Wave Optics Level 1 1. Huygen’s conception of secondary wavelets A. allows us to find the focal length of a lens B.is used to determine the velocity of sound C. is used to explain polarisation D. is a geometrical method to find a wavefront 2. 3. 4. The wavefront due to a lined source is A. plane A. plane B. spherical B. spherical C. cylindrical C. cylindrical D. of complicated shape D. of complicated shape The wavefront due to a source situated at infinity is Choose the correct statement(s) from the following A. a wave travelling in a rarer medium, on refraction into a denser medium, undergoes no phase change B. a wave travelling in a rarer medium, on reflection from the boundary of a denser medium, undergoes a phase change of π rad C. a wave travelling in a denser medium, on reflection from the boundary of a rarer medium, undergoes a phase change of π rad D. a wave travelling in a denser medium, on refraction into a denser medium, undergoes a phase change of π rad 5. Light wave propagates a distance of 10 cm in glass of refractive index 1.5 in time t. In the same time light waves propagates a distance 12 cm in a medium. The refractive index of the medium is A. 1.25 6. B. 1.33 C. 1.50 D. 1.60 Two coherent monochromatic beams of light of intensities I and 9I are superimposed. The ratio of maximum and minimum possible intensities are A. 1.25 7. B. 2.5 C. 3 D. 4 In Q. 6, if the beams are non coherent, then the ratio of maximum to minimum possible intensities will be A. 1.25 8. B. 2.5 C. 3 D. 4 Young’s double slit experiment establishes that A. light consists of particles D. light is neither particles nor waves 9. The phenomenon of interference is shown by A. longitudinal mechanical waves only C. non mechanical transverse waves only 10. 11. A. scattering B. interference B. transverse mechanical waves only D. all the above type of waves C. diffraction D. dispersion B. light consists of waves C. light sometimes behaves as particle and sometimes as waves When viewed in white light, soap bubbles show beautiful colouration because of Interference pattern is not observed in excessively thick film because A. most of the incident light is absorbed y the film B. there is too much overlapping of colours washing out the interference pattern C. the maximas of interference pattern are far from minimas D. a thick film has a high coefficient of reflection 12. A single slit diffraction pattern is obtained using a violet colour. What happens if the violet light is replaced by the red light? A. diffraction fringes becomes narrower and crowded together B. diffraction fringes becomes broader and farther apart C. there is no change in the diffraction pattern D. the diffraction pattern disappears 13. In Young’s double slit experiment, if the separation between the two coherent silts is reduced to half and the distance between the silts and the screen is doubled, the fringe width A. becomes four times C. is doubled 14. B. is halved D. remains unchanged In a Young’s double slit experiment, the fringe width with a light of 4800 Å is found to be 0.03 mm. What will be the fringe width if the light of wavelength 6400 Å is used? A. 0.01 mm B. 0.02 mm C. 0.03 mm D. 0.04 mm In Q. 14, what will be the fringe width using light of wavelength 4800 Å if the entire apparatus is immersed in water of refractive index 4/3? A. 0.01 mm B. 0.02 mm C. 0.03 mm D. none of these In Young’s experiment, the intensity of maxima is I. If the width of each slit is doubled, the intensity of maxima will be A. I B. 4I C. 2I D. I/2 In an interference pattern, at a point we observe the 12th order maximum for λ 1 = 6000 Å. What order will be visible here, if the source is replaced by light of wavelength λ 2 = 4800 Å? A. 10 B. 12 C. 13 D. 15 The ratio of intensity at the centre of a bright fringe in an interference pattern to the intensity at a point one quarter of the distance between two fringes from the centre is A. 1 B. 2 C. 3 D. 4 In a two slit interference pattern, the intensity of light at a point on the screen when