List of Figures Page Figure 8.1 Figure 8.2: Figure 8.3: Figure 8.4: Figure 8.5: Figure 8.6: Figure 8.7: Figure 8.8: Figure 8.9: Figure 8.

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1 Contents Page Chapter 08 Wave and Geometric Optics Introduction Transverse and Longitudinal Waves Parameters of Wave The speed of a Travelling Wave Wave Speed on a Stretched String Interference of Waves Energy and Power of a Traveling String Wave Phasors Sound Wave Traveling Sound Wave Interference of Sound Wave Doppler Effect Supersonic Speeds Geometric Optics Mirror Spherical Refracting Surfaces Thin Lenses Two Lens Systems Optical Instruments Simple Magnifying Lens Compound Microscope Refracting Telescope Tutorials i -

2 List of Figures Page Figure 8.1: Illustration of transverse wave Figure 8.2: Illustration of longitudinal wave Figure 8.3: A moving beetle creates both transverse and longitudinal waves Figure 8.4: Waveform of the wave function y(x, t) ym sin(kx t) Figure 8.5: Illustrations of the interference displacement of two waves of different phases Figure 8.6: Phasor representation of two waves of small frequency and the displacement result of interference Figure 8.7: Illustration of a sound wave travels from a point source S Figure 8.8: Plots of (a) displacement and (b) air pressure at t = Figure 8.9: Two point sources S 1 and S 2 emit spherical sound waves in phase Figure 8.10: Illustration of Doppler Effect Figure 8.11: Illustration of a source moving at speed v equal to speed v s of sound showing it moving as fast as the wavefronts Figure 8.12: Illustration of a source moving at speed v faster than speed v s of sound showing Mach cone of wave Figure 8.13: An extended object O and its virtual image l in a plane mirror Figure 8.14: An illustration of image formed by a concave mirror Figure 8.15: An illustration of image formed by a convex mirror Figure 8.16: Real image formed by concave mirror Figure 8.17: Focal point F and length of a concave mirror Figure 8.18: Focal point F and focal length of a convex mirror Figure 8.19: Images formed by spherical refracting surfaces Figure 8.20: (a) Convergence of light by convex lens and (b) divergence of light by concave lens Figure 8.21: Image formed by convex lens and concave lens Figure 8.22: Object placed on near point of human eye Figure 8.23: Object placed closer to the human eye, a distance shorted than near point Figure 8.24: Correction for fussy image formed by object placed shorter than Pn closer to eye and closed to the focal point Figure 8.25: Correction for fussy image formed by object placed shorter than P n closer to eye and is inside the focal point Figure 8.26: The structure of a compound microscope Figure 8.27: Structure of a refracting telescope Figure 8.28: The height of image and angle made by the parallel rays and observer ii -

3 Chapter 08 Wave and Geometric Optics 8.0 Introduction Wave and geometric optics is a study of behavior and properties of wave transmitting in the medium or vacuum and certainly we will study the applications of light wave for mankind. Wave can be classified into three types, which are mechanical wave, electromagnetic wave, and matter wave. Mechanical wave is mechanically generated wave such as water wave, sound wave, and seismic wave. All these waves are governed by Newton s laws and they can exist only with material medium like water, air, and rock etc. Electromagnetic wave is wave that does not require medium for transmission. Wave such as visible light, x-ray, and radar wave are examples of this type of wave. All electromagnetic waves travel in vacuum with the speed of x10 8 m/s. Matter wave is commonly used in modern technology. The wave is associated with electron, proton, and other fundamental particles including atoms and molecules. These particles and molecules constitute matter. This is the reason the wave is termed as matter wave. Light wave is commonly known as light. Geometric optics, a study of how light is transmitted through medium such as lens and uses its properties through these mediums to design instruments such as microscope and telescope to extend human visual capability. In this chapter, we begin to discuss the types of mechanical waves like transverse, longitudinal wave, and sound wave, in the aspects of their properties and interference. We will also discuss how is form image, reflection due to mirror, refraction and refraction from lenses, and basic design of optical instruments

4 8.1 Transverse and Longitudinal Waves 08 Wave and Geometric Optics Transverse wave has its displacement perpendicular to the direction of traveling wave. An example of generating transverse wave is by moving the tied string up and down continuously as shown in Fig Figure 8.1: Illustration of transverse wave Longitudinal wave is the wave type that has its displacement parallel to the direction of the moving wave. Moving a piston back and forth in the air filled pipe will create longitudinal wave as shown in Fig Figure 8.2: Illustration of longitudinal wave A moving beetle on the surface of sand will both create transverse and longitudinal waves as illustrated in Fig Each of these waves has different speed. It allows its predator like scorpion to catch

