Class XII Physics (Theory)

DATE : 5/3/9 SET-3 Code No. 55//3 egd. Office : Aakash Tower, 8, Pusa oad, New Delhi-5. Ph.: -4763456 Class XII Physics (Theory) Time : 3 Hrs. Max. Marks : 7 (CBSE 9) GENEAL INSTUCTIONS :. All questions are compulsory. There are 7 questions in all.. This question paper has four sections: Section A, Section B, Section C and Section D. 3. Section A contains five questions of one mark each, Section B contains seven questions of two marks each, Section C contains twelve questions of three marks each, and Section D contains three questions of five marks each. 4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions. 5. You may use the following values of physical constants wherever necessary. c = 3 8 m/s h = 6.63 34 Js e =.6 9 C μ o = 4π 7 T m A ε = 8.854 C N m 4 9 N m C 9 Mass of electron (m e ) = 9. 3 kg Mass of neutron =.675 7 kg Mass of proton =.673 7 kg Avogadro s number = 6.3 3 per gram mole Boltzmann constant =.38 8 JK

Section-A. Distinguish between unpolarized and linearly polarized light. [] Sol. Unpolarized light Polarized light The light in which Electric field oscillations are in every direction perpendicular to propagation of light. The light in which Electric field oscillations are confined in a specific direction perpendicular to propagation of light.. Define the term threshold frequency, in the context of photoelectric emission. [] Sol. Threshold frequency The minimum frequency of incident light below which the photoelectric phenomenon does not occur. Define the term Intensity in photon picture of electromagnetic radiation. O Sol. Intensity of adiation Intensity is defined as the radiation energy incident per unit area per unit time. The SI unit of intensity is Wm. 3. How is the drift velocity in a conductor affected with the rise in temperature? [] ee Sol. Drift velocity in a conductor is related as v d =, where is relaxation time. m Due to rise in temperature, relaxation time decreases, hence drift velocity decreases. 4. In sky wave mode of propagation, why is the frequency range of transmitting signals restricted to less than 3 MHz? [] Sol. In sky wave mode of propagation, the ionospheric layer of atmosphere acts as a reflector for range frequencies below 3 MHz. Electromagnetic waves of frequencies higher than 3 MHz penetrate the ionosphere and escape. On what factors does the range of coverage in ground wave propagation depend? Sol. The range of coverage in ground wave propagation depends on following factors. O (i) Transmitted power (ii) Frequency of signal 5. Draw the pattern of electric field lines when a point charge +q is kept near an uncharged conducting plate. [] Sol. q E

Section-B 6. (a) Define the terms, (i) threshold frequency and (ii) stopping potential in photoelectric effect. [] (b) Plot a graph of photocurrent versus anode potential for a radiation of frequency and intensities I and I (I < I ) Sol. (a) In the photoelectric effect the term, (i) Threshold frequency The minimum frequency of incident radiation below which the photoelectric phenomenon does not take place, for a given metallic surface. (ii) Stopping potential The minimum negative potential of anode terminal due to which the highest kinetic energy photoelectron will stop before reaching anode, therefore photoelectric current becomes zero. i.e., K max = ev (b) Photoelectric current [] I (I > I ) I Stopping potential V etarding potential Anode potential 7. Why a signal transmitted from a TV tower cannot be received beyond a certain distance? Write the expression for the optimum separation between the receiving and the transmitting antenna. [] Sol. At frequencies above 4 MHz, communication is essentially limited to the path of line of sight. These direct waves get blocked at some point by the curvature of earth. [] d ange h T h = adius of earth h = Height of antenna ange h T h [] 3