the path difference is λ is I’. What will be the intensity of light at a point where the path difference is λ /3, λ being the wavelength of light used? A. I’/2 B. I’/3 C. 3I’ D. I’/4 White light is used to illuminate the two slit in Young’s double slit experiment. The separation between the slits is d and the distance between the screen and the slit i D(D < < D). At a point on the screen directly infront of one of the slits, certain wavelengths are missing. The missing wavelength are A. ?? B. ?? C. ?? D. ?? A light beam of wavelength 5500 Å falls normally on a slit of width 2.2 10 -3 mm. The angular position of the first minima on either side of the central maxima is A. ?? 15. 16. 17. 18. 19. 20. 21. 22. The wavelength effective in visual observation is 5890 Å. What is the smallest angular separation (in radian) between two point stars which a telescope of 0.20 m diameter objective can resolve? A. 0.9 × 106 B. 1.8 × 10-6 C. 2.7 × 10-6 D. 3.6 × 10-6 In a double slit experiment, seventh dark fringe is formed opposite to one of the slits, the wavelength of light is A. ?? C. A parallel beam of light of intensity I is incident on a thin film having reflection coefficient 0.25 for the upper surface and 0.50 for the lower surface. The ratio of intensities of maxima to that of minima in the interference region of reflected beam is A. 95 B. 75 C. 65 D. 45 In the double slit arrangement shown in the figure, S1O = S2O A monochromatic light of wavelength λ is incident normally on the slits S1 S2 of equal width. At xP point P on the screen one observes A. first order bright fringe B. first order dark fringe C. second order dark fringe D. third order dark fringe d x S2 D x S1 3λ D/2d O 23. 24. 25. 26. 27. In Q. 25, the ratio of intensity at points O to that of intensity at point P on the screen is A. 4 : 1 B. 2 : 1 C. zero D. infinity In a Young’s double experiment shown in figure, the length of the overlapping region of interference is 2 mm. If the fringe width is 0.035 mm, the number of fringes is A. 57 B. 58 C. 115 D. 116 x S2 Q -5 x S1 P 28. Plane wave (λ = 6 × 10 cm) falls normally on a straight slit of width 0.2 mm. The total angular width of the central diffraction maximum is A. 10-3 radian B. 2 × 10-3 radian 2. 7. 12. 17. 22. 27. C A B D D C 3. 8. 13. 18. 23. 28. C. 4 × 10-3 radian Answers A B A B C D 4. 9. 14. 19. 24. D. 6 × 10-3 radian AB D D D A 5. 10. 15. 20. 25. A B D B B 1. 6. 11. 16. 21. 26. D C B C A D 2mm 1mm Answers 2. 3. Cylindrical wavefronts is produced if the light source is in the form of a lit (or lined source). Plane wavefront is produced if the light source is situated at infinity. Since, the direction of propagation of light wave i.e., the direct of light ray is always normal to the plane of the wavefront. When a wave travelling in a rare medium incidents on the boundary of a denser medium, the reflected wave suffers a phase change of π radian while the refracted ray remains in phase with the incident ray. Such a phase change does not occur when wave travelling in a denser medium reflects from the boundary of rare medium. In a given time, the optical path µ x travelled by the light waves remains the same Thus, µ 1x1 = µ 2x2 6. Given I1 = I and I2 = 9I Resultant intensity after superimposition ( I1 + I2 ) 2 Imax = Imin ( I1 ~ I2 )V I = I1 + I2 + 2 I1 I2 . cos φ 4. 5. or µ2 = µ 1 x1 1.5 × 10 5 = = = 1.25 x2 12 4 = ( I + 3 I )2 ( I1 ~ 3 I ) 2 = 16 I 4I =4 7. For the incoherent beams of intensities I1 and I2, the maximum and minimum intensities after superimposition are given by (I1 + I2) and (I1 – I2), respectively Thus, Imax = I1 + I2 = I + 9I = 10 I and Imin = I ~ 9I = 8I ∴ 9. 11. Im ax 10I = = 1.25 Im in 8I Interference is the result of superimposition of two coherent waves of all types. When white light incidents on a thick film, the coherent beams reflected from the upper and lower surfaces of the film have large path difference. The condition of getting maximas is that the path difference between the two coherent beams reflected from the upper and lower λ surfaces of the film must be (2n − 1) where n = 1, 2, 3, . . . etc. 2 For large path difference, we have 21 λ red 23λ yellow 25λ violet ≈ ≈= 2 2 2 which means that 11th order of red, 12 order of yellow and 13 th order of violet colour appears at the same point and that is why overlapping of colours washes out the interference pattern. 