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https://docs.google.com/document/d/1hw1wxVePrMpUnXETMXWhWo62PGcaHOuA/edit?usp=share_link&ouid=109474854956598892099&rtpof=true&sd=true Unit 20 - Wave Optics WAVE NATURE OF LIGHT Huygens’ Wave Theory (i) Each point on a wavefront acts as a source of new disturbance and emits its own set of spherical waves called secondary wavelets. The secondary wavelets travel in all directions with the velocity of light so long as they move in the same medium. (ii) The envelope or the locus of these wavelets in the forward direction gives the position of new wavefront at any subsequent time. A surface on which the wave disturbance is in the same phase at all points is called a wavefront. Wave optics involves effects that depend on the wave nature of light. In fact, it is the results of interference and diffraction that prove that light behaves as a wave rather than a stream of particles (as Newton believed). Like other waves, light waves are also associated with a disturbance, which one consists of oscillating electric and magnetic field. The electric field associated with a plane wave propagating along the x-direction can be expressed in the form: = o[sin(ωt - kx + φo)] where ω, k and φo bearing their usual meanings. Points to remember regarding Interference When two waves with amplitude A1 and A2 superimpose at a point, the amplitude of resultant wave is given by A = Where φ is the phase difference between the two waves at that point. Intensity (I) = . C = speed of light, E0 = electric field amplitude Intensity (I) = I1 + I2 + 2cosφ. Hence for I to be constant, φ must be constant. When φ changes randomly with time, the intensity = I1 + I2. When φ does not change with time, we get an intensity pattern and the sources are said to be coherent. Coherent sources have a constant phase relationship i.e. one that does not change with time. The intensity at a point becomes a maximum when φ = 2nπ (n = 0, 1, 2, . .) and there is constructive interference. If φ = (2n − 1)π, there is destructive interference. (Here n is a non-negative integer) Determination of Phase Difference The phase difference between two waves at a point will depend upon (a) the difference in path lengths of the two waves from their respective sources. (b) the refractive index of the medium (c) initial phase difference, between the source, if any. (d) Reflections, if any, in the path followed by waves. In case of light waves, the phase difference on account of path difference = = where λ is the wavelength in free space. In case of reflection, the reflected disturbance differs in phase by π with respect to the incident one if the wave is incident on a denser medium from a rarer medium. No such change of phase occurs when the wave is reflected in going from a denser medium to a rarer medium. YOUNG’S DOUBLE SLIT EXPERIMENT A train of plane light waves is incident on a barrier containing two narrow slits separated by a distance 'd'. The widths of the slits are small compared with wavelength of the light used, so that interference occurs in the region where the light from S1 overlaps that from S2. A series of alternately bright and dark bands can be observed on a screen placed in this region of overlap. The variation in light intensity along the screen near the centre O shown in the figure Now consider a point P on the screen. The phase difference between the waves at P is φ, where φ= (where ΔPo is optical path difference, ΔPo=μΔPg; ΔPg being the geometrical path difference.) (here μ = 1 in air) As As, D >> d, S2P - S1P ≈ d sinθ sin θ ≈ tanθ( = y/D). [for very small θ] Thus, φ = For constructive interference, φ = 2nπ (n = 0, 1, 2...) ⇒ ⇒ y = n Similarly for destructive interference, y = (2n − 1) (n = 1, 2, .........) Fringe Width W It is the separation of two consecutive maxima or two consecutive minima. Near the centre O [where θ is very small], W = yn+1 - yn [yn gives the position of nth maxima on screen] = Intensity Variation on Screen. If A and Io represent amplitude of each wave and the associated intensity on screen, then, the resultant intensity at a point on the screen corresponding to the angular position θ as in above figure, is given by I = Io + Io + 2 cosφ, when φ = = 4Io cos2 Illustration 1: A beam of light consisting of two