8. State Bohr's quantization condition of angular momentum. Calculate the shortest wavelength of the Brackett series and state to which part of the electromagnetic spectrum does it belong. [] Calculate the orbital period of the electron in the first excited state of hydrogen atom. O Sol. (i) According to Bohr's quantization condition. h The angular momentum of orbiting electron is integral multiple of, where h is the Plank's constant. Thus the angular momentum (L) of the orbiting electron is quantised. L n h = [] n = number of orbit. (ii) In Brackett series for minimum wavelength. n = 4, n = = 4 min 6 = min = ydberg constant.9 m 7 = 9Å Then, min = 4,59Å This is associated with infrared region of electromagnetic spectrum. O In the first excited state n = r T n = v n n n.53 z = s 6 z.9 n 6 n T th.5 s n z 3 6 8.5 s 5 T. s nd [] 4

9. Why is wave theory of electromagnetic radiation not able to explain photoelectric effect? How does photon picture resolve this problem? [] Sol. The wave theory of electromagnetic radiation is not able to explain photoelectric effect due to the following reasons () According to wave theory, electromagnetic wave of any frequency should be able to cause photoelectric effect over a sufficient time. However, it is observed that there is a threshold frequency below which photoelctric effect is not observed. () It is observed that even with low intensity radiation of frequency above threshold frequency, emission of photoelectrons is almost immediate. This is contrary to prediction of wave theory. According to the photon picture if frequency of photon f is greater than h ( = work function) the photoelectric emission will happen almost instantaneously even if intensity is low.. Two bulbs are rated (P, V) and (P, V). If they are connected (i) in series and (ii) in parallel across a supply V, find the power dissipated in the two combinations in terms of P and P. [] Sol. Let (P, V) is rating of bulb B and (P, V) is for bulb B. Then, from relation V P V P V P...(i)...(ii) In series combination, V V eq P P P Series V eq V V P P P Series PP P P P P PP P Series PP P P 5

In Parallel combination eq V V V P P P eq V V P P Parallel P P P Parallel. Obtain the expression for the ratio of the de-broglie wavelengths associated with the electron orbiting in the second and third excited states of hydrogen atom. [] Sol. The de-broglie wavelength associated with the momentum P is given as h...(i) P Since momentum associated with electron revolving in n th orbit is given as P P...(ii) n Now for nd excited state, n = 3 h h Hence (n3) 3 P P 3...(iii) For 3 rd excited state, n = 4 h h Hence (n4) 4 P P 4 From equation (iii) and (iv)...(iv) h 3 (n 3) P 3 h (n4) 4 4 P (n3) (n4) 3 4 6

. A charged particle q is moving in the presence of a magnetic field B which is inclined to an angle 3 with the direction of the motion of the particle. Draw the trajectory followed by the particle in the presence of the field and explain how the particle describes this path. [] Sol. Let magnetic field B is along x-direction as shown in figure. Now charge particle enters in magnetic field at an angle 3. Hence velocity component vcos3 will be along the magnetic field and vsin3 will be perpendicular to the magnetic field. vsin3 q = 3 vcos3 Now, from equation of force B y z x m(v sin3 ) qbv sin3n mvsin3 qb where is radius of circle made by the charge particle called as radius of helix. Now, time taken to complete one cycle is t m (vsin3 ) qb Then distance travelled along the x axis (called as pitch) is given as mvcos3 Pitch vcos3t qb Hence, path traced by the particle will be helical. Section-C 3. (a) Explain briefly how utherford scattering of -particle by a target nucleus can provide information on the size of the nucleus. [3] (b) Show that density of nucleus is independent of its mass number A. Sol. (a) On the basis of utherford scattering experiment following observations were made: () Most of the -particles passed through the foil without any deviation. () About.4% of incident -particles scattered by more than (3) Deflection of more than 9 was observed in about.5% of the incident -particles. [] On the basis of above observation, it was concluded that () Most of the space in an atom is unoccupied. () There must be an extremely small region of concentrated positive charge at the centre of atom. This small region is called nucleus. [] (b) The volume of the nucleus is directly proportional to the number of nucleons (mass number) constituting the nucleus r = r A /3... (i) As nuclear density A r 3 (ii) From equation (i) & (ii) Density of nucleus is independent of its mass number A 7