5 Figure 8.3: A moving beetle creates both transverse and longitudinal waves Example 8.1 The movement of a beetle sends out two pulses, which are longitudinal and transverse pulses. The speed of longitudinal pulse is 150ms -1, while the speed of transverse pulse is 50m/s. A scorpion has eight legs spread roughly in circle about 5cm in diameter intercepts the faster longitudinal pulse first and learns the direction of beetle and then sense the time difference t between first intercept and second intercept of slow transverse pulse to determine the distance d to the beetle. Find the distance d if the time difference is 5.0ms. Solution The time difference t is equal to Thus, the distance d is equal to Parameters of Wave d d t. v T v L d t / = 5.0x10 / vt v L = 37.5cm. A description of a wave on a string and the motion of any element along the length require a function that provides the shape of the wave. This means that it needs a relation in the form y = h(x, t). It is a function of transverse replacement, which depends on time t and position x. Mathematically, it can be expressed by equation (8.1). y(x,t) ym sin(kx t) (8.1)

6 where y m is the amplitude, sin( kx t) is oscillation term, k is the angular wave number, x is the position, and is the angular frequency. Based on equation (8.1), the waveforms are shown in Fig (a) (b) Figure 8.4: Waveform of the wave function y(x, t) ym sin(kx t) The phase of the wave is the argument ( kx t). The wave sweeps through a string element at a particular position x. The phase changes linearly with time t. This shall mean that the sine function also oscillates between +1 and -1, which is corresponded to a peak and a valley. The peak is called amplitude y m. The wavelength of a wave is the distance parallel to the direction of the wave s travel, between repetitions of the shape of the wave. The illustration is shown in Fig. 8.4(a). At time t = 0, the wave function from equation (8.1) is y(x,0) ym sin(kx). Based on this equation, the displacement is the same at position x = x 1 and x = x 1 + as shown in Fig. 8.4(a), which is ym sin(kx1) = y m sink x 1. This implies that k = 2. Thus, the angular wave number k is equal to k 2 (8.2) Period of the oscillation T of a wave is the time taken at any string element to move through one complete oscillation. If one fixes the position say x = 0, the wave equation is equal to y(0,t) ym sin( t). It is also equal to y(0,t) ym sin( t) At time t = t 1 and t = t+t, the displacements are the same as shown in Fig. 8.4(b). Thus, function y(0,t 1) ym sin( t1) = ym sin(t 1 T). This implies that T = 2. This shall mean that the angular frequency of the wave is equal to 2 T. (8.3)

7 The frequency of the oscillation f is equal to 1/T. Thus, the angular frequency is also equal to 2f The speed of a Travelling Wave If the wave is travelling in the position x-direction, after a small time t, it travels a distance x. The ratio of x/t is the wave speed. As the wave moves at a fixed point, the displacement is retained. This shall mean the phase (kx-t) is constant. Thus, wave speed can be determined from (kx-t) = constant. Thus, dx k 0 (8.4) dt This implies that the wave speed is dx dt k (8.5) Example 8.2 A wave traveling along a string is described by y(x,t) sin(72.1x 2.72t) (a) What is the amplitude of this wave? (b) What are the wavelength, period, and frequency of this wave? (c) What is the velocity of this wave? (d) What is the displacement at x = 22.5cm and t = 18.9s?. Solution The amplitude of this wave is m. The angular wave number is 72.1/m. The wavelength of the wave is 2/72.1 = 87.15mm. The angular frequency of the wave is 2.72Rad/s. Thus, the frequency of the wave is 2.72/2 = 0.432Hz. The period of this wave is 1/0.432 = 2.31s The velocity of this wave is =3.77cm/s. k 72.1 The displacement is y(x,t) sin(72.1x 2.72t) = sin(72.1x22.5x x18.9) = sin( 35.18rad ) = sin( 35.18Rad ) = x0.583 = 1.91mm

8 8.1.3 Wave Speed on a Stretched String The speed of wave defined by equation (8.5) is set by the properties of the medium it travels. If the wave is traveled through a medium such as water, air, steel, or stretched string, it causes that medium to oscillate. For this to happen, the medium must possess both mass and elasticity. For a string that has mass m and length l, we define the linear density as equal to the ratio of m and l, which is m l (8.6) We cannot send a wave along a string unless it is under tension. Thus, the tension in the string is equal to the common magnitude of the two forces tying the ends of the string. The speed of the wave traveling in the stretched string can then be defined as (8.7) Interference of Waves Suppose we send two sinusoidal waves of same wavelength and amplitude in the same direction along a stretched string. The superposition principles applied, which states that the displacement of the string when the waves overlap is the algebraic sum of the displacement of the individual wave. The waves can combine which we call interference. Supposing one wave is traveling along a stretched string is given by y1(x,t) ym sin(kx t) and the second one is given by y2(x,t) ym sin(kx t ), where is the phase constant. When these two waves interference, the resultant displacement is given by ' y (x,t) y sin(kx t) y sin(kx t ) (8.8) m m ' which is equal to y (x, t) 2y cos sin kx t. Note that 1 sin sin 2sin 2 interference is y cos 2 m cos. From this equation, the amplitude at 2 m and oscillation term is sin kx t. If is equal to zero then at interference, the magnitude is 2y m. This is a case of fully