wavelengths 6500 oA and 5200 oA is used to obtain interference fringes in YDE. The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm. (a) Find the distance of the third bright fringe on the screen from the central maxima for the wavelength 6500 oA. (b) What is the least distance from the central maxima where the bright fringes due to both the wavelengths coincide? Solution: Let nth maxima of light with wavelength 6500 Å coincides with that of mth maxima of 5200Å. = Illustration 2: The intensity of the light coming from one of the slits in a Young's double slit experiment is double the intensity from the other slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed. Solution : Displacement of Fringes When a film of thickness 't' and refractive index 'μ' is introduced in the path of one of the sources, then fringe shift occurs as the optical path difference changes. Optical path difference at P = S2P − [S1P+ μt − t] = S2P − S1P − (μ − 1)t = y.d/D − (μ − 1) Illustration 3: Monochromatic light of wavelength of 600 nm is used in a YDSE. One of the slits is covered by a transparent sheet of thickness 1.8 x 10-5 m made of a material of refractive index 1.6. How many fringes will shift due to the introduction of the sheet? Solution: As derived earlier, the total fringe shift = . As each fringe width = w, Illustration 4: In the YDSE conducted with white light (4000Å-7000Å), consider two points P1 and P2 on the screen at y1=0.2mm and y2=1.6mm, respectively. Determine the wavelengths which form maxima at these points Solution: The optical path difference at P1 is p1 = In the visible range 4000 - 7000Å n1 = The only integer between 0.714 and 1.25 is 1 ∴ The wavelength which forms maxima at P is λ = 5000Å For the point P2, p2 = Here n1 = The integers between 5.71 and 10 are 6, 7, 8, 9 and 10 ∴ The wavelengths which form maxima at P2 are λ1 = 4000Å for n = 10 λ2 = 4444Å for n = 9 λ3 = 5000Å for n = 8 λ4 = 5714Å for n = 7 λ5 = 6666Å for n = 6 Illustration 5: A transparent paper (μ = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young’s double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed? Solution: Due to pasting the fringes shift which will restore its position after removal. Path difference will be S2P − S1P − t + μt = t (μ − 1) + ⇒ for a bright fringe Δx = nλ t(μ − 1) + yd/D = nλ. y = (nλ − t(μ − 1)). Again after removing y′ = nλ ⇒ y′ − y = t(μ − 1) No of fringes shifted will be n = = = 14.5 INTERFERENCE BY THIN FILM A ray of light incident on a thin film of thickness ‘t’ gets partially reflected and refracted at A at surface I and thereafter it gets reflected and refracted at B of surface II. The rays after emerging in the first medium interfere. Now the inference will depend upon the path difference between AD and ABC, as beyond CD path difference is zero. Δ x = μ (AB + BC) – AD = μ (AE + EB + BL) – AD = μ AE–AD+μ(EB+BC) (I) In Δ LBF = BC = BF (II) ⇒ Δ x = μ AE – AD + μ EF (III) In Δ ECF EF = CP cos r = 2 t cos r (IV) In Δ ADC and Δ AEC ⇒ ⇒ μ AE = AD (V) Putting (IV) and (V) in (III) we obtain Δ x = 2 μt cosr IAD as In reflected at a denser medium it suffers an additional path difference λ/2 Total path difference the taken place is 2 μ t cosr – For constructive interference 2 μ t cosr – = μ n. ⇒ 2 μ t cosr = nλ + λ/2 = (2n + 1)λ/2 maxima For normal incidence r = 0 ⇒ 2 μ t = (2n + 1) = μ n 2 μ t cosr – = (2n – 1) λ / 2 ⇒ 2 μ t cosr = nλ For normal incidence 2 μ t = n λ Illustration 6: White light is incident normally on a glass plate of thickness 0.50 x 10-6 m and index of refraction 1.50. Which wavelengths in the visible region (400 nm - 700 nm) are strongly reflected by the plate? Solution : The light of wavelength λ is strongly reflected if the light rays reflected are interfering constructively. As we know the condition for constructive interference 2μt = λ. Here 2μt = 2 x 1.5 x (0.5 x 10-6)m = 1.5 x 10-6 m. Putting λ = 400 nm, 1.5 x 10-6 = 400 x 10-9 ⇒ n = 3.25 Similarly, by putting λ = 700 nm. 1.5 x 10-6 = (700 x 10-9) ⇒ n = 1.66 Thus, within 400 nm to 700 nm, for integral values of n = 2 and 3. Now, DIFFRACTION Diffraction is the bending or spreading of waves that encounter an object ( a barrier or an opening) in their path. In Fresnel class of diffraction, the source and/or screen are at a finite distance from the