4. State the underlying principle of a cyclotron. Explain its working with the help of a schematic diagram. Obtain the expression for cyclotron frequency. [3] Sol. Cyclotron : It is a device by which charged particles like protons, deuterons etc. can be accelerated to high speed. Principle : A charged particle can be accelerated by making it to cross the same electric field repeatedly with the help of magnetic field. Diagram : H.F. Oscillator N D [] D Target S Working and theory : Cyclotron consists of two semi-circular disc D and D (called Dee s) enclosed in an evacuated chamber as shown in figure. The magnetic field, acts on the particle and makes it go round in a circular path inside a dee. Every time the particle moves from one dee to the other it is speeded up by the electric field. The sign of the electric field is changed alternately in tune with the circular motion of the particle. At a certain instant, let D be positive and D be negative. The radius of the circular path is given by mv qvb r r mv qb As frequency of revolutions, v = rf v mvf qb qb f m f is independent of both v and r. 5. Two infinitely long straight wires A and A carrying currents I and I flowing in the same directions are kept d distance apart. Where should a third straight wire A 3 carrying current.5i be placed between A and A so that it experiences no net force due to A and A? Does the net force acting on A 3 depend on the current flowing through it? [3] Sol. A A I A 3.5I I x F F d 8

Magnetic field due to wire A at a perpendicular distance x is I B x Force per unit length on wire A 3 due to wire A is F =.5IB.5II F = (i) x Magnetic field due to wire A at a perpendicular distance (d x) is I B = (d x) Force per unit length on wire A 3 due to wire A is F =.5IB =.5I I (dx) As net force on wire A 3 is zero. F F.5I I.5I I x (dx) x d x d x = x x = d 3 (ii) [] Distance is d 3 from wire A. Since net force per unit length on wire A 3 is given as F =I (B B ) net 3 Hence net force acting on wire A 3 depends on the current flowing through it and it will be independent only when net magnetic field is zero. 6. (a) Draw the equipotential surfaces due to an electric dipole. [] Sol. (a) (b) Derive an expression for the electric field due to a dipole of dipole moment p at a point on its perpendicular bisector. [] [] + q q Equipotential surface Equipotential surfaces due to an electric dipole 9

(b) Let distance of point P where field has to be calculate be y from axial line. E +q E E q Y P X y q +q a q E 4 y a q q E 4 y a q EE E q q Due to symmetry electric field in y direction will cancel out. E E cos ( ˆi) q [] qa ( ˆi) E 4 (a y ) for y >> a 3/ qa E ( ˆi) 3 4y p E [] 3 4 y 7. Using Kirchhoff s rules, calculate the current through the 4 and resistors in the following circuit. [3] A 8 V + B D 4 C Sol. 8 V A B E + 4 V F I D I 4 C E (I + I ) 4 V Apply KVL in FEDC (I + I ) + 4 4I = 4 5I = I F []

Apply KVL in CDAB 4 I + 8 I = I + 4 = I Solving we get I =, I = 4 A [] [] So, current through resistor is 4 A and current through 4 resistor is zero O What is end error in a metre bridge? How is it overcome? The resistances in the two arms of the metre bridge are = 5 and S respectively. When the resistance S is shunted with an equal resistance, the new balance length found to be.5l, where l is the initial balancing length. Calculate the value of S..B. S A G P B Q l cm ( l ) cm ( ) E K Sol. End Error : Shifting of zero of scale at different points as well as the stray resistance give rise to the end error in meter bridge wire. It can be overcome by interchanging resistance in resistance box or can be overcome by using end correction in measurement C l = S l.5l = S.5l.5l l =.5 l l 5.5l = 3l.5l = 5 S = = [] 8. (a) Describe briefly the functions of the three segments of n-p-n transistor. [3] (b) Draw the circuit arrangement for studying the output characteristics of n-p-n transistor in CE configuration. Explain how the output characteristics is obtained. Sol. (a) The function of three segments of a transistor are : Emitter : Emits the majority charge carriers. Collector : Collects the majority charge carriers. Base : Base provides the interaction between the collector and emitter. []