9 constructive interference. If is equal to then at interference, the magnitude is equal to zero. This is a case of fully destructive interference. If is equal to 60 0 then the amplitude is y m. This is a case of intermediate interference. The illustrations are shown in Fig (a) = 0 (b) = rad (c) = 2/3 rad Figure 8.5: Illustrations of the interference displacement of two waves of different phases Energy and Power of a Traveling String Wave As the wave travels in the stretched string, it transports both kinetic and elastic potential energies. An element of string has mass dm oscillating transversely in simple harmonic motion as the wave pass through it. It has kinetic energy associated with its transverse velocity u. When the element is rushing through its y = 0 position, its transverse velocity is maximum. When the element is at y = y m, its transverse velocity is zero. Thus, at y = 0, the kinetic energy of the element is maximum, while at y = y m, the kinetic energy is zero. The string must be stretched in order to send the sinusoidal wave. As the string element of length dx oscillates transversely, its length must increase and decrease in a periodic way if the string element is to fit sinusoidal waveform. When the string element is at y = y m, its potential energy is equal to zero, while at y = 0, its potential energy is maximum because it stretches to maximum. The string element has maximum elastic and kinetic energies at y = 0 and zero energy at y = y m

10 by The kinetic energy dk associated with a string element of mass dm is given 1 2 dk u dm (8.9) 2 The transverse speed of the oscillating string can be determined by differentiating equation (8.1) with respect time t. while keep to position x constant, which is y(x, t) dy sin(kx t d m u = y m cos(kx t) dt dt (8.10) Equation (8.9) becomes equation (8.11) after substituting equation (8.10) and dm = dx from equation (8.6). dk y cos kx tdx (8.11) m Diving equation (8.11) with dt, it yields the rate of energy transmission equation, which is shown in equation (8.12). dk 2 m dt 2 dx dt 2 2 y cos kx t (8.12) dx Note that is equal to, the traveling speed of the wave. The average rate at dt which kinetic energy is transported is the average over an integral number of wavelength and using the fact that the average value of the cosine square function over an integral number of period is 1/2. Hence, dk dt avg 1 2 y 4 2 m (8.13) The average rate of elastic potential energy carried along with the wave is same as the average rate of kinetic energy carried along with the wave. Thus, the average power P avg transmitted by the wave is the sum of average rate of kinetic energy and average rate of elastic potential energy transmitted. i.e. P avg ym (8.14)

11 8.1.6 Phasors Supposing two sinusoidal waves of functions y1(x,t) ym 1 sin(kx t) and y2(x,t) ym2 sin(kx t ), when they interfere each other, the displacement of interference is equation (8.15). (x,t) y sin(kx t) + sin(kx t ) ' y m1 y m2 (8.15) ' ' The displacement of the interference is also equal to y (x,t) ym sin(kx t ). ' The amplitude y m and phase constant can be calculated from phasor representation of the waves, which are illustrated in Fig The amplitude y can be calculated using equation (8.16). ' m y ' m 2 y cos y y sin 2 (8.16) m2 m1 The phase constant can be calculated using equation (8.17). y ym2 sin cos y m2 1 tan (8.17) m2 m1 Figure 8.6: Phasor representation of two waves of small frequency and the displacement result of interference 8.2 Sound Wave Sound wave is defined roughly as longitudinal wave. Seismic engineer uses such wave to probe Earth s crust for fossil oil. Submarine uses sound wave to stalk other submarine by listening for the characteristic noise produced by the propulsion system. Point source sound represents a tiny sound source that emits sound in all directions. Figure 8.7 illustrates a sound wave travels from a point source S. The

12 wavefronts and rays indicate the directions of travel and the spread of the sound waves. Wavefronts are surfaces over which the oscillation of the air due to the sound wave has the same value. This surface is represented by whole or partial circles in two dimensional drawing for point source. Rays are directed lines perpendicular to the wavefronts that indicate the direction of travel of the wavefronts. The double short arrows indicate longitudinal oscillations of the air are parallel to the rays. Near point source the wavefronts are spherical and spread out in three dimensions. The wave is termed as spherical wave. As the wavefronts move outward and their radii become larger and their curvature decreases. Far from the source, we can approximate the wavefronts as planes or lines in two dimensional drawing. Thus, this wave is called planar wave. Figure 8.7: Illustration of a sound wave travels from a point source S In stretched string, potential energy is associated with the periodic stretching of the string element as the wave passes through them. Sound wave passes through air, potential energy is associated with periodic compression and expansion of small volume of air. Thus, the pressure on the medium is changed to bulk modulus B, which is defined as B p V / V (8.18) Here V / V is the fractional change in volume due to change in pressure p. Bulk modulus has dimension pascal, which has unit kgm -1 s -2. In order to the dimensional unit for the speed of sound wave traveling in medium, linear density has to be changed to density of the medium. Thus, the speed of sound in a medium is

13 B (8.19) Traveling Sound Wave The air element of the sound wave oscillates longitudinally. The displacement s(x, t) of the sound wave is s(x,t) sm cos(kx t) (8.20) or it can be s(x,t) sm sin(kx t). As the wave moves, the air pressure at position x varies sinusoidally. Thus, the description of air pressure variation is p(x,t) pm cos(kx t) (8.21) where p m is the pressure amplitude. The pressure amplitude is normally very much smaller than the pressure p when there is no wave. The relationship between displacement s(x, t) and pressure amplitude is shown in equation (8.22). p ( ) (8.22) m s m Figure 8.8 shows plots of displacement and air pressure variation at t = 0. The result shows the displacement s and pressure variation p are 90 0 out of phase. The pressure variation p at any point along the wave is zero when the displacement is at maximum. (a) (b) Figure 8.8: Plots of (a) displacement and (b) air pressure at t =