aperture. In Fraunhoffer class of diffraction, the source and screen are at infinite distance from the diffracting aperture. Fraunhoffer is a special case of Fresnel diffraction. Single Slit Fraunhoffer Diffraction In order to find the intensity at point P on the screen as shown in the figure the slit of width 'a' is divided into N parallel strips of width Δx. Each strip then acts as a radiator of Huygen's wavelets and produces a characteristic wave disturbance at P, whose position on the screen for a particular arrangement of apparatus can be described by the angle θ. The amplitudes ΔEo of the wave disturbances at P from the various strips may be taken as equal if θis not too large. The intensity is proportional to the square of the amplitude. If Im represents the intensity at O, its value at P is A minimum occurs when, sin α = 0 and α ≠ 0, so α = nπ, n = 1, 2, 3... Angular width of central maxima of diffraction pattern = 2θ1 = 2 sin1(λ/a) [ θ1 gives the angular position of first minima] The concept of diffraction is also useful in deciding the resolving power of optical instruments. Illustration 7: Light of wavelength 6 × 10-5cm falls on a screen at a distance of 100 cm from a narrow slit. Find the width of the slit if the first minima lies 1mm on either side of the central maximum. Solution: Here n = 1, λ = 6 × 10-5 cm. Distance of screen from slit = 100 cm. Distance of first minimum from central maxima = 0.1 cm. sinθ = θ1 = We know that asinθ = nλ a = = 0.06 cm. Illustration 8: In YDSE if the source consists of two wavelengths λ1 = 4000Å and λ2 = 4002Å . Find the distance from the centre where the fringes disappear, if d=1cm ; D=1 m . Solution: The fringes disappear when the maxima of λ1 fall over the minima of λ2. That is Where p is the optical path difference at that point. or p = Here λ1 = 4000Å, λ2 = 4002Å ∴ p = 0.04 cm In YDSE, p = dy/D ∴ y = Illustration 9: A beam of light consisting of two wavelengths 6500A°and 5200A° is used to obtain interference fringes in a Young’s double slit experiment (i) Find the distance of the third fringe on the screen from the central maximum for the wavelength 6500A°. (ii) What is the least distance from the central maximum where the bright fringes due to both wavelength coincide? (iii) The distance between the slits is 2mm and the distance between the plane of the slits and screen is 120cm. What is the fringe width for λ = 6500A°? Solution: (i) The width of the fringe Then distance of the third fringe 3w = = 0.117 cm (ii) Let mth and nth bright fringe of the wavelength coincide. Now position mth bright fringe is ym = nλ1 and yn = nλ2 ⇒ Now (iii) Fringe width w = = 3900 × 10−7m = 0.039 cm Illustration 10: In YDSE light of two wavelengths of 700 nm and 500 nm. If D/d = 103 find the minimum distance from central maxima where the maxima of two wavelength coincide again. Solution: ⇒ y= 3.5 mm POLARISATION Polarisation of two interfering wave must be same state of polarisation or two source of light should be unpolarised. BREWSTER LAW According to this law when unpolarised light is incident at polarising angle (i) on an interface separating a rarer medium from a denser medium, of refractive index μ as shown in Fig., below such that μ = tan i then light reflected in the rarer medium is completely polarised. Reflected and refractive rays are perpendicular to each other. REDUCTION IN INTENSITY Intensity of polarised light is 50% of that of the unpolarised light, i.e., where = Intensity of polarised light and = Intensity of unpolarised light. ASSIGNMENT 1. In Young's double slit experiment the light emited from source has l = 6.5 × 10–7 m and the distance between the two slits is 1 mm. Distance between the screen and slit is 1 metre. Distance between third dark and fifth birth fringe will be (a) 3.2 mm (b) 1.63 mm (c) 0.585 mm (d) 2.31 mm Solution- (b) x5 - x3 1.63 mm. 2. Light is incident normally on a diffraction grating through which first order diffraction is seen at 32o. The second order diffraction will be seen at (a) 84o (b) 48o (c) 64o (d) None of these Solution- (d) For second order diffraction, = 2 sin 320 > 1 Which is not possible. Hence there is no second order diffraction. 