devices are all based scientifically on: 1. Two or more points inside the same liquid and affected by equal pressures must be at the same horizontal level. 2. Inside the same liquid P = ρ g h where p α h and is independent of the shape and area of the container. 1- What is meant by: 1. The density of mercury is kg/m kg is the mass of mercury per unit volume ( 1 m 3 ) 2. The specific gravity of aluminium is is the ratio between the density of Al & that of water at the same temperature. 3. The pressure on a point is 50 N/m N is the force acting normally per unit area ( 1 m 2 ) around that point. 4. The atmospheric pressure is equal to 76 cm Hg. - The atmospheric pressure is equal to the pressure exerted by a mercury column of unit area and height equal to 76 cm at 0 o C at sea level. ~ 100 ~

102 5. The atmospheric pressure is equal to 2 bar. - The atmospheric pressure is equivalent to the pressure exerted by a Hg column of unit area and height equal to 1.5 m.hg. 6. The atmospheric pressure is equal to Pascal. - The atmospheric pressure is equivalent to the pressure exerted by a Hg column of unit area and height equal to 1.52 m.hg x10 5 N is the force (weight) of atmospheric air acting normally per unit area. 7. The reading of a manometer is 10 cm. Hg. - The absolute pressure of the gas enclosed in the manometer is 86 cm. Hg. 8. The absolute pressure of a gas enclosed in a manometer is 56 cm.hg. - The reading of the manometer is 20 cm.hg. - The pressure difference between the pressure of the enclosed gas and P a is 20 cm.hg. 2-Give reasons for: 1. Density is a characteristic for the substance. - Because density is the mass of the substance per unit volume which depends on the chemical structure ( type & number of atoms & bonds ) which differ from a substance to another. 2. Fluids can flow while solids can't. - Because Fluids have moderate or weak intermolecular attractive forces & wide spaces while solids have closely packed molecules with strong intermolecular attractive forces & narrow spaces so the have definite shapes. 3. Solids are more dense than liquids. - Because in solids molecules are closely packed so the mass per unit volume is greater than in liquids & density = mass / volume. 4. The edge of a knife is thin. - Because P = F / A & as area decreases, pressure increases to become high enough to cut an object. ~ 101 ~

103 5. Camel has broad pads. - Because P = F / A & as area increases, pressure decreases to prevent it from sinking in the loose sand of the desert. 6. Pressures exerted by two different liquids having the same heights are different. - Because P = ρgh, & different liquids have different densities. 7. In a homogeneous liquid all points found at the same level have equal pressure. - Because P = ρgh, & for the same liquid both density & g are constants & Pα h & all points at a similar level have equal pressure. 8. It is better to use water in manometers while measuring small pressure differences. - Because H 2 O has a low density & as P =ρ g h then at constant P, h = P/g ρ & h α 1/ρ so for slight P. changes the change in height will be observed. 9. A liquid rises in the connected vessels of different shapes similarly. - Because P = ρgh & is independent of shape & area of the container so different points affected by the same P must be at a similar level. 10. A barometric tube has been filled completely with mercury then inverted into a basin containing mercury & the Torricellian space hasn t been observed. - Because the tube used may be short or inclined, the temperature may be high decreasing the density increasing the height of Hg or the pressure is very high. 11. Mercury was used in Torricellian barometer. - Because Hg has a high density & as h α 1/ρ, this allows us to use a short barometric tube as the height of the Hg column will be suitable, it also has high cohesive forces so it doesn't wet glass & is opaque. 12. As the density of the liquid increases its height decreases to keep the pressure constant. - Because P = ρgh & for a constant P, ρ α 1/h. ~ 102 ~

104 * Solved examples: 1. A Hg manometer is used to measure the pressure of a gas inside a container. The Hg level in the free side was higher than the one in the side attached to the container by 36 cm. Find the pressure of the gas enclosed in cm.hg, P a and N/m 2. P a = x 10 5 N/m 2 = 76 cm.hg a) P = P a + h = = 112 cm.hg. b) P = P in cm Hg = P a. c) P = P in P a x x 10 5 = x x 10 5 = x 10 5 N/m If the relative density of the liquid in the tank joined to the Hg manometer is 0.8.Find the pressure at the point A. 4 m 20 cm (g = 10m/s 2,P a = 10 5 N/m 2, ρ Hg = kg/m 3 ). A* P P g h g a 5 Pg Pa g h 10 ( ) N m PA Pg g h (800104) N m 2 2 ~ 103 ~

105 *Density: mass vol. -Specific gravity :( Relative density):- m -In mixtures and alloys: sub sub sub h2 o mof vol. ofh2 o Wtof v. ofh2o Guide rules for problems Wt m m m OR Vol Vol Vol t 1 2 t t *Pressure: F Wt m g vol g gh -Absolute pressure:- P P g h t a P 76 h in cm t of a Hg column P P P P P *In case of many immiscible liquids:- t a Atmospheric pressure:- -U-tube:- gh a Hg Hg h h Manometers:- P P P gh g a P P incm Hg P incm Hg h incm. g a Hg ~ 104 ~

106 Homework 1- Choose the correct answer: 1. The relative density of a substance is 0.1, then its density is... ( ) 2. The volume of 2 gms of a liquid whose density is 0.8 gm/cm 3 is... ( 0.4 cm cm 3-2 cm cm 3 ) 3. The unit used to measure the pressure is... ( N - J/m - N/m - N/m 2 - N m 2 ) 4. The U shape tube can be used to determine the.. ( pressure - atmospheric pressure relative density of a liquid force ) 5. In the U tube the higher liquid is... ( denser less dense similar density as the other no correct answer ) 6. In the Torricellian barometer the difference in the mercury level in & out decreases when the... ( t o increases - barometer is taken at the top of a mountain - A wide tube is used. ) 7. The height of mercury in a barometer depends on the following factors except... (the density of mercury the acceleration due to gravity - the area of the tube the P a ) 8. The total pressure at the bottom of a liquid depends on the following factors except... ( The P a - The area of the bottom - The density of the liquid - The depth of the liquid ) 9. The pressure exerted by a liquid column weighting 2 N on the base of a container whose area is 2 m 2 equals... ( Pascal 2 bar 1 Pascal ) 10. The atmospheric pressure is equal to. ( bar 760 Torr Pascal N/m 2 all the previous) 11. When the mercury level in the free branch of the manometer is lower than the one in the branch connected to the gas container Δ P is ( positive negative equal zero ) ~ 105 ~