(b) I C C B I b E V CC V CE V BB I e Output characteristics : A graph showing the variation of collector current (I C ) with collector emitter voltage V CE at constant base current I b is called the output characteristic of transistor. [] A study of these curve reveals the following (i) Initially the collector current I C increases rapidly with increase in V CE and becomes constant afterwards. (ii) Saturation collector current depends entirely on base current I b. I (ma) C I= b 3 A I= b A I= b A V CE (volts) Output resistance (r o ) : This is defined as the ratio of change in collector emitter voltage to the change in collector current at constant base current. VCE ro I [] c I b O Draw the circuit diagram of a full wave rectifier and explain its working. Also, give the input and output waveforms. Sol. Full wave rectifier Centre-Tap Transformer A Diode Centre Tap D X [] B Diode D L Output Y A full wave rectifier consists of two diodes connected in parallel across the ends of secondary winding of a centre tapped step down transformer. The load resistance L is connected across secondary winding and the diodes between A and B as shown in the circuit. During positive half cycle of input a.c., end A of the secondary winding becomes positive and end B negative. Thus diode D becomes forward biased, whereas diode D reverse biased. So, diode D allows the current to flow through it, while diode D does not, and current in the circuit flows from D and through load L from X to Y.

During negative half cycle of input a.c., end A of the secondary winding becomes negative and end B positive, thus diode D becomes reverse biased, whereas diode D forward biased. So, diode D does not allow the current to flow through it but diode D does, and current in the circuit flows from D and through load L from X to Y. V in O t V out O t Input-Output waveforms Since, in both the half cycles of input a.c., electric current through load L flows in the same direction, so d.c. is obtained across L. Although direction of electric current through L remains same, but its magnitude changes with time, so it is called pulsating d.c. [] 9. Define the term wavefront. Using Huygen s wave theory, verify the law of reflection. [3] Sol. In a region where wave is propagating, geometrical structure formed by joining all the points which are vibrating in same (absolute) phase are said to be wavefront. [] eflection of plane wave : A i F B B E C r C D Consider a plane wave ABC incident at an angle i on a reflecting surface AD. If v represents the speed of the wave in medium t is the time taken to move from ABC to B C and also from B C to point D. So, at the time when disturbance of point C reaches to point D disturbance of point A must have formed a hemispherical shape of radius vt and at point B it must have formed a hemispherical shape of radius vt. Drawing a common tangent from point D to both of the above mentioned spherical envelope we shall have DEF as reflected wavefront. Now in ACD and in DFA C F A i D A r D Side AD = AD and side CD = AF = vt And ACD = DFA = 9 so by HS rule both the triangles are congruent. Hence each corresponding parameters must be same. Hence CAD = i (angle of incidence) Must be same as FDA = r (angle of reflection) i = r [] 3

O Define the term, refractive index of a medium. Verify Snell s law of refraction when a plane wavefront is propagating from a denser to a rarer medium. Sol. The factor by which speed of light gets reduced with respect to the speed of light in vacuum is said to be the refractive index for the medium. c v or c v Here v is the speed of light in medium, c is the speed of light in vacuum and is the defined refractive index for the medium. [] > A > VT II A VT r i B I B > > In the figure above Let B B is the incident wavefront and A A is the refracted wavefront in other medium with V as the wave speed. When the point B traverse a distance B A = V T in medium I, the wavefront originated from point B must have travelled a distance of V T in medium II. Now, consider the incident wavefront B B and refracted wavefront A A BA sin i = AB sin r = AB AB sin i VT V sin r V T V sin i sin r. (a) Identify the part of the electromagnetic spectrum used in (i) radar and (ii) eye surgery. Write their frequency range. [] (b) Prove that the average energy density of the oscillating electric field is equal to that of the oscillating magnetic field. [] Sol. (a) (i) Microwaves are used for adar system. Frequency range of microwaves are from GHz to GHz. (ii) Ultraviolet rays (UV rays) are used in eye surgery. Frequency range of UV rays are from 5 to 7 Hz (b) For a travelling electromagnetic wave equations can be written as E = E sin(kx t) and B = B sin(kx t) Two important results are [] E C and C B 4