3. In Young's double slit experiment, if the widths of the slit are in the ratio 4:9, ratio of intensity of maxima to intensity of minima will be (a) 25:1 (b) 9:4 (c) 3:2 (d) 81:16 Solution- (a) As ratio of slit widths = Ratio of intensities amax = a1 + a2 = 3 + 2 =5; amin. = 3 - 2 = 1 4. A light source approaches the observer with velocity 0.5 cm. Doppler shift for light of wavelength 550 Å is (a) 616 Å (b) 1833 Å (c) 5500 Å (d) 6160 Å Solution- (b) In case of light, or 5. Angular width of a central max. is 30o when the slit is illuminated by light of wavelength 6000 Å. Then width of the slit will be approx. (a) 12 × 10–6 m (b) 12 × 10–7 m (c) 12 × 10–8 m (d) 12 × 10–9 m Solution- (b) Angular width = 2θ = 300 = 12 x 10–7 m 6. Light of wavelength 6000 12 × 10–6 m is incident on a single slit. First minimum is obtained at a distance of 0.4 cm from the centre. If width of the slit is 0.3 mm, then distance between slit and screen will be (a) 1.0 m (b) 1.5 m (c) 2.0 m (d) 2.3 m Solution- (c) Using a sin θ = nλ or 7. If velocity of a galaxy relative to earth is 1.2 × 106 ms–2 then % increase in wavelength of light from galaxy as compared to the similar source on earth will be (a) 0.3 % (b) 0.4 % (c) 0.5 % (d) 0.6 % Solution- (b) Since or 8. Doppler shift for the light of wavelength 6000 Å emitted from the sun is 0.04 Å. If radius of the sun is 7 × 108 m then time period of rotation of the sun will be (a) 30 days (b) 365 days (c) 24 hours (d) 25 days. Solution- (d) Doppler shift, or Substituting values for all parameter, we get or T = 25 days 9. On introducing a thin nuca sgeet if tgucjbess 2 × 10–6 m and refractive index 1.5 in the path of one of the waves, central bright maxima shifts by n fringes. Wavelength of the wave used is 5000 Å, then n is (a) 1 (b) 2 (c) 5 (d) 10 Solution- (d) shift in no. of fringes is given by nλ = (u-1) t 10. Two beams of light having intensities I and 4I interfere to produce a fringe patternon the screen. Phase difference between the beams is at point a and π at sities at A and B is (a) 3 I (b) 4 I (c) 5 I (d) 6 I Solution- (b) Given I1 = I, I2 = 4I; IA = I1 + I2 + 11. White light is used to illuminate the two slits in Young's double slit experiment, separation between the slits is b and the screen is at a distance d(>>b) from the slits. At a point on the screen, directly in front of the slits, certain wavelengths are missing. Some of these missing wavelengths are (a) (b) (c) (d) Solution- (a,d) Path difference = For missing wavelengths, If n = 1, If n = 2, 3 x 12. A radar operates at wavelength 50.0 cm. If the beat freqency between the transmitted singal and the singal reflected from aircraft (Δ ν) is 1 kHz, then velocity of the aircraft will be (a) 800 km/hr (b) 900 km/hr (c) 1000 km/hr (d) 1032 km/hr Solution- (b) when source is fixed and observer is moving towards it when source is moving towards observer at rest = 900 km/hr 13. A plane electromagnetic wave of frequency wo falls normally on the surface of a mirror approaching with a relativisitic velocity ν. Then frequency of the reflected wave will be (a) (b) (c) (d) Solution- (c) Frequency of Em waves going towards the approaching mirror, w'= Frequency of waves reflected from mirror and movi9ng towards source, or 14. A spectral line of wavelength 0.59 mm is observed in the directions to the opposite edges of the solar disc along its equator. A difference in wavelength equal to (Δλ) 8 picometer is observed. Period of Sun's revolution around its own axis will be about (Radius of sun = 6.95 × 108 m) (a) 30 days (b) 24 hours (c) 25 days (d) 365 days Solution- (c) Change in wavelength for two edges = Total change, and T = 24.8 days 25 days 15. If light with wavelength 0.50 mm falls on a slit of width 10 mm and at an angle θo = 30o to its normal. Then angular position of first minima located on right sides of the central Fraunhoffer's diffraction will be at (a) 33.4o (b) 26.8o (c) 39.8o (d) None of these Solution- (a) For first diffraction minima at angle d(sin - sin 0) = For right of C.M., = 0.5 + = 0.55 i.e., 16. Angular width of central maximum in the Fraunhoffer's diffraction pattern is measured. Slit is illuminated by the light of another wavelength, angular width decreases by 30%. Wavelength of light used is (a) 3500 Å (b) 4200 Å (c) 4700 Å (d) 6000 Å Solution- (b) For first diffraction min. d sin= and if angle is small, sin i.e. Half angular width, Full angular width w = 