107 2- Complete: 1. Density is measured in.. & its dimensional formula is. 2. The relative density of a liquid is half that of another then its mass is... that of the other. 3. The density of copper is 8900 kg/m 3, then its relative density is The ratio between the mass of a liquid & the one of an equal volume of water at the same temperature is called.. 5. At constant force the pressure when the area increases 3 times, while the pressure is. When the area is decreased to half. 6. The pressure that a woman exerts on the ground increases as the heels of her shoes are.. 7. The pressure of a liquid in a container is independent of the. &. of the container. 8. All the points found at the same depth in a liquid are affected by the... pressure. 9.The pressure exerted by a liquid layer only in a container equals. or..while the absolute pressure exerted by this layer equals. 10. The... is used to measure the atmospheric pressure, while the... is used to calculate the pressure of a gas in a container. 11. The part above Hg in the barometer is called.. & the pressure inside it equals Pascal =.. bar =.. N/m 2 =. Torr. 13. While the barometric tube is inclined, the of Hg remains constant while its changes in the tube. 3- What is meant by: 1. The specific gravity of aluminum is The density of mercury is kg/m ~ 106 ~

108 3. The pressure on a point is 50 N/m The atmospheric pressure is equal to 76 cm Hg The atmospheric pressure is 5 10 Pascal The pressure of a trapped gas= 3 Atm The atmospheric pressure is 1520 torr Atmospheric pressure is 1 bar The reading of a manometer is 6 cm.hg The reading of a manometer is -12 cm.hg Give reasons for: 1. Solids are more dense than liquids. ~ 107 ~

109 2. The edge of a knife is thin Camel has broad pads Pressures exerted by two different liquids having the same heights are different In a homogeneous liquid all points found at the same level have equal pressure. 6. A liquid rises in the connected vessels of different shapes similarly Water level remains equal in the branches of a U tube whatever are their diameters As the density of the liquid increases its height decreases to keep the pressure constant A barometric tube has been filled completely with mercury then inverted into a basin containing mercury & the Torricellian space hasn t been observed Mercury was used in Torricellian barometer.... ~ 108 ~

110 11. It is better to use water in manometers while measuring small pressure differences Mention one use for each of: 1. The U shaped tube. 2. The barometer. 3. The manometer. 6- What are the factors controlling each of the following: 1. Absolute pressure at a point inside a liquid. 2. Density of a substance. 7- What is the scientific base of: 1. The barometer. 2. The manometer. 3. Systolic and diastolic blood pressure measurements. 4. Identification of the accumulator status. 5. Diseases diagnosis through blood or urine. ~ 109 ~

111 8- Derive the relation used to: 1. Find the pressure exerted by a liquid on the base of a container. 2. Find the relative density of a liquid using a U tube. 9- Draw a diagram to show: 1. The U shaped tube & show how it can be used to find the relation between the density & the height of a liquid. 2. The Torricellian barometer showing the Torricellian space. ~ 110 ~

112 3. The manometer when the gas pressure is less than & greater than the atmospheric pressure. 10- Essay questions: 1. State two applications for density. 2. The opposite graphs indicates the relation between the mass and the volume for two different liquids A and B:- m A 1) which liquid is denser? B Vol 3. Prove that the pressure at a point inside a liquid is directly proportional to its depth from the surface. 6. Prove that both F/A & gh have the same dimensional formula. ~ 111 ~

113 5. The three vessels shown are filled with the same height of water =150 cm & have each a base of area ( 30 x 60 cm ) 1. Find the pressure at A, B & C. 2. What is the total force acting upon each base? A B C ( Pa = x 10 5 N/m 2, ρ = 1000 kg/m 3, g = 9.8 m/s 2 ) 6. Comment on the following figure. a b c 7. The opposite graphs indicates the relation between the pressure and the depth of two different liquids A and B in two containers:- P B 1) Which liquid is denser? Why? A 2) Is the container opened or closed? why? h 8. Which of the following graphs represents the relation between the pressure ( P ) at a point inside a liquid & the depth ( h ) of that point from the liquid s surface in each of the following cases. Mention what does each of ( A ) & ( B ) represent in each case P P P P P A A h b h h h b h 1. Case ( 1 ) is represented by graph Case ( 2 ) is represented by graph... h h h 3. Case ( 3 ) is represented by graph... ( 1 ) ( 2 ) ( 3 ) ~ 112 ~

114 9. Prove that the density of a liquid is inversely proportional to its height in a U tube. 10. If the atmospheric pressure is equal to 76 cm Hg. What is the height of the Hg column whose pressure 10cm 8 cm is equal to the pressure of the enclosed gas. * Problems: ( ρ of water = 1000 kg/ m 3, ρ of Hg = kg/m 3 ) 1. A small flask of mass 20 g when empty is used for measuring density, If the mass becomes 22 g when filled with water and g when filled with benzene. Find the 3 density of benzene and the mass of air that fills the flask.( 1.3Kg m ). air 3 ( 880 kg m, 2.6 m Kg ) 2. If the density of a metallic sphere is 2762 Kg /m 3, & its weight is 196 N. Find its diameter.(g = 9.8m/s 2 ) (0.24 m. ) ~ 113 ~