14 8.2.2 Interference of Sound Wave It is like the transverse wave, sound wave can undergo interference. Let s consider two identical point sources as shown in Fig. 8.9 emitting two identical sound waves that are in phase and of identical wavelength. The waves will be in phase if they travel with identical paths to reach point P. If they are in phase at point P, like the transverse waves, they will undergo constructive interference at point P. However, if the path traveled by wave from S 2 is longer than the wave from S 1 or vice versa then the waves may not be in phase when they reach point P. Figure 8.9: Two point sources S 1 and S 2 emit spherical sound waves in phase The difference in path length is L = L 2 L 1. To relate phase difference with different path length L, their relationship is shown in equation (8.23). 2L (8.23) For constructive interference, the phase difference should be either 0, 2, 4,.. It would mean that the phase difference is integral multiple of 2. i.e. m, where m = 0, 1, 2, 3,. (8.24) 2 L This is also meant that = 0, 1, 2, 3,.. For destructive interference, the phase difference should be, 5,... It will mean that the phase difference is odd multiple of. i.e. 2 m, where m = 0, 1, 2, 3,. (8.25) 1 3,

15 L This is also meant that = 0.5, 1.5, 2.5, 3.5,.. Two waves at point P may neither constructive interference nor destructive L interference. It can be intermediate interference. This shall mean that has value not specified by either constructive or destructive interference Intensity and Sound Level The intensity l of a sound wave at a surface is the average rate per unit area at which energy is transferred by wave through or onto the surface. Thus, P A l (8.26) P is the time rate of transfer of energy of the sound wave, which is power. A is area of surface intercepting sound. The intensity is related with displacement s illustrated by equation (8.27). 1 2 l (8.27) 2 2 sm The intensity varies with distance from a real sound source is often complex. Like loudspeaker may transmit sound only in a particular direction and environment usually produces echo that overlap the direct sound wave. If we assume that the source is point source, which produces isotropically sound then the energy emitted from the source must pass through the surface of the sphere that has surface are 4r 2. Thus, the time rate at which energy is transferred through the surface by the sound waves must be equal the time rate at which energy is emitted from the source. From equation (8.26), the intensity of the sound wave is P 4r l (8.28) S 2 P S is power of the source. From equation (8.28), we notice that the intensity l decreases with the square of the distance r

16 Wave and Geometric Optics Instead of deal with intensity of sound, it is better and convenient to use sound level, which is defined as l l 10dBlog (8.29) 0 Here db is denoted as decibel, which is unit for sound. l 0 is the standard reference intensity that has value W/m Doppler Effect The Doppler Effect is observed whenever the source of waves is moving with respect to an observer. The Doppler Effect can be described as the effect produced by a moving source of waves in which there is an apparent upward shift in frequency for observers towards whom the source is approaching and an apparent downward shift in frequency for observers from whom the source is receding. It is important to note that the effect does not result because of an actual change in the frequency of the source. Doppler Effect was proposed y in 1842 by Austrian physicist Johann Christian Doppler. Astronomers who use the shift in frequency of electromagnetic waves produced by moving stars in our galaxy and beyond in order to derive information about these stars and galaxies. The belief that the universe is expanding is based in part upon observations of electromagnetic waves emitted by stars in distant galaxies. Furthermore, specific information about stars within galaxies can be determined by application of the Doppler Effect. Galaxies are clusters of stars that typically rotate about some center of mass point. Electromagnetic radiation emitted by such stars in a distant galaxy would appear to be shifted downward in frequency (a red shift) if the star is rotating in its cluster in a direction that is away from the Earth. On the other hand, there is an upward shift in frequency (a blue shift) of such observed radiation if the star is rotating in a direction that is towards the Earth. The Doppler Effect can be observed for any type of wave - water wave, sound wave, light wave, radio wave etc. We are most familiar with the Doppler Effect because of our experiences with sound waves. As illustrated in Fig. 8.10, when a police car or emergency vehicle is traveling towards you on the highway, as the car approached with its siren blasting, the pitch of the siren sound is high and after the car pass by the pitch of the siren sound is low. This is a typically example illustrating the Doppler Effect. There is an apparent shift in frequency for a sound wave produced by a moving source.