2 = Also w' = = 6000 x 0.7 = 17. A parallel beam of white light falls on a thin film whose refractive index is 1.33. If angle of incidence is 52o then thickness of the film for the reflected light to be coloured yellow (λ = 6000 Å) most intensively must be (a) 14 (2n + 1) μm (b) 1.4 (2n + 1) μm (c) 0.14 (2n + 1) μm (d) 142 (2n + 1) μm Solution- (c) For constructive interference on reflection =0.14 (22+1) μm 18. A plane monochromatic light falls normally on a diaphragm with two narrow slits separated by a distance d = 2.5 mm. A fringe pattern is formed on the screen placed at D = 100 cm behind the diaphragm. If one of the slits is covered by a glass plate of thickness 10 μ m, then distance by which these fringes will be shifted will be (a) 2 mm (b) 3 mm (c) 4 mm (d) 5 mm Solution- (a) or x = 19. A two slit Young's experiment is done with monochromatic light of wavelength 6000 Å. Slits are 2 mm apart and fringes are observed on a screen placed 10 cm away from the slits. If a transparent plate of thickness 0.5 mm is placed in front of one of the slit, interference pattern shifts by 5 mm. Then refractive index of transparent plate should be (a) 1.1 (b) 1.2 (c) 1.3 (d) 1.5 Solution- (b) or 20. In Young's double slit experiment, using monochromatic light, fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thichness 1.964 mm is introduced in the path of one of the two waves. If now mica sheet is removed and distance between slit and screen is doubled, distance between successive max. or min. remains unchanged. The wavelength of the monochromatic light used in the experiment is (a) 4000 Å (b) 5500 Å (c) 5892 Å (d) 6071 Å Solution- (c) when distance between screen and slit is doubled, them fringe width As or = 0.5892 x 10–6 m = 21. In a two slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If screen is moved by 5 × 10–2 m towards the slits, then change in fringe width is 3 × 10–5 m. If the distance between slits is 10–3 m then wavelength of the light used will be (a) 4000 Å (b) 6000 Å (c) 5890 Å (d) 8000 Å Solution- (b) or 22. White light is used to illuminate two slits in Young's double slit experiment. Separation between the slits is b and the screen is at a distance d (>>b) from the slits. Then wavelengths missing at a point on the screen directly in front of one of the slit are (a) (b) (c) (d) Solution- (a) For missing wavelength 23. Interference fringes from sodium light (1 = 5890 Å) in a double slit experiment have an angular width 0.20o. To increase the fringe width by 10%, wavelength of light used should be (a) 5892 Å (b) 4000 Å (c) 8000 Å (d) 6479 Å Solution- (d) and angular fringe width, 24. In a Young's double slit experiment, angula width of a fringe formed on a distant screen is 0.1o. If wavelength of light used is 6000 Å , then distance between the slits will be (a) 0.241 mm (b) 0.344 mm (c) 0.519 mm (d) 0.413 mm Solution- (b) or 25. In a Young's double slit experiment two narrow slit 0.8 mm apart are illuminated by the same source of yellow light (l = 5893 Å). If distance between slits and screen is 2m then separation between adjacent bright lines will be (a) 14.73 mm (b) 14.73 cm (c) 1.473 mm (d) 147.3 mm Solution- (a) 26. Monochromatic light of wavelength 5000 Å is incident on two slits separated by a distance of 5 × 10–4 m. Interference pattern is seen on the screen placed at a distance of 1 m from the slits. A thin glass plate of thickness 1.5 × 10–6 m and refractive index 1.5 is laced between one of the slits and the screen. If intensity in the absence of plate was Io then new intensity at the centre of the screen will be (a) I0 (b) 2 I0 (c) (d) Solution- (a) or = 1.5 x 10–3 m 27. In Young's double slit experiment, separation between the slits is 2 × 10–3 m and distance of the screen from the slit is 2.5 m. Light in the range of 2000–8000 Å is allowed o fall on the slits. Wavelength in the visible region that will be present on the screen at 10–3 m from the central maxima will be (a) 4000 Å (b) 5000 Å (c) 6000 Å (d) 8000 Å Solution- (a) For bright line 8000 = If i.e., second bright line for visible region is present. 