Now electric field energy density E E E Esin (kxt) Average of sin (kx t ) will come out to be over a long time. so E 4 E Now magnetic field energy density (Average) B B Bsin (kx t)...() B B B 4...() E As B C E E B 4 C 4 B E...(3) 4 So from equation () & (3) avarage electric field energy density and average magnetic field energy density are equal.. Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position. A giant refracting telescope at an observatory has an objective lens of focal length 5 m and an eyepiece of focal length. cm. If this telescope is used to view the Moon, find the diameter of the image of the Moon formed by the objective lens. The diameter of the Moon is 3.48 6 m, and the radius of lunar orbit is 3.8 8 m. [3] Sol. Tube length [] Objective lens Image D Diagram of Astronomical Telescope Eye piece Now, f = focal length of objective lens = 5 m f e = focal length of eyepiece lens = cm Now, the image formed by objective lens is in its focal plane. Moon Objective lens f Image 5

d r moon moon d r image focal plane [] 3.48 3.8 8 6 d 5 image d image = 3.74 m d image = 3.74 cm []. (a) If A and B represent the maximum and minimum amplitudes of an amplitude modulated wave, write the expression for the modulation index in terms of A and B. Sol. (a) Let the amplitude modulated signal be C m (t) = (A c + A m sin m t) sin c t A = A c + A m...(i) B = A c A m...(ii) From equations (I) and (II) A B Ac AB Am The modulation index is given as A AB A A B m c (b) A message signal of frequency khz and peak voltage V is used to modulate a carrier of frequency MHz and peak voltage of 5 V. Calculate the modulation index. Why the modulation index is generally kept less than one? Sol. (b) Modulation index, Here, A m = V A A m c A c = 5 V [] 5 3 Modulation index, is generally kept less than one to avoid distortion. When is greater than one, over modulation occurs and the modulating signal being of greater amplitude, a part of its information is lost in the process of modulation. 3. (a) State Gauss s law for magnetism. Explain its significance. [] Sol. (a) Gauss's law of magnetism states that net magnetic flux through a closed surface (Gaussian surface) is zero. Mathematically Bds = [] Gauss s Law for magnetism tells us that magnetic monopoles do not exist. 6

(b) Write the four important properties of the magnetic field lines due to a bar magnet. [] Sol. (b) Important properties of magnetic field lines due to a bar magnet: (i) Field lines form close loop (ii) Field lines do not intersect (iii) Degree of closeness of lines represents relative strength of field in regions (iv) Direction of magnetic field at a point is along the tangent to a line at that point O Write three points of differences between para-, dia- and ferro- magnetic materials, giving one example for each. [3] Sol. S. No. Paramagnetic Material Diamagnetic Material Ferromagnetic Material (i) (ii) The materials which get weakly attracted by a magnet are called Paramagnetic materials. The magnetic susceptibility of paramagnetic material is positive and small. Field lines concentration increases inside the material. The materials which get weakly repelled by a magnet are called Diamagnetic materials. The magnetic susceptibility of diamagnetic material is negative and small. The concentration of field lines inside the material decreases. The materials which gets strongly mangetised when placed in external magnetic field (strongly attracted) are called Ferromagnetic materials. The magnetic susceptibility of ferromagnetic material is positive and very large. The concentration of magnetic field lines increases very much inside the material. (iii) Paramagnetic nature decreases with increase in temperature. Ex. Aluminium, calcium, sodium, etc. Diamagnetism is found in every material. At low temperature, they behave as super conductors. Ex : Bismuth, Copper, Silicon, etc. Ferromagnetism also decreases with temperature. At Curie temperature, they become paramagnetic. Ex : Iron, Cobalt, Nickel 4. (a) Three photo diodes D, D and D 3 are made of semiconductors having band gaps of.5 ev, ev and 3 ev respectively. Which of them will not be able to detect light of wavelength 6 nm? Sol. (a) Given that: Band gap of diode D =.5 ev Band gap of diode D = ev Band gap of diode D 3 = 3 ev Wavelength of light incidented = 6 nm Energy associated with a photon of incident light, E hc 4 ev 6 E =. ev To detect the light, the band gap should be less than E Band gap < E Only diode D will be able to detect the light. 7 []