17 Figure 8.10: Illustration of Doppler Effect If either the detector or source is moving, or both are moving, the emitted frequency f and the detected frequency f are related by equation (8.30). ' D f f (8.30) S where is the speed of sound through air, D is the detector s speed relative to air and S is the source s speed relative to the air. The choice of plus or minus signs is set by the rule, which is stated here. When motion of detector or source is moved toward other, the sign on its speed must give an upward shift in frequency. When the motion of detector or source is way from the other, the sign on its speed must give a downward shift in frequency. With the above understanding, let s apply equation (8.30) for the case with stationery source and moving detector. For a stationery source, S is equal to zero. Thus, equation (8.30) becomes equation (8.31). ' D f f (8.31) The plus sign denotes the detector is moving toward the source, while minus sign denotes the detector is moving away from the source. For the case of moving source and stationery detector, D is equal to zero. Thus, equation (8.30) becomes ' f f (8.32) S

18 The minus sign denotes the source is moving toward the detector, while plus sign denotes the source is moving away from the detector. Example 8.3 A rocket moves at a speed of 242m/s. directly toward a stationery pole while emitting sound wave at frequency f = 1,250Hz. (a). What is the frequency f measured by a detector that is mounted on the pole? (b). Some of the sounds reach the pole reflected back to rocket as an echo. What is the frequency detected by the detected mounted on the rocket? Let s look at the general equation f ' f D S and apply it to this scenario. To the detector mounted on the pole, the rocket (source) is approaching it, thus, the relative speed is v+v D. To the rocket, the detector is closer as time lapsed. Thus, the relative speed is v-v S. Thus, the Doppler frequency equation applying to this scenario is f ' f D S Since v D is equal to zero, the f measured by the detector mounted on the pole is f ' f S 343 1,250x = 4,250Hz. v S is equal to zero, the frequency f measured by the detector mounted on the rocket is 8.4 Supersonic Speeds D f " f 4250x 343 = 7,240Hz. If a source is moving toward stationery detector at the speed of sound, which is speed of sound v equal to speed of source v s, then from equation (8.32) the detected frequency will be infinitely large. This means that the source is moving so fast that its keeps pace with its own spherical wavefronts as shown in Fig

19 Figure 8.11: Illustration of a source moving at speed v equal to speed v s of sound showing it moving as fast as the wavefronts If the speed of sound v is exceeded the speed of sound v s, which is at supersonic speed then all the spherical wavefronts bunch along a V-shaped cone envelope as illustrated in Fig This V-shaped cone is called Mach cone. A shock wave is said to exist along the surface of this cone because the bunching of wavefornts causes an abrupt rise and fall of air pressure as the surface passes through any point. W 1 is the wavefront when the source is at S 1. Similarly W 6 is the wavefront when the source is at S 6. Figure 8.12: Illustration of a source moving at speed v faster than speed v s of sound showing Mach cone of wave

20 The radius of any wavefront in Fig is vt, where v is the speed of sound after and t is the time that has lapsed since the source emitted that wavefront. From Fig. 8.12, the half angle of the cone is called Mach cone angle and it is given by equation (8.33). sin vt v v t s vs (8.33) where v s t is the distance travelled by the source in time t. The ratio of v s /v is called Mach number. 8.5 Geometric Optics We shall begin with the study of image formed by mirror and lens. There are two types of image, which are real image and virtual image. Real image can be formed on a surface such as on a card or movie screen. Virtual image is only exist within the brain but nevertheless is said to be in perceived location. When a person stands in front of a mirror, virtual image is formed behind the mirror. This image of course is not real image. We shall explore several ways in which real and virtual images are formed by reflection with mirror and refraction with lenses. Finally we shall discuss the use of lenses for designing optical instruments Mirror We shall discuss three types of mirror namely plane mirror, concave mirror, and convex mirror pertaining their reflection, formation of image, central axis, focal point, and formulae to calculate their lateral magnifications. The image formed by a plane mirror from an extended object as illustrated in Fig is a virtual image. The distance between the object and the plane mirror p is equal to the distance between the virtual image and the mirror is the same. Thus, p is equal to i. If the object has height h and the height of virtual image is h then the lateral magnification m is defined as h' m (8.34) h For the case of plane mirror, the lateral magnification is equal to -1 since the height of the object and image is the same. It can also be shown that the lateral magnification is also equal to

21 i p m (8.35) For plane mirror i = -p since i is at the other side of mirror. This implies that m is equal to +1 for plane mirror. The + means that the object and image have same orientations. Figure 8.13: An extended object O and its virtual image l in a plane mirror Concave and convex mirrors are spherical mirrors. Unlike the plane mirror, they have a finite radius of curvature. Concave mirror has the following characteristics. 1. The center of curvature C is closer and is in front of the mirror. 2. The field view is the extent of the scene that is reflected to the observer. It is decreased. 3. The image of the object is farther behind the concave mirror. 4. The image is large. This is the feature used for making makeup mirror and shaving mirror. The illustration of the image formed by concave mirror is shown in Fig The image is a virtual image formed at the opposite of the mirror

22 Figure 8.14: An illustration of image formed by a concave mirror Convex mirror has the following characteristics. 1. The center of curvature C is closer and is behind the mirror. 2. The field view is the extent of the scene that is reflected to the observer. It is increased. 3. The image of the object is closer behind the convex mirror. 4. The image is small. This is the feature used for mirror placed at road junction and mirror placed at store to have more view. The illustration of the image formed by convex mirror is shown in Fig The image is a virtual formed at the opposite of the mirror. Figure 8.15: An illustration of image formed by a convex mirror