28. A double slit experiment is immersed in a liquid of refractive index 1.33. Separation between the slits is 1.0 mm and he distance between slit and screen is 1.33 m. If slits are illuminated by a parallel beam of light whose wavelength is 6300 Å, then fringe width will be (a) 6.3 mm (b) 63 mm (c) 0.63 mm (d) None of these Solution- (c) 29. In a Young's interference experimental arrangement incident yellow light is composed of two wavelength 5890 Å and 5895 Å. between the slits is 1 mm and the screen is placed 1 m away. Order upto which fringes can be seen on the screen will be (a) 384 (b) 486 (c) 512 (d) 589 Solution- (d) or 30. Ratio of intensities between a point A and that of central fringe is 0.853. Then path difference between two waves at point A will be (a) (b) (c) (d) λ Solution- (c) R2 = a2 + b2 + 2ab cosφ IR = I + I0 + 2I cosφ = 2I (1 + cosφ) =0.853 x 4I For Test 1. When light is incident on a soap film of thickness 5 × 10–5 cm, wavelength reflected maximum in the visible region is 5320 Å. Refractive index of the film will be (a) 1.22 (b) 1.33 (c) 1.51 (d) 1.83. Solution- (b) or 2. A ray of unpolarised light is incident on a glass plate of refractive index 1.54 at polarising angle, then angle of refraction is (a) 33o (b) 44o (c) 57o (d) 90o Solution- (a) i.e., ip = tan–1 1.54 = 570, But r + ip = 900 r = 900 - ip = 900 - 57 = 330. 3. Two waves of same intensity produce interference. If intensity at maximum is 4I, then intensity at the minimum will be (a) 0 (b) 2I (c) 3I (d) 4I. Solution - (a) I1 = I2 = I = a2 Imax. = (a + a)2 + (2a)2 = 4a2 = 4I Imin. = (a - a)2 = 0 4. First diffraction minima due to a single slit of width 1.0 x 10–5 cm is at 300. Then wavelength of light used is (a) (b) (c) (d) Solution - (b) As a sin = or = 0.5 x 10–5 = 5 x 10–6 cm = 5. Light of wavelength is normally incident on a slit. Angular position of second minimum from central maximum us 300. Width of the slit should be (a) 12 x 10–5 cm (b) 18 x 10–5 cm (c) 24 x 10–5 cm (d) 36 x 10–5 cm Solution - (c) = 24 x 10–7 m = 24 x 10–5 cm 6. Light of wavelength is incident normally on a slit of width 0.2 mm. Angular width of the central maximum on the screen will be (a) 0.90 (b) 0.180 (c) 0.540 (d) 0.360 Solution - (d) = 3164 x 10–6 rad = 0.003164 radian = 2 x 0.003164 x 7. Light of wavelength is incident normally on a slit. First minimum of diffraction pattern is formed at a distance of 5 mm from the central maximum. If slit width is 0.2 mm, then distance between slit and screen will be (a) 1 m (b) 1.5 m (c) 2.0 m (d) 2.5 m Solution - (c) or 8. Light of wavelength is incident on a slit. First minima of the diffraction pattern is found to lie at a distance of 6 mm from the central maximum on a screen placed at a distance of 2 m from the slit. If slit width is 0.2 mm, then wavelength of the light used will be (a) (b) (c) (d) Solution - (b) or 9. A slit 5 cm wide when irradiated by waves of wavelength 10 mm results in the angular spread of the central maxima on either side of incident light by about (a) 1/2 radian (b) 1/4 radian (c) 3 radian (d) 1/5 radian. Solution - (d) Angular spread on either side is given by 10. In Young's double slit experiment, ten slits separated by a distance of 1 mm are illuminated by a monochromatic light source of wavelength 5 x 10–7 m. If the distance between slit and screen is 2 metre, then separation of bright lines in the interference pattern will be (a) 0.5 mm (b) 1.0 mm (c) 1.5 mm (d) 1.75 mm Solution - (b) = 1 mm. 11. A star is moving away from earth and shift in spectral line of wavelength 5700 is 1.90. Velocity of the star is (a) 50 km s–1 (b) 70 km s–1 (c) 80 km s–1 (d) 100 km s–1 Solution - (d) or 12. Intensity of central bright fringe due to interference of two identical coherent monochromatic sources is I. If one of the source is switched off, then intensity of central bright fringe becomes (a) (b) (c) (d) I. Solution - (c) I = (a+a)2 = 4a2, I'=(a)2 = a2 13. In two separate set ups of Young's double slit experiment, fringes of equal width are observed when lights of wavelengths in the ratio 1:2 are used. If the ratio of slit separation in two cases is 2:1, then ratio of distances between the plane of slits and the screen in the two set ups is (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1 Solution - (d) 14. In an experiment to demonstrate interference of flight using Young's slits, separation of two slits is doubled. In order to maintain same spacing of fringes, distance D of screen from slits must be