(b) Why photodiodes are required to operate in reverse bias? Explain. Sol. (b) The photodiodes are operated in reverse biased. In this case, the depletion region is large, electric field strength is very high and current through the circuit is very small. All these things, facilitate the easy detection of light. [] p n When light photons get incidented on depletion region, they generate new pairs of holes and electrons. Due to strong electric field, they separate out and constitute rise in electric current in the circuit, which indicates detection. Section-D 5. (a) Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor. [½] Sol. (a) When plates A and B of a parallel plate capacitor are connected to the terminals of a battery, the battery sets up electric field in the connecting wires. The electric field drives charge Q from plate B, connected to the negative terminal of the battery to plate A connected to the positive terminal. The charge transfer takes place till the potential difference across the capacitor plates equals the emf of the battery. V Q C +Q Q A B [] or Q = CV Energy stored in the capacitor V Q q Q U Vdq dq C C q V= C Q U C (b) A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor. [½] 8

Sol. (b) Let charge q be transferred from one capacitor to the other. Then equating potential difference across the two capacitors, CV q q C C +CV CV + S C q CV Now, Energy stored in the combination, q CV Uf C 4 [] Initial energy, Ui CV CV q (CV q) q q C Uf CV U 4 CV i [] O (a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole. Sol. (a) Let P be an equatorial point for a dipole consisting of charges q and +q with a separation a (see figure). Then E Electric field due to the charge +q q E 4 a r E Electric field due to the charge q E q 4 a r 9

By superposition principle E E E net By symmetry the electric field in y-direction will cancel out. E q a E cos 4 (a r ) a r net In vector form, qa 4 (a r ) E 3/ p 4 (a r ) net 3/ [] For a short dipole, r >> a then p Enet 3 4 r (b) Two identical point charges, q each, are kept m apart in air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q. Sol. (b) To overcome the electrostatic repulsion between the point charges q each, Q must be at the centre of the line segment joining the two. The sign of Q must be opposite to that of q. [] kq kqq Also Q q m m q Q q 4 Q must be at a distance of m from each charge q. 6. (a) In a series LC circuit connected across an ac source of variable frequency, obtain the expression for its impedance and draw a plot showing its variation with frequency of the ac source. [5] Sol. (a) i L C (i) (ii) as source of variable frequency In series current will be same across all components. Voltage drop across resistance V = i Voltage drop across inductor V L = ix L Voltage drop across capacitor V C = ix C

(iii) Voltage across resistance is in phase with current Voltage across inductor leads current by / Voltage across capacitor lags current by / So phasor diagram V i V L = V V L VC V C V L C V V (V V ) L C iz i (ix ix ) Z (X L X C) Z L C Variation of impedance (Z) with frequency of source Z f f (frequency) f esonance frequency at this frequency impedance will be minimum and equal to the resistance in the circuit. (b) What is the phase difference between the voltages across inductor and the capacitor at resonance in the LC circuit? Sol. (b) Phase difference between voltage across inductor and capacitor is in series LC circuit. [] (c) When an inductor is connected to a V dc voltage, a current of A flows through it. When the same inductor is connected to a V, 5 Hz ac source, only.5 A current flows. Explain, why? Also, calculate the self inductance of the inductor. Sol. (c) A real inductor has some resistance i.e., L eal inductor Impedance of real inductor V V i Z = L Z X, for dc source X L =