115 3. If the capacity of a barrel is 90 Kg of water or 60 Kg of kerosene. Find the rel. density of kerosene & its volume, as well as the volume of the barrel in liter. ( m 3-90 L) 4. A jar whose capacity is 1000cm 3, has been filled with 2 liquids (A) & (B). Their rel. density together is 1.5 while that of (A) is 1.8 & that of (B) is 0.8. Knowing that by mixing both liquids the volume isn t changed. Find the volume of each liquid. ( cm 3 ) 5. Calculate the force acting on the base of a cylindrical vessel of diameter 6 m, containing oil exerting a pressure of 2.5x10 2 N/m 2 (ρ = 800kg/m 3 ) ( N. ) 6. If the height of water in a barometer is 10m. Find its reading.(g = 9.8 m/s 2 ) ( 9.8x10 4 N/m 2 ) ~ 114 ~

116 7. Calculate the water pressure on the base of a pool of 10 6 cm 2 containing water weighting 4x10 4 N. ( 400 N/m 2.) 8. Calculate the absolute air pressure in N/m 2 & in atm. inside a car tire, if the pressure difference is 3.039x10 5 N/m 2 & P a = x 10 5 N/ m 2. ( 4.052x10 5 N/m 2-4 atm.) 9. Find the pressure at the bottom of a closed vessel of depth 76cm, when it is full of: a) Mercury. b) Water. (g=9.8 m/s 2 ) ( 1.013x N/m 2 ) 10. Calculate the height to which water can rise through the pipes of a building given that the pressure at the ground floor is 2.94x10 5 N/m 2. ( 30m.) ~ 115 ~

117 11. When a submarine dives to a depth, where water is expelled from its blast tank under a pressure of N/m 2. Find the depth at which the submarine is found. ( density of salt water = 1030 Kg /m 3 P a = bar & g =9.8m/s 2 ) ( 80m.) 12. Calculate the water pressure only, exerted on the door of a submarine found at 50m below salt water surface of density 1030 Kg / m 3. ( g =9.8 m/s 2 ) (504700N/m 2 ) 13. A submarine resisting a water pressure of N/m 2 has a door of area 2m 2. Find the maximum depth it can reach & the force exerted on its door neglecting P a. (density of salt water = 1030 Kg /m 3 & g =9.8m/s 2 ) (130 m N.) 14. A cylinder of height 10cm & radius 2cm dipped in a liquid of rel. density 1.3 in such a way that its upper end becomes 15 cm below the liquid surface. Find the force acting on both upper and lower surfaces due the liquid.(g =9.8 m/s 2 ) ( N.) ~ 116 ~

118 15. A building made of several floors, each of 4 m height, carries a cubic container of 2 m height, completely filled with water. If the pressure at the horizontal pipe ( X ) found at a height of 16 m from the ground is Pascal, find the number of floors of the building.(g = 10 m/s 2 ) X ( 10 floors ) 16. In a given tank containing water of density 1000 Kg/m 3, Find the pressure at point (A) 2m 6m Knowing that p a = 10 5 N/m 2 & g= 10 m/s 2. 4m A ( N/m 2 ) 17. A U shape tube with one branch having twice the cross-section of the other and of height 66cm has been filled to its half with water. If the wide branch has been filled with oil of rel. density 0.8, what will be the height of oil required. ( 45cm.) ~ 117 ~

119 18. A U-Shape tube of similar cross sectional area and of height = 28 cm. water is poured till it reaches 17 cm, then oil is poured till a branch is completely filled. Find the 3 change in water column height. (ρ of oil = 900 Kg m ) (9 cm.) 11 cm X 28 cm x 17 cm 19. If the atmospheric pressure at a mountain s base is 75cm Hg & at its top is 50 cm Hg. Find the height of the mountain if the density of Hg is & that of air is 1.29 Kg /m 3. ( m.) 20. A submarine is located horizontally in sea water, the interior of the submarine is maintained at sea level atmospheric pressure. Find the force acting on one of the submarine s windows of circular shape of radius 21 cm and whose center is at depth of 50 m from sea level if the density of sea water is 1030 kg/m 3, g = 9.8 m/s 2. ( N) ~ 118 ~

120 21. If the pressure at the surface of a water basin is 1 atm. & at its bottom is 3 atm. Calculate the depth of the basin.(g =9.8 m/s 2 ) ( 20.67m) 22. A manometer is connected to a gas supply, the difference in the Hg levels in its branches is + 25 cm. Calculate the pressure difference & the absolute pressure of the trapped air in N/m 2, given that the P a = x 10 5 N /m 2 & g = 9.8m/s 2. ( x 10 5 N/m x 10 5 N/m 2 ) 23. A Hg manometer is used to measure the pressure of a gas inside a container. The surface of Hg in the free branch is lower than that in the one connected to the gas container by 20 cm. Find the absolute pressure of the enclosed gas in Bar, given that the atmospheric pressure is 10 5 Pascal, the Hg density kg/m 3 & g =10 ms -2 ) ( Bar ) ~ 119 ~

121 24. If a gas (x) causes the rise of the Hg level in a manometer by 15cm, while another gas (y) drops the original level by 5cm. Find the difference in pressure between both gases.( density of Hg = Kg /m 3 & g =9.8 m/s 2 ) ( N/m 2 ) 25. A car tire whose pressure difference is 2.026x10 5 Pascal. Find its absolute pressure in Pascal and atm. ( P a = x 10 5 Pascal ) ( 3.039x atm.) 26. A water layer of 1m depth floats on a Hg layer of 0.5m depth. Find the difference in pressure between two points one on the water surface & the other at the Hg bottom. (g =9.8 m/s 2 ) ( N/m 2.) ~ 120 ~

122 Pascal s principle: If any pressure is applied to an enclosed liquid, it is transferred undiminished to all the points of the liquid as well as to the walls of container. - If the liquid compressed below the piston is exposed to Fg a pressure ( P1 ) due to weight of the piston then the pressure at a any point in the liquid becomes: P t = P a + P 1 + g h - When the pressure is increased by ( P ) on the piston, the pressure on any point in the liquid will increase by the same value to become P t = P a + g h + P 1 + P, as liquids are incompressible. - N.B: The increase can t be observed since the liquid can t flow. * One of the applications on Pascal s principle is: The hydraulic press: - Used to magnify a small force and carry heavy loads. - A small force is exerted on the small piston of area ( a ) changes the pressure by ( P ) which is then transmitted undiminished to the large piston of area ( A ) to produce a big force ( F ) which is enough to raise a car up. F f f A P F N A a a - The little force is then magnified to become a strong one. ~ 121 ~