changed to (a) D (b) (c) 2D (d) Solution- (c) i.e., D' = 2D. 15. In an interference pattern by two identical slits, intensity of central maxima is I. If one slit is closed, intensity of central maxima changes to I0. Then I and I0 are related by (a) I = I0 (b) I = 2I0 (c) I = 3I0 (d) I = 4I0 Solution - (d) I1 = I2 = a2 Imax. = (a+a)2 = 4a2 = I If one slit is closed, Intensity, I0 = (a)2 16. In a young's double slit experiment, two slits are illuminated by a mixture of two wavelengths and . At 6.0 mm from the common central bright fringe on a screen 2 m away from the slits, a bright fringe of one interference pattern coincides with the bright fringe of other. Distance between the slits should be (a) 1.0 mm (b) 1.5 mm (c) 2.0 mm (d) 2.5 mm Solution - (c) n x 12000 = (n+1) x 10000 or n = 5 and x = = 2 x 10–3 m = 2 mm 17. In a young's double slit experiment, slits are illuminated by a monochromatic source of wavelength and fringes are obtained. If screen is moved by a distance of 5 cm towards slits, change in fringe width is 3 x 10–5 m. Then separation between the slits will be (a) 1 mm (b) 1.2 mm (c) 1.5 mm (d) 1.63 mm Solution - (a) and or 18. An unpolarized beam of light is incident on a group of three polarizing sheets which are arranged in such a way that plane of rotation of one make an angle of 400 with the adjacent one. The 0% of incident light transmitted by first polarizer will be (a) 33% (b) 16.6% (c) 50% (d) 25% Solution - (c) First polariser, polarises the light and hence intensity of light reduces by 50%. 19. 80 gm of impure sugar when dissolved in a litre of water gives an optical rotation of 9.90 when placed in a tube of length 20 cm. If concentration of sugar solution is 75 gm/litre then specific rotation of sugar is (a) 440 (b) 550 (c) 660 (d) 730 Solution - (c) or 20. In Young's double slit experiment, distance between the slits S1 and S2 is d and the distance between slits and screen is D. Then first missing wavelength on the screen in front of S1 is (a) (b) (c) (d) None of these Solution - (a) Path difference, For first missing wavelength, 21. In Young's double slit experiment, 12 fringes are obtained to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by (a) 18 (b) 24 (c) 30 (d) 36. Solution - (a) 22. In Young's double slit experiment, interference pattern is found to have an intensity ratio between bright and dark fringes as 9. Then amplitude ratio will be (a) 1 (b) 2 (c) 9 (d) 3. Solution - (b) 23. Interference fringes from sodium light in a double slit experiment have an angular width 0.200, To increase the fringe width by 10%, new wavelength should be (a) (b) (c) (d) Solution - (d) or 24. If A is the amplitude of the wave coming from a point source at distance r then (A) (B) (C) (D) Sol (B) For a point source, and 25. If A is the amplitude of the wave coming from a line source at a distance r, then (A) (B) (C) (D) Sol (B) For a line source, and 26. Phase difference between two waves having same frequency (v) and same amplitude (A) is 2π/3. If these waves superimpose each other, then resultant amplitude will be (A) 2A (B) 0 (C) A (D) A2. Sol (C) 27. Ratio of amplitudes of the waves coming from two slits having widths in the ratio 4 : 1 will be (A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 4 : 1 Sol (B) Intensity ∝ slit width 28. Two slits S1 and S2 illuminated by a white light source give a white central maxima. A transparent sheet of refractive index 1.25 and thickness t1 is placed in front of S1. Another transparent sheet of refractive index 1.50 and thickness t2 is placed in front of S2. If central maxima is not effected, then ratio of the thickness of the two sheets will be (A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 4 : 1 Sol (B) Since there is no shift in central maxima. Therefore path difference introduced by the two sheets are equal i.e. where and are refraction index i.e. 29. Two coherent waves are represented by y1 = a1 cos ωt and . Their resultant intensity after interference will be (A) (B) (C) (D) Sol (D) Since the two wave differ in phase by . 30. If a thin film of thickness t and refractive index μ is placed in the path of light coming from a source S, then increase in length of optical path is (A) μ t (B) μ / t (C) (μ – 1) t (D) None of these. Sol (C) Increase in optical path