23 For concave mirror, a real image would be formed if the object is placed larger than the focal length from the center point c as illustrated in Fig The lateral magnification m has a negative value since the object and image are in same orientation. Based on equation (8.34) both p and i have are positive value. This is true because the image is a real image. Thus, the distance i between real image and central point c has a positive value. Figure 8.16: Real image formed by concave mirror When the parallel rays reach a concave mirror as illustrated in Fig. 8.17, the ray near the central axis reflected through a common point F called focal point F or focus of the mirror. If one placed a card board at this point, an image of distance point object would be formed. The distance from focal point F to the center of mirror c is called focal length f. This focal point is a real focal point that has a positive value since real image can be formed. When the parallel rays reach a convex mirror as illustrated in Fig. 8.18, the rays near the central axis diverged away. If one s eyes intercept some of these rays, one would perceive the rays as originated from point source. This perceived point source is the focal point F. The distance from focal point F to the center of mirror c is called focal length f. This focal point is a virtual focal point that has a negative value since virtual image can only be formed

24 Figure 8.17: Focal point F and length of a concave mirror Figure 8.18: Focal point F and focal length of a convex mirror For mirror of both types, it can be shown that the focal length f is equal half of radius of the mirror, which is f 1 r (8.36) 2 A simple equation relating the object distance p, image distance i to the center of spherical mirror, and focal length f of the mirror is shown in equation (8.37). 1 f 1 1 p i (8.37)

25 Example 8.4 A tarantula (spider) of height h sits cautiously before a spherical mirror whose focal length has absolute value f = 40cm. The image form the tarantula produced by the mirror has same orientation as the tarantula and has height h = 0.20h. (a). Is the image real or virtual, and is it on the same side or opposite side? (b). Is the mirror concave or convex and what is the focal length f? Solution The image is a virtual image since it has same orientation. It must be at the opposite side of the mirror. The image and object have same orientation. It means that the lateral magnification has positive value. Thus, from equation (8.34), i = p. The focal length f follows equation (8.36), which is f p i.i.e f p 0.2p This implies that focal length f is equal to p/4. In another word the focal length has negative value, which is -40cm. Negative focal length means the mirror is a convex mirror Spherical Refracting Surfaces Let s look at the images formed by reflections to images formed by refraction through surfaces of transparent materials. Spherical surface with radius of curvature r and center of curvature C shall be considered in this study. The light emitted through a point object O in a medium with refractive index n 1 will refract through a spherical surface into a medium of refractive index n 2. After refracting through the surface, the question is whether a real image or virtual image would be formed. The answer depends on the relative value of n 1, n 2 and the geometry of the situation. Let s consider six situations as illustrated in Fig Figure 8.19(a) and Fig. 8.19(b) show that real images are formed. These situations would happen if the refracted light diverges toward the central axis. Situations shown in Fig. 8.19(c), Fig. 81.9(d), Fig. 8.19(e), and Fig. 8.19(f) would form virtual images since the refracted light diverges away from central axis

26 Based on above scenario, real images form on the side of refracting surface that is opposite the object and virtual images form on the same side as object. (a) (b) (c) (d) (e) (f) Figure 8.19: Images formed by spherical refracting surfaces The equation governing the spherical refracting surface is shown in equation (8.38). n1 n n2 n1 p i r 2 (8.38) Like in the case of mirror, p is positive for object. i is positive for real image, whereas i is negative for virtual image. r is the radius of curvature. However, rule must be applied for radius of curvature. For object facing convex surface the radius of curvature is positive. For object facing concave surface, the radius of curvature is negative

27 Example 8.5 A Jurassic mosquito is discovered embedded in a chunk of amber which has refractive index of 1.6. One surface of the amber is spherical convex with radius of curvature 3.0mm. The mosquito head happened to be on the central of axis of the surface viewing along axis appears to be buried 5.0mm into amber. What is the actual depth of the mosquito head? Solution The virtual image is formed from this scenario. The illustration is shown in figure below. Based on equation n1 n n2 n1 p i r 2, we need to calculate the value of p. The image is virtual image then i = -5.0mm. The object is facing a concave surface. Thus, radius of curvature is 3.0mm. Thus, depth of the mosquito head p is 4.0mm Thin Lenses p The actual A lens is a transparent object with two refracting surfaces whose central axes coincide. A lens that causes parallel light to converge to a common point is a converging lens or convex lens. A lens that causes parallel light to diverge is a diverging lens or concave lens. We shall only discuss thin lens, which is the lens that has its thickest part thin as compared with the object distance p, image distance i, and radii curvature r 1 and r 2 of the two surfaces of the lens. We also consider only light rays that make small angle with central axis.