123 - The mechanical advantage of a press: is the ratio between the resulting force & the applied one. * can be calculated as: - the ratio between the mass carried by the large piston & the one acting on the small one = the ratio between the surface area of the large piston & that of the small one = the ratio between the square of the radius of the large piston & that of the small one = the ratio between the square of the diameter of the large piston & that of the small one = the ratio between the displacement of the small piston to that of the large one. = The ratio between the speed of the small piston to that of the large one. - Since the volume of the liquid displaced from the small piston ( s ) is equal to the one a A 1 displaced to the large one ( S ) then Vol a s A S where and A S s S 2 2 F M g A R D s v 2 2 f m g a r d S V F η - For a group of presses gr. 1 2 F F final initial f Woutput FS - Efficiency of a press w f s input - For an ideal press, efficiency = 100 % since work output = work input F P W f p w - N.B:- For an ideal press 1, 1, 1 - A press may not respect Pascal s principle if it contains some air bubbles, as gases are compressible, or if it leaks leading to a liquid outflow. - For a press to be at equilibrium (pistons stop moving) the pressure at two points at the same horizontal level inside the liquid must be equal, and it is not a must that both pistons be at the same level to achieve this. ~ 122 ~

124 Applications on Pascal s principle 1. Hydraulic press is used to magnify a small force to carry heavy loads. 2. Hydraulic brakes are used in cars. 3. Powerful hydraulic pistons are used to compress & shape metals. 1- What is meant by: 1. The mechanical advantage of a press is is the ratio between the resultant force & the applied one. 2. The efficiency of a press is 80 % is the ratio between the output work to the input work of the press. 2- Give reasons: 1. Hydraulic presses can lift large loads while acting upon them with small forces. - Because according to Pascal's principle, pressure is transmitted undiminished through liquids and f/a = F/A so by increasing the area, the resultant force is magnified to keep the pressure constant. 2. In the hydraulic press, the motion of the small piston is faster than that of the large one. - Because the liquid is incompressible so the volume displaced from the small piston = to the one reaching the large piston & as A/a = s /S & A α 1/ S The displacement of the small piston is greater as its area is smaller. 3. A press with well sealed, frictionless pistons doesn t respect Pascal s principle. - Because the press may contain gas bubbles which are compressible. ~ 123 ~

125 * Solved examples: 1. The area of the small piston of a press is 10 cm 2. If a force of 100 N acts on it & the area of the large piston is 800 cm 2 & g = 10 m/s 2. Calculate: a) The mechanical advantage of the press. b) The maximum mass carried by the large piston. c) The distance moved by the small piston to move the large one along a displacement of 1 cm. 800 a) A 80. a 10 b) f A F 8000 N, a 10 F 8000 M 800 Kg. g 10 c) FS s 80 cm. f For the shown hydraulic press, the mass of the large piston is 1300 Kg and its area 0.2 m 2. The small piston has a negligible mass & area 30 cm 2. Calculate the force applied on the small piston necessary to maintain equilibrium. 3 ( oil 780 Kg m, P a P f F gh a A b 2 g 9.8m s ) f N f a 3.5 m b ~ 124 ~

126 *Pascal s principle: -Hydraulic press:- F f P A a Guide rules for problems -The mechanical advantage of the press:- F A Wt 2 R y s f a wt 2 r Y S F f out -For group:- t 1 2 in FY work output -Efficiency of press:- 100% 100% f work input y ~ 125 ~

127 1- Choose the correct answer: Homework 1. In the hydraulic press the ratio between the resultant force to the applied one is.. ( 1 less than 1 greater than 1 ) 2. In the hydraulic press the ratio between the resultant pressure to the applied one is... ( 1 less than 1 greater than 1 ) 3. In the hydraulic press the ratio between the output work to the input one is. ( 1 less than 1 greater than 1 ) 4. The ratio between the diameters of a hydraulic press is 10 1 then the mechanical advantage is.( 2- Complete: ) Pascal s principle states that In the hydraulic press the ratio between the resultant work & the applied one equal if the ratio between the area of the large and the small one is 25:1,the force required to lift a car of weight N is.... The speed of the big piston The speed of the small piston 3- What is meant by: 1. The mechanical advantage of the press is The efficiency of a press is 100%. 3. The efficiency of a press is 70%. 4- State the scientific base of: 1. Hydraulic press:. ~ 126 ~

128 5- Give reasons for: 1. Hydraulic presses can lift large loads while acting upon them with small forces. 2. In the hydraulic press, the motion of the small piston is faster than that of the large one. 3. A hydraulic press with well sealed, frictionless pistons doesn t respect Pascal s principle. 6- State one use for each of the following: 1. The hydraulic press. 2. Pascal s principle in cars. 7- Essay: A 1. Which point will be affected by the highest pressure in the container, taking in consideration that it is opened, & if it c is exposed to the pressure of a piston, will it remain the the same why d b ~ 127 ~

129 * Problems: 1. Find the force required by a press of areas 20 & 800cm 2 for its small & large pistons respectively to lift a car of 2000 Kg.(g= 9.8 m/s 2 ) ( 490 N.) 2. Calculate the force acting on the small piston of a press, whose diameter is 2cm, to allow the large piston of diameter 28cm to lift a weight of N. ( 10.2 N.) 3. If the area of the large piston of a press is 100 times that of the small one, Calculate the force required to lift a load of 3000 N. ( 30 N.) ~ 128 ~