12. λ where e is the slit e width. Therefore, the diffraction fringes become broader and farther apart when λ is increased. Since, λ red >λ violet and the angular half width of diffraction maximas is 13. 14. 15. Fringe width w = w∝λ Dλ . Here, d is halved and D is doubled, therefore w increases to 4-times. d w2 λ 2 = w1 λ1 ⇒ air w2 = w1 λ2 6400 = 0.03 × λ1 4800 = 0.04 mm Wavelength of light in air, λ = 4800 Å wavelength of light in water, λ water = 16. 17. v vwater 4 a µ g = = air v 3 vwater ' Imax = ( 2I1 + 2I2 ) 2 3 v air 3 = λ air 4 v 4 = 2I1 + 2I2 + 4 I1 I2 = 3 × 4800 = 3600Å 4 Imax = ( I1 + I2 )2 = I (initially). If I1 and I2 are doubled, then = 2[I1 + I2 + 2 I1 I2 ] = 2I In double slit interference, the distance y of a bright fringe from the centre (zero-order bright fringe) y = nλ D d 2 where n = 0, 1, 2, 3, . . . . . etc. n λ = constant Thus, at a given point, we have n1λ 1 = n2λ n2 = 18. n1 λ 1 12 × 6000 = = 15 λ2 4800 If ‘a’ be the amplitude of each of the two interfering waves, then the resultant intensity at a point on the screen where they meet in a phase difference φ is given by I = a2 + a2 + 2a . a cos φ I0 = 4a2 = 4I The phase difference between successive fringes is 2π . Thus, the phase difference at a point distance one quarter of the distance between two fringes from the centre will be π /2. The intensity at this point is I1 = 2a2[1 + cos π /2] = 2a2 = 2I = 2a2 [1 + cos φ ] As the centre of bright fringe φ = 0 so that I0 = 4a2 = 4I I0 4I = =2 I1 2I 19. Phase difference = φ= 2π .Δ λ 2π × path difference λ At ∆ = λ , φ = 2π and resultant intensity is I' = I1 + I2 + 2 I1 I2 . cos φ = I + I + 2I cos 2π [Let the two coherent sources are of equal intensity I] I’ = 4I At Δ= . . . . (i) and I" = + + I c s I I 2 o 2π 3 λ 2π , φ= 3 3 2π = 2I 1 + cos 3 1 2I (1 – cos 600) = 2I1 − 2 I ' I" == I 4 20. The nth order dark fringe is at a distance yn from the central zero order bright fringe where 1 D yn = n + λ here n = 0, 1, 2, 3, . . . etc. 2 d or 1 D yn = n − λ here n = 1, 2, 3, . . . etc. 2 d For the point on the screen directly infront of one of the slit, yn = So, 21. d 1 D = n − λ 2 2 d d 2 or λ= d2 D(2n −1) For a parallel beam of light falling on a single slit, the position of minima, on either side of the central maxima are given by mλ sin θ = ± e sin θ = ± m λ or e where m = 1, 2, 3, . . . . etc. For the first order minima, m = 1 =± 5500 × 10 −7 2.2 × 10 −3 sin θ = ± 1 θ = sin −1 ± 4 λ e =± 1 4 22. If d is the diameter of telescope objective, θ= 1.22 λ radian d = 1.22 × 5890 × 10 −10 0.20 = 3.59 × 10-6 radian 23. The distance yn of nth order dark fringe from the central bright fringe is 1 λD yn = n + 2 d or yn d 1 = n + λ where n = 0, 1, 2, 3, . . . . . etc. D 2 Here at yn = ∴ d ,n=6 2 (for seventh dark fringe) or λ= d2 13 D d2 1 13 = 6 + λ = λ 2D 2 2 24. The intensity of light reflected from the upper surface of the film is given by I 1 = 0.25 I and that refracted inside the film is I – 0.25I = 0.75I. Thus the intensity of the film is 0.50 × (0.75I) I2 = 0.375 I 2 2 Now Imax = ( I1 + I2 ) = ( 0.25 I + 0.375 I ) = 0.2 I +0.3 5 I +2 0.2 ×0.3 5 . I 5 7 5 7 = (0.625 + 0.612) I Also, Imin = ( I1 − I2 ) 2 = 1.237 I = (0.625 – 0.612) I = 0.0131 I Imax 1.237I = ≈ 95 Imin 0.013I 25. Let there be a nth order bright (or dark) fringe at P and OP = yn = 3λD 2d yn = nλ D for bright fringe d or 3λ D nλ D = 2d d ⇒ n= 3 2 3λD 1 λD = n + 2d 2 d Since n must be an integer hence this condition is no satisfied. For the dark fringe at P 1 λD yn = n + 2 d or n+ 1 3 = ⇒ n =1 2 2 27. In the overlapping region OP, the number of fringes is OP 2.0 n= = = 57.14 or n = 57 fringes 0.035 0.035 This includes central zero order fringe at