28 Equation (8.37) used for spherical mirror is also true for thin lens calculation. However, the equation for calculating the focal length f of thin lens of refractive index n surrounded by air is given by 1 f 1 1 n 1 r1 r2 (8.39) Equation (8.39) is also called the lens maker s equation. Here r 1 is the radius of curvature of the lens surface nearer to the object and r 2 is that of other surface. The sign of radius follows the rule mention earlier Section If the surrounding medium is not air then the refractive index cannot be 1. Equation (8.39) needs to be modified to include the refractive index n medium of the medium. Thus, equation (8.39) will become 1 f n n medium r1 r2 (8.40) Light rays parallel to central axis passing through convex lens would converge to a real focal point F 2 as shown in Fig. 8.20(a). Light rays parallel to central axis of concave lens would diverge away. The extension of diverged light rays pass through a virtual focal point F 2 as shown in Fig. 8.20(b). (a)

29 (b) Figure 8.20: (a) Convergence of light by convex lens and (b) divergence of light by concave lens Image formed by thin lens can be real or virtual types. Real image formed at the opposite side of the object, while virtual image formed at the same side where the object is situated. Figure 8.21 illustrates the image formed by convex and concave lenses. (a) (b) (c) Figure 8.21: Image formed by convex lens and concave lens In Fig 8.21(a), the object is placed at the point beyond the focal point F 1 of the convex lens. The image formed at the opposite side of the lens is real, enlarged, and inverted. Figure 8.21(b) shows that an enlarged virtual upright image is formed when the objective is placed at the point less than the focal length F 1 of convex lens. Figure 8.21(c) shows that an upright diminished virtual image is formed at the same side as the object Two Lens Systems When an object O is placed in front of a system of two lenses whose central axes coincide, one can locate the final image by working in step. Let lens 1 be nearer to the object and lens 2 located further away from object

30 Step 1: Let p 1 be the distance of object from lens 1. One can then final the i 1 value from equation 1 f 1 1 p i or drawing the ray diagram. Step 2: Ignore the presence of lens 1 and treat the image formed by lens 1 as the object for lens 2. If this object is located beyond lens 2, the object distance p 2 is taken as negative. Otherwise, p 2 is taken as positive. To find the location of the final image, one can use equation drawing the ray diagram f 1 1 p i or Step 1 and step 2 procedure can be used for multiple lens systems. The overall lateral magnification M is the product of m 1 and m 2 produced by the two lenses. i.e. M = m 1 m Optical Instruments Human eye is a remarked effective organ. Its range can be extended in many ways by optical instruments such as using a microscope to view micro-object, seeing distance star using telescope etc. Satellite-borne infrared camera and x- ray microscope are another two examples that extended human vision. Mirror and thin lens equations derived earlier will still be used in our studies of optical instruments, although we know well that some of these instruments do use thick lens. It is for simplicity of the study that we adopt this approach. We will study three optical instruments here. They are simple magnifying lens, compound microscope, and refracting telescope Simple Magnifying Lens Human eye can focus sharp image of an object on the retina if the objective is located from infinity to a point called near point P n. If one moves the object to the eye closer than the near point, the retina perceived a fussy image. The location of near point P n normally depends on the age of the person. An old age person would have a near point further away from the eye than a young person. For the purpose of study, we take near point of 25cm for human eye. An object O of height h is placed at the near point P n of human eye will occupy an angle in the eye s view as shown in Fig

31 Figure 8.22: Object placed on near point of human eye For small angle, tan, thus is approximately equal to h 25cm (8.41) As illustrated in Fig. 8.23, the object is placed at location P n is less than 25cm. The image on the retina would be fussy because the human eye cannot bring the image to be focus on retina. Figure 8.23: Object placed closer to the human eye, a distance shorted than near point To correct the fussy image problem, convex lens is placed in front of eye and placed the object just inside the focal length of the lens. The image produced would be an enlarged virtual type and occupied a large angle situated at infinity than what is done by the object in the case shown in Fig The illustration is shown in Fig The object distance from the central point is equal to focal length F 1 of the convex lens

32 Figure 8.24: Correction for fussy image formed by object placed shorter than Pn closer to eye and closed to the focal point From Fig. 8.24, one can obtain equation (8.42), which is h f ' (8.42) The angular magnification m of what is seen is ' m (8.43) Substituting equation (8.41) and (8.42) into equation (8.43), it yields equation (8.44). 25cm m f (8.44) For object placed inside the focal point as illustrated in Fig then the objective distance p is less than the focal length and the image distance i is equal to 25cm. Thus, according to lens equation, the objective distance p from central axis is equal to 25 f f 25cm p (8.45) The lateral magnification is equal to

33 i 25 f 25 25/ 1 p f 25cm f m cm (8.46) Figure 8.25: Correction for fussy image formed by object placed shorter than P n closer to eye and is inside the focal point Compound Microscope A compound microscope is shown in Fig It consists of an objective of focal length f ob and an eyepiece of focal length f ey. The tube length s can be approximated as the distance i between the objective and the image I. Figure 8.26: The structure of a compound microscope The lateral magnification m is defined earlier as and is i p s m (8.47) f ob