130 4- A car service station uses a hydraulic press. The diameters of its compressed air tubes are 2 & 32 cm. Calculate the pressure of air required to raise a car of mass 1800 kg. (g=9.8 m/s 2 ) ( N/m 2 ) 5. In a press the area of the small press is 2cm 2 & that of the large one is 50cm 2. Calculate: a) The mechanical advantage of the press. b) The distance moved by the small piston to move the large one 3cm. ( 25-75cm ) 6. Find the absolute pressure of a gas captured in a vertical cylinder of area 10cm 2, by a piston of 100 N in weight. ( N/m 2 ) ~ 129 ~

131 7. In a press, the diameters of the small & large pistons are 2 & 24cm respectively. Find: a) the force required to lift a load of 2880 N. b) The distance moved by the small piston if the large one moves 1cm. ( 20 N - 144cm.) 8. The cross sections of the narrow & wide branches B A are 20 & 400cm 2. The mass of the large piston 5m. is 200 Kg while that of the small one is neglected. If the relative density of oil is 0.8.Find the force needed to keep the pistons at equilibrium. (g =9.8 m/s 2 ) ( 19.6 N.) ~ 130 ~

132 9. The two ideal presses shown in the fig. are joined 10 N. with a lever AB with fulcrum E at the middle. the ratio of both branches of the 1 st is 1:40. & that of the 2 nd is 1:50. Find : a) The mechanical advantage of the group. b )The mass raised by the group. ( g = 10 m /s 2 ) (2000 Kg ) 10. Two pistons, each of area 0.2 m 2 are used to carry a car of mass 1 ton, if they are connected to the small one as in the fig. where a force of 100 N acts. Calculate the area of the small piston is g = 10 m/s 2. ( m 2.) ~ 131 ~

133 11. While using a hydraulic press the following results were obtained:- Force on the small piston (N) Force on the large piston (N) Draw the graphical relation between the force acting on the large piston on the (Y) axis and the one acting on the small piston on the (X) axis, then find :- a) The mechanical advantage of the press b) The force required to act on the large piston to keep the balance with a force of 60 N acting on the small piston. 2. If the radius of the small piston is 5 cm, find the radius of the large piston. ~ 132 ~

134 ~ 133 ~

135 CHAPTER ( 4 ) CHARACTERISTICS OF FLUIDS IN MOTION (Dynamic fluids) * Steady & turbulent flows: - Steady flow: - Is the transfer of a liquid from a point to another easily & smoothly due to a P & is expressed by a specific number of streamlines which never intersect. - Steady or laminar flow is characterized by: 1. Liquid fills the tube completely. 2. The quantity of liquid flowing in at a certain time equals the one flowing out at the same time. 3. The flow is not rotational. ( not a vortex flow containing vortices ) 4. The flow is represented by imaginary stream lines that do not intersect. 5. The rate of flow is represented by the number of stream lines passing per unit area at a point. N.B: - If the rate of flow increases than a certain limit the flow is changed to turbulent, characterized by the presence of vortices. This takes place in gases too while flowing from a high pressure to a low pressure region. - In the fig. the volume of the liquid ( Vol 1 ) flowing per unit time ( t ) along the length ( L 1 ) of cross section ( A 1 ) constitutes the volumetric rate of flow ( Q vol 1 ), represented as: Vol t A L Q vol 1 t while the rate of flow of the liquid along ( L 2 ) of area ( A 2 ) is represented as: Vol t A L Q vol 2 t ~ 134 ~

136 - At steady flow, the rate of flow is constant. Q Vol 1 = Q Vol 2 and since velocity. L t A L t A L t x where (Δx) is the distance covered by the liquid per unit time or its t the relation becomes A1 V1 A2 V2 known as the equation of continuity A Showing that during a steady flow: 1 A Q Vol to keep ( Q V Vol ) constant. 1 V - N.B: Q Vol 3 QVol A L m AV ( ) and t t s V A r QVol V A r The massive rate of flow : mass of the liquid flowing in A(V) or out per second. Q m m QVol AL kg AV ( ) t t t s - The total rate of flow in case of several outlets Q Q Q Q A V A V A V t In case of branching Q1 Q2 Q3 or A1V 1 A2 V2 A3V3 - If the branches are all equal: Q1 n Q2 or A1V 1 n A2 V2 ((n) is the number of branches). ~ 135 ~

137 - Blood flows at a low speed through blood capillaries? because the total cross section of the capillaries is greater than that of the artery and the velocity is inversely proportional to the cross section, so this allows the gases exchange as well as it prevents the explosion of the capillaries. * Solved examples: 1. A main artery where blood flows at 0.08 m/s, distributes the blood in 150 blood capillaries each of diameter 1/8 that of the artery. Find the velocity of blood flow through each capillary. A No A A (8 r) 64r No A1 ( r) 150 r ms. * Viscosity: - Is the resistance to the flow of a fluid due to the frictional forces between the fluid layers during sliding. - Is the property causing the presence of frictional forces between the different layers of a liquid trying to prevent their sliding past each other. - Experiments to observe viscosity: 1. Try to pour alcohol & glycerine from two similar beakers. You observe that the rate of flow of alcohol is higher. 2. Try to stir two similar volumes of water & honey with a glass rod. You observe that stirring is easier in case of water while honey stops moving when you just stop stirring. 3. Drop two similar metallic spheres in honey & water & measure the time taken by each to reach the bottom. You observe that the one falling in water reaches the bottom faster. ~ 136 ~

138 - Determination of the coefficient of viscosity ( ) - Suppose that a liquid is kept between two glass sheets. - The lower one fixed & the upper is moving at a uniform velocity ( v ). - The velocity of the liquid will vary from the upper layer to the lower one, from v to 0. - Considering that the liquid is made of several mobile layers, we conclude that: - There is a frictional force between the static sheet and the liquid, this is high and is able to stop the liquid. - This force becomes weaker as we rise up towards the mobile sheet. - This force applied keeps a layer moving at a uniform velocity ( v ). F A( area ) F V ( velocity) F 1 d ( dis tan ce) AV AV F and F Vs N d d F d N Vs AV m.sec (.sec. Pascal kg m s ) M L T 1 1 -The coefficient of viscosity ( vs ): is the tangential force acting on a liquid layer of unit area to produce a difference in velocity of 1m/s between it and another layer distant by a unit distance. F Vs AV d F Vs V d A F Vs A d V F Vs AV 1 d * Applications on viscosity: 1. Lubrication oil is used in machines to reduce the corrosion between their parts, because oil has a high viscosity so it sticks against the different parts of the machine reducing friction & heating. ~ 137 ~