for ac source V i Z.5 Z Z = 4 so for ac source impedance is more hence current is less as Z X L 4 X L XL 3 X L = L L 3 fl 3 3 L 5 3 L henry O (a) Draw the diagram of a device which is used to decrease high ac voltage into a low ac voltage and state its working principle. Write four sources of energy loss in this device. [5] Sol. (a) The device is stepdown transformer. It works on the principle of mutual Induction. [] ~ Primary winding Secondary winding L O A D Laminated core [] Losses: (i) Flux linkage loss (ii) Copper loss in primary and secondary windings (iii) Eddy current loss in the core (iv) Hysteresis loss in the core [] (b) A small town with a demand of kw of electric power at V is situated km away from an electric plant generating power at 44 V. The resistance of the two wire line carrying power is.5 per km. The town gets the power from the line through a 4- V step-down transformer at a sub-station in the town. Estimate the line power loss in the form of heat.

Sol. (b) Total electric power required, P = kw = 3 W Supply voltage V = V Voltage at which electric plant is generating power V = 44 V Distance between the town and power generation station, d = km esistance of the two wire lines carrying power =.5 /km Total esistance of the wires = ( + ).5 = [] A step-down transformer of rating 4 V is used in the sub-station. Input voltage V = 4 V Output voltage V = V rms current in the wire lines is given as P I 3 A V 4 3 Line power loss i = 3 = 8 kw Line power loss in the form of heat = 8 kw [] 7. (a) Describe any two characteristic features which distinguish between interference and diffraction phenomena. Derive the expression for the intensity at a point of the interference pattern in Young s double slit experiment. [5] Sol. (a) Difference between Interference and diffraction: Interference () Interference is caused by superposition of finite number of waves. () All fringes are of equal width. Diffraction () D i f fra c ti o n i s ca u se d b y superposition of infinite number of waves. () All fringes are not of equal width. [] Expression for the intensity : Let displacement produced by S and S are y = acos t and y = acos (t + ) then amplitude of resultant displacement is y = y + y = a[cos t + cos (t + )] a cos cos t A acos Thus intensity at any point is I = 4I cos (since I A ) [] (b) In the diffraction due to a single slit experiment, the aperture of the slit is 3 mm. If monochromatic light of wavelength 6 nm is incident normally on the slit, calculate the separation between the first order minima and the 3 rd order maxima on one side of the screen. The distance between the slit and the screen is.5 m. 3

Sol. (b) Given wavelength = 6 nm = 6. 7 m Slit width a = 3 mm = 3 3 m Distance of screen from slit is D =.5 m Fringe width = = D a Now separation between st minimum and 3 rd order maxima is 7 D D 5 D y a a a 7 5 6..5 4 y 7.75 m 3 3 [] 4 y 7.75 m.775 mm O [] (a) Under what conditions is the phenomenon of total internal reflection of light observed? Obtain the relation between the critical angle of incidence and the refractive index of the medium. [5] Sol. (a) When light travel from denser medium to rarer medium it moves away from the normal. If we keep on increasing the angle of incidence then at a particular angle of incidence angle of refraction become 9, this angle of incidence is known as critical angle. Beyond this angle of incidence light ray does not move to other medium it gets reflected to the same medium. This phenomenon known as Total Internal eflection (TI) [] Snell's Law μ sin i = μ sin r μ sinθ c = μ sin9 sin c μ efractive index of denser medium μ efractive index of rarer medium [] (b) Three lenses of focal lengths + cm, cm and +3 cm are arranged coaxially as in the figure given below. Find the position of the final image formed by the combination. O + cm cm +3 cm 3 cm 5 cm cm 4

Sol. (b) O + cm cm +3 cm 3 cm 5 cm cm From lens formula v u f v 3 v 3 v 5 cm It will act as an object for middle lens u = + cm v =? f = v u f v v = ays will become parallel to principal axis So for third lens object is at infinite so final image v u f v = 3 cm ight of the last lens v 3 5