34 ' Since the image I is located just inside focal point F1 of the eyepiece, the eyepiece acts as a simple magnifying lens and the observer see a final virtual and inverted image I through it. The overall magnification M of the instrument is the product of lateral magnification of the instrument m produced by the objective given by equation (8.44) and the angular magnification m produced by the eyepiece shown in equation (8.47). Thus, s 25cm M mm (8.48) f f ob ey Refracting Telescope Telescope comes in variety of forms. The one described here is the refracting type that consists of an objective and an eyepiece, which is shown in Fig Unlike the microscope, telescope is designed to view large object such as galaxy, star, planet etc at large distance. This makes the difference between a microscope and a telescope. The second focal point F 2 of the objective is made ' to coincide with the focal point F 1 of eyepiece, whereas in microscope design, the points are separated by a distance s. Figure 8.27: Structure of a refracting telescope The rays from distant object are parallel rays strike the objective at an angle ob with the axis of telescope and forming a real and inverted image at the common ' focal point F 2 and F 1. This image acts as object for eyepiece, though which an observer sees a distant and inverted virtual image I. The rays defining the image make an angle ey with telescope axis

35 The angular magnification m of the telescope is ey / ob. From Fig. 8.28, the rays close to the central axis such that ob = h /f ob and ey h /f ey. Thus, the angular magnification m is given by f ob f ey m (8.49) where negative sign indicates that I is inverted. Figure 8.28: The height of image and angle made by the parallel rays and observer Tutorials 8.1. Two identical sinusoidal waves moving in same direction along a stretched string interfere with each other. The amplitude y m of each string is 9.8mm and the phase difference between them is What is the amplitude of the resultant wave and what is the type of interference occurs? 8.2. A stretch string has linear density = 525g/m and is under tension = 450N. A sinusoidal wave of frequency f = 120Hz and amplitude y m = 8.5mm is sent at one end of the string. What is the average wave transport energy? 8.3. Two sinusoidal waves y 1 (x, t) and y 2 (x, t) have same wavelength and travel together in same direction. Their amplitudes are y m1 = 4.0mm and y m2 = 3.0mm, and their phase constants are zero and /3 rad. What are the amplitude and phase constant of the resultant wave? Write the wave equation of the resultant wave

36 8.4. The maximum pressure amplitude p m that a human ear can tolerate in loud sound is about 28Pa. What is the displacement amplitude s m for such a sound in air density = 1.21kg/m 3, at frequency of 1,000Hz and speed of 343ms -1? 8.5. In the figure below, it shows two sound source S 1 and S 2, which are in phase and separated by distance D = 1.5 emit identical sound waves of wavelength. (a). What is the path length difference of the waves from S 1 and S 2 at point P 1, which lies on the perpendicular bisector of distance D at a distance greater than D from the sources? Name the type of interference. (b). What is the path length difference and the type of interference at point P 2? 8.6. Human brain used to determine the direction of a source of sound is the time delay t between the arrival of the sound at the right ear closer to the source and arrival at the further left ear. Assume that the source is distinct so that a wave front from it is approximately planar when reaching you, and let D represent the separation between your ears. Given that the speed of sound in air and water are respectively equal to 343m/s and 1,472m/s respectively

37 (a). Find the time delay in terms of D. the speed of sound, and angle between the direction of the source and the forward direction. (b). Suppose that you are submerged in water at 20 0 C when the wavefront arrives directly to your right ear. Based on time delay, at what angle from the forward direction does the source seem to be? 8.7. An electric spark jumps along a straight line of l = 10m, emitting a pulse of sound that travels radially outward. The power of the emission is P s = 1.6x10 4 W. Note that the spark is said to be a line source of sound. (a). What is the intensity of the sound when it reaches a distance r = 12m? (b). What is power P d sound energy intercepted by an acoustic detector of area A d = 2.0cm 2, aimed at the spark and located at distance r = 12m from the spark? 8.8. The sound level 46m in front of the speaker was β 2 = 120dB. What is the ratio of the intensity I 2 of the band at that spot to the intensity I 1 of a jackhammer operating at sound level β 1 = 92dB? 8.9. A trooper is chasing a speeder along a straight stretch of road at 160km/h. Both have same speed. Thus, the trooper sounds a siren of frequency 500Hz. What is the frequency heard by the speeder? You may take the speed of sound to be 343m/s The 16,000Hz whine of turbine engine in the jet plane moving with speed 200m/s. What is the frequency heard by the pilot of a second plane trying to overtake it at a speed of 250m/s?

38 8.11. A bullet is fired with a speed of 685m/s. Find the angle made by the shock cone with the line of motion of the bullet. You may use 343m/s for the speed of sound A jet plane passes over you at height of 5,000m and a speed of Mach 1.5. You may use 343m/s for the speed of sound. (a). Find the Mach cone angle. (b). How long after the jet passes directly overhead does the shock wave reach you. You may use 331m/s for the speed of sound A praying mantis preys along the central axis of a thin symmetric lens 20cm from the lens. The lateral magnification of the mantis provided by the lens is m = and the refractive index of the lens material is Determine (a). The type of image produced by the lens. (b). The type of lens (c). Whether the mantis is inside or outside the focal point. (d). Which side of the lens the imaged appears. (e). Whether the image is upright or inverted. (f). What is the radii of the curvature of the lens? The jalapeno seed O is placed in front of two thin symmetrical coaxial lens 1 and lens 2, in which the focal lengths are f 1 = +24cm and f 2 = +9.0cm respectively with the lens separation of L = 10cm. The seed is 6.0cm from lens 1, where is the final image located? And state its type A photographer has 8X magnifier for examining the negative. What is the focal length of the magnifier?