139 2. The shapes of cars are designed glidingly, to reduce their friction against air to be able to overcome the air resistance ( viscosity ) since at low speeds viscosity velocity, while at high speeds viscosity velocity 2, & this increases the fuel consumption. 3. The rate of sedimentation of blood is used to identify some diseases, since the rate of sedimentation of R.B.Cs the square of their radii. In normal cases it is 15 mm/ hr. If the rate increases, this proves that the size of R.B.Cs increased, usually due to their adherence together in case of Rheumatic fever. If the rate decreases this proves that the size of R.B.Cs decreased usually due to Anemia. *Remember:- 1- What is meant by: 1. The flow rate of a liquid is 6 kg/sec. - 6 kg is the mass of the liquid flowing in or out per second. 2.The coefficient of viscosity of glycerine is 6.1 kg /m.s N is the tangential force acting on a liquid layer of unit area to produce a difference in velocity of 1 m/s between it & another layer distant by a unit distance. 2- Give reasons for: 1. The speed at which blood flows through capillaries is less than that at which it flows through artery providing them with blood. - Bec. the total area of blood capillaries is greater than that of the artery supplying them with blood as A 1 v 1 = A 2 v 2 so Aα 1/ v & the velocity of blood flow decreases as the area increases. 2. Firemen use wide rubber tubes ending with narrow valves. - Bec. at steady flow the rate of flow in = the rate of flow out & A 1 v 1 = A 2 v 2 so Aα 1/ v so they use narrow valves to increase the velocity of water getting out to reach long distances. ~ 138 ~

140 3.Oil is used in machines. - Bec. it has a high viscosity so it is used to lubricate machines as it sticks against their parts for a long time reducing friction, heating & corrosion. 4. At high speeds, cars consume much fuel. - Bec. at high speed air viscosity α v 2 so the car's engine consumes much fuel to do much work to overcome the viscosity of air. 5. The rate of blood sedimentation is used in medicine. - Bec. the rate of blood sedimentation α radius 2 of R.B.C. the is 15 mm/hr. If it increases this means an attack by a rheumatic fever while its decrease means the attack by anemia which destroys the cells reducing their size. 6. A metallic ball falls slower in glycerine that in water. - Bec. glycerine has a higher viscosity which resists the flow of bodies through it due to frictional forces. 7. The diameter of a water column flowing out of a hose increases when directing its tip upwards. - Bec. water flowing out is decelerated due to gravity decreasing its velocity, and as the flow is steady A 1 according to the equation of continuity and so the diameter of the water column increases. V ~ 139 ~

141 * Solved examples: 1. Two metallic sheets are separated by a liquid layer 6 mm thick & whose coefficient of viscosity is 4 kg m - 1 s - 1. If the area of the upper sheet is 0.03 m 2 & it was affected by a horizontal force of 5 N. Calculate its velocity. 3 FS ms. A vs 2. Two parallel plate of distance 2.5 cm apart. The fluid between them is glycerin with Vs kgm s 1 1. Calculate the force necessary to move a thin plate of 0.75 m 2 area between them with velocity 0.5 m/s in the following two cases: a) If the plate is in the middle between the two fixed plates. b) If the plate is 1 cm apart from one of the two fixed plates. a) F1 F2 F A F F1 F2 2 F 2( vs ) 2(0.785 ) 47 N 2 d A A F F F (0.785 ) (0.785 ) 49 N b) 1 2 vs vs 2 2 d1 d *Steady flow: -Equation of continuity:- AV A V Volumetric rate of flow:-. Q -Massive rate of flow:- Q m = Q vol -Total rate of flow:- vol AL AV t OR t t Vol AL t t Q Q Q Q Vol t Vol Vol Vol t t t t Guide rules for problems Branching tubes:- AV 1 1 A2V 2 NO of branches Fd Fd1 Fd2 *viscosity:- vs or VS AV AV AV ~ 140 ~

142 Homework 1- Complete: 1. The rate of flow of a liquid at a certain point is determined by the number of... that pass perpendicular to.. at the point. 2. The volume of a liquid flowing at a velocity ( v ) through a tube of cross section ( A ) per second is calculated from the relation Some liquids as alcohol & water have a rate of flow & a. resistance to the motion of bodies inside them, so they are said to have viscosity. 2- Choose the correct answer: 1. When the diameter of a pipe is doubled, the volumetric rate of flow of the liquid flowing through it steadily is.. ( halved doubled unchanged quartered ) 2. The coefficient of viscosity is measured in... ( J/m - kg/m 2 - kg m - 1 s N/m) 3. From the diseases where the volume, radius, & rate of sedimentation of R.B.Cs decrease, is the... ( Anemia - Rheumatic fever - Aids ) 3- What is meant by: 1. The flow rate of a liquid = 0.9 kg/sec. 2. The flow rate of a liquid = 4 m 3 /sec. 3.The coefficient of viscosity of glycerine is 8.3 kg /m.s. 4- Give reasons for: 1.Oil is used in machines. ~ 141 ~

143 2. Firemen use wide rubber tubes ending with narrow valves. 3. The speed at which blood flows through capillaries is less than that at which it flows through artery providing them with blood At high speeds, cars consume much fuel The rate of blood sedimentation is used in medicine The diameter of a water column flowing out of a hose increases when directing its tip upwards A metallic ball falls slower in glycerine than in water Define: 1. Steady flow. 2. Viscosity. 3. Coefficient of viscosity. ~ 142 ~

144 6- What are the factors controlling each of the following: 1. Coefficient of viscosity Essay: 1. What are the conditions necessary for the steady flow of a liquid. 2. Draw a diagram to derive the equation of continuity for steady flow. 3. Draw a diagram to derive the relation used to calculate the coefficient of viscosity. * Problems: 1. If the velocity of water flowing through the wide part of a tube whose diameter is 1.2 cm is 3 m/s. Calculate the diameter of its narrow end where water s velocity becomes 27 m/s. ( 0.4 cm ) ~ 143 ~

145 2. The rate of water flow through a tube of area 2 cm 2 is m 3 /min. Calculate the velocity of water. ( 2 m /s ) 3. Calculate the rate of flow of oil through a tube of diameter 4 cm, when flowing at a velocity of 12 m /s. ( m 3 /s ) 4. If the velocity of blood flowing through an artery of radius 0.5 cm is 0.04 m/s & the artery provides blood to 100 veins, each of radius 0.1 cm. Find the velocity of blood flowing through veins. ( 0.01 m/s ) 5. A liquid flows through a tube of non-uniform cross section. If the velocity at a point (A) is twice the velocity at a point (B) where the cross section of the tube is 4 cm 2. Find its cross section at (A). ( 2x10-4 m 2 ) ~ 144 ~

146 6. If the diameter of the tube at (A) is 45 cm, at (B) 30 cm, at (C) 20 cm, & at (D) 15 cm. Calculate: A B D a) The rate of flow at (A) given that water flows at 2 m /s. C b) The velocity at each of (B) & (D) if its velocity at (C)is 4 m /s. ( = 3.14 ) ( m 3 /s - 4.5m /s m /s ) 7. Water flows through a tube of diameter 2 cm at a velocity of 5 m/s. Calculate the volume of water flowing per minute & the time needed to fill a tank of 20 m 3. ( m 3 /min min. ) 8. In the shown figure: given that, the radius of the tube at (a) C is 30 cm and the velocity of water entering at the same point d b a is 2 m/s, the velocity of the water flow at (c) is 4m/s and at e (e) is 3 m/s where the radius of the tube at (b) is 20 cm, at (c) is 15 cm, at (d) is 10 cm and at (e) is 5 cm. Calculate:- a)the rate of volume of water entering at (a). b) The velocity of the water flow at (b) and (d). ( 0.56 m 3 /s 4.5 m/s 8.25 m/s ) ~ 145 ~

147 9. Three valves can fill a tank, the first fills the tank in ¼ hour, the second fills that tank in ½ hour, the third fills that tank in 1 hour, so if they are opened together. Find the time taken to fill the tank. ( 8.57 min ) 10. A water pump, pumps water from a fresh water lake at a rate of 600 L/min., through a tube of diameter 4 cm & expels it in air at a height of 20 m from the water surface. Find the water velocity at the point of discharge. ( 7.96 m /s ) 11. A water pipe entering a home has a radius of 1.5 cm and the speed of the water is 0.2 m/s. If the pipe becomes of radius of 0.5 cm at its end, Calculate: a) The speed of water at the narrow end b) The volume of water that flows per minute across any cross sectional area ( = 3.14 ) ( 1.8 m/s m 3 /min. ) ~ 146 ~

148 12. A squared metallic sheet of side 0.2 m long, separated from another by a liquid of thickness 0.4 cm & is affected by a force of 20 N to move at a velocity of 1 m/s. Calculate the coefficient of viscosity. ( 2 kg m -1 s -1 ) 13. A squared metallic sheet of side 10 cm long, moves at a velocity of 10 cm/s, parallel to another one. Knowing that they are separated by water whose coefficient of viscosity is 0.01 gm/cm.s, and that the force acting upon is of 200 dynes. Calculate the distance between the two sheets. ( 0.05 cm ) 14. A viscous liquid layer 8 cm thick is found between two parallel large planes. If the coefficient of viscosity of the liquid is 0.8 kg/m.s. Calculate the force needed to move a thin sheet of 0.5 m 2 at a velocity of 2 m/s between both planes at a distance of 2 cm apart from one of them. Then find the pressure due to this force on the lower plane. ( N Zero ) ~ 147 ~

149 School book revision questions ~ 148 ~

150 Answers 18) 1. Liquid fills the tube completely. 2. The quantity of liquid flowing in at a certain time equals the one flowing out at the same time. 3. The flow is not rotational. ( not a vortex flow containing vortices ) 4. The flow is represented by imaginary stream lines that do not intersect. 5. The rate of flow is represented by the number of stream lines passing per unit area at a point. * - At steady flow, the rate of flow is constant. & Q Vol 1 = Q Vol 2 L t A L t A L t x where (Δx) is the distance covered by the liquid per unit time (velocity). t the relation becomes A1 V1 A2 V2 known as the equation of continuity & A A V V 1 1 where A V ~ 149 ~

151 ~ 150 ~

152 UNIT ( 3 ) HEAT CHAPTER ( 5 ) GAS LAWS * Brownian movement :- (random movement). - Illuminate a glass box by a strong light. - Introduce smoke in it and observe by the aid of a microscope the motion of graphite particles. We observe that they move randomly. - The particles of air moving in all directions in a vibratory and translational motion at different velocities collide against the graphite particles. When collision on one side is stronger than the other the last moves. - Gaseous molecules always move randomly ( in a haphazard motion ) colliding against each other and against the walls of container elastically occupying its volume. - How to prove the presence of intermolecular spaces:- - A jar containing ammonia gas of low density is inverted on the one containing hydrogen chloride gas of higher density. - A white cloud of NH 4 Cl (ammonium chloride) is formed along the tubes. - Although HCl is more dense it moves up diffusing in the intermolecular spaces of NH 3 and vice versa so they combine to form NH 4 Cl. Intermolecular spaces are found between the molecules of gases and weak intermolecular forces allow them to move randomly in all directions. ~ 151 ~

153 To study the relation bet volume (vol), pressure ( P ) of temperature ( t ) of a gas, one of the three variables must be fixed at a time. 1- Relation between Vol. and P. at constant ( t o ) 2- Relation between Vol. and t o at constant ( P. ) 3- Relation between P. and t o at constant ( Vol. ) 1- Relation between Vol. & P. at constant t:- Boyle s law: * Exp:- 1- Open the tap of the burette B after filling the tube A with Hg. 2- Move A up & down until the Hg level becomes equal in both A and B, then measure the volume of air in the burette Vol 1 whose pressure P 1 = P a after closing the tap. 3- Raise the tube A up and measure the new air volume Vol 2 of pressure P 2 = Pa + g h. 4- Move the tube A down & measure the volume of air in the burette V 3 of pressure P 3 = Pa - g h. 5- Repeat step 3 & 4 for several times & draw the relation Vol Vs 1 P 1 Vol & P Vol cont. P P P Vol Vol or Boyle s law :- For a certain mass of gas, its volume is inversely proportional to its pressure provided that its temperature is constant. N.B:- CaCl 2 is added over Hg to dry air. - The motion of the tubes must be slow to keep t constant. P 2- The effect of temperature change on the volume at constant P: Exp :- To show the effect of t on equal Vol. of different Gases :- - Get two flasks of similar volumes fitted with 2 narrow glass tubes each containing a thread of Hg. Vol. ~ 152 ~

154 - Fill one with O 2 & The other with CO 2. - Dip both flasks in a hot water bath. - We observes that the Hg threads move similarly in both. - This proves that equal volumes of different gases kept under constant pressure expand similarly for a similar t rise. - These are said to have the same coefficient of cubic expansivity. * Coefficient of cubic expansivity ( vol ): Is the amount of increase in a unit volume of a gas at 0C if its t is raised by 1C at constant pressure - The coefficient of cubic expansivity vol :- V depends on :- t (change of t o.) V o ol 0 c ( Volume of the gas at 0C) and it was found experimentally that :- V ol V ol 0 o & V ol t V ol V ol 0 o t ( vol is the constant ) Vol Vol Vol t 0 * Determination of vol practically:- Charles s law ( Relation between the volume and temperature of a gas at a constant pressure): - get a glass tube 30 cm. Long & 1 mm in diameter closed from its lower end. - A Hg ( H 2 SO 4 ) drop is introduced to trap air, dry it & keep its pressure constant. - The tube is inserted in a jar full of ice starting to melt (at 0C). - Measure the length of air column, this is proportional to its volume ( V o ol 0 c) since its cross section is uniform. - Remove ice & water & start passing vapor at 100C through the jar. ~ 153 ~

155 - Wait for a while & then measure the length of the air column. (V ol 100 o c ). V V V V V AL AL V (100 0) V AL ol 100 ol 0 ol 100 ol ol 0 ol For all gases was found to be equal per degree. N.B: Precautions while performing Charles experiment: 1. The narrow tube must be of a uniform cross sectional area, to be able to substitute the volume by length and facilitate the calculation of Vol. 2. Use a drying agent to remove water vapour which is not an ideal gas from air. 3. Do not measure the volume (length) unless the liquid drop rests at equilibrium completely to grant that the pressure is constant = P a. Charles s law:- For a certain mass of gas, its volume increases by of its volume at 0 o C for each degree rise in temperature provided that its pressure is kept constant. 3- The effect of temperature change on pressure at a constant Volume:- Exp :- To show the effect of t on P. at constant Vol. - A glass flask containing air is fitted with a U shaped tube containing Hg. - Hg in both branches of the tube has the same level. (fig. a) - Put the flask in a hot water bath, the gas expands and pushes the Hg which rises up in branch B ( fig. b). - Add Hg to get the Hg in A back to its initial level. ~ 154 ~

156 - Now the Hg level at B is higher than that at A by ( h ) cm. which proves that the pressure of gas increases by heating ( fig. c). - Repeat the experiment several times using equal volumes of different gases. - We find that different gases increase their pressures similarly, for similar t rise provided that their volumes are kept constant. * Coefficient of pressure expansivity at constant Vol. ( P ) Is the amount of increase in the unit pressure of a gas at 0C if its t increases by 1C at constant volume. - The coefficient β P :- P depends on Po & t. Where P P o & P t P Po t p P Po t * Practical determination of P Pressure s law :- ( Relation between the pressure and temperature of a gas at a constant volume): - Jolly s instrument: ( Gaseous thermometer ) - It consists of: - A flask filled to 1 7 of its volume with Hg to overcome the expansion of glass. - The flask is connected to a tube (B) which is connected through a rubber tube to another barometric tube (C) filled with Hg. - A vertical graduated scale & a thermometer. Procedure :- 1- Dip the flask in melting crushed ice 2- Wait for a while to allow the t of the dry air inside the tube to reach 0 C. 3-Move the tube C up & down until the Hg in tube B reaches a certain point ( x ). ~ 155 ~

157 4-Measure the difference of Hg levels in B & C, this is called ( h ). 5-Pressure of air in the flask is calculated as: Po = P a - g h or ( P a h ) 6-Place the flask in boiling water & wait for a while until air reaches 100 C. 7-Move C until the level of Hg in B returns to the same point ( x ) to keep the volume of the gas constant. 8- Measure the difference in Hg levels between B & C call it ( h 1 ). 9-The pressure of air is now calculated as :- P 100 = P a + g h 1 or ( P a + h 1 ) The pressure expansivity of a gas at constant volume β p is given as :- P P100 Po P 100 o N.B :- P for all gases = Vol = P for all gases. * Pressure s law :- for each degree rise in t. For a certain mass of gas, the pressure increases by of its value at 0 C, for each degree rise in temperature, provided that the volume is kept constant. * Graphs representing the relations between ( P & t o ) & ( Vol & t o ) & their uses to find the absolute zero (zero Kelvin) - By plotting the results obtained in the previous experiments we obtain the graphs from which we can find that :- ~ 156 ~