O. This total number of fringes in the region PQ. N = 2n + 1 = 2 × 57 + 1 = 115 Level 2 1. A beam of light consisting of two wavelengths 6500 Å and 5200 Å is used to obtain interference fringes in Young’s double slit experiment. The distance of third bright fringe from the central maximum for the wavelength 650 Å is (Distance between the slit = 2 mm and distance between the plane of slits and the screen = 120 cm) A. 1.17 mm 2. B. 1.28 mm C. 1.56 mm D. 1.73 mm In Q. 1, what is the least distance from the central maximum when the bright fringes due to both the wavelengths coincide? A. 1.17 mm 3. B. 1.28 mm C. 1.56 mm D. 1.73 mm A double slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of 1 mm and distance between the plane of slits and screen 1.33 m. The slits are illuminated by a parallel beam of light whose wavelength in air is 6300Å. What is the fringe width? A. (1.33 × 0.63) mm B. ?? 4. C. D. 0.63 mm In Young’s experiment, the fringe width was found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index 4/3, the new fringe width in mm is A. 0.25 5. B. 0.30 C. 0.40 D. 0.53 In Young’s double slit experiment, 12 fringes are observed to be formed in a certain region of the screen when light of wavelength 600 nm is used. If the light of wavelength 400 nm is used, the number of fringes observed in the same region will be A. 12 6. B. 18 C. 24 D. 8 A parallel beam of monochromatic light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit is A. zero 7. B. π /2 C. π D. 2π Yellow light is used in a single slit diffraction experiment with a slit of width 0.6 mm. If yellow light is replaced by x-rays, then the observed pattern would reveal A. that the central maximum is narrower B. more number of fringes C. less number of fringes 8. D. no diffraction pattern To demonstrate the phenomenon of interference we require two sources which emit radiations of A. nearly the same frequency C. different wavelength relationship 9. B. the same frequency D. the same frequency and having a definite phase The angle of incident at which reflected light is totally polarised for reflection from air to glass (refractive index µ ), is A. sin-1 (µ ) B. ?? D. The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young’s double slit experiment is A. infinite B. five C. three D. zero A young’s double slit experiment used a monochromatic source. The shape of the interference fringes formed on a screen is 10. 11. A. hyperbola 12. B. circle C. straight line D. parabola Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm. Approximately what is the maximum distance at which these dots can be resolved by the eye? (Take wavelength of light = 500 nm) A. 5 m B. 1 m C. 6 m D. 3 m When an unpolarised light of intensity I0 is incident on a polarising sheet. The intensity of light which does not get transmitted is A. ?? C. zero D. I0 If I0 is the intensity of principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled? A. 2I0 B. 4I0 C. I0 D. I0/2 2 13. 14. 15. Wavelength of light used in an optical instrument are λ 1 = 4000 Å and λ of their respective resolving power (corresponding to λ 1 and λ 2) A. 16 : 25 B. 9 : 1 C. 4 : 5 D. 5 : 4 = 5000 Å, then ratio 16. In a Young’s double slit experiment the intensity at a point where the path difference is ?? (λ being the wavelength of the light used) is I. If I0 denotes the maximum intensity, then (I/I0) is equal to A. ?? C. 1/2 Answers 2. 7. 12. C A A 3. 8. 13. D D A 4. 9. 14. D. 3/4 B D C 5. 10. 15. B B D 1. 6. 11. 16. A D A D Answers 1. The position of nth bright fringe is given by yn = nλD d = 3 × 6500 × 10 −10 × 120 × 10 −2 2 × 10 −3 = 1.17 × 10-3 m = 1.17 mm 2. The condition for least distance when the bright fringe due to two wavelengths coincide is that the difference of their fringe order must be 1. Therefore nλ1D (n + 1)λ 2D = d d (n + 1) × 5200 yn = This gives Hence, 1.56 mm 5. yn = nλD d or nλ 1 = (n + 1) λ 2 or n × 6500 = n=4 yn = nλ1 D d = 4 × 6500 × 10 −10 × 120 × 10 −2 2 × 10 −3 = 1.56 × 10-3 m = or nλ = yn d = constant D ∴ n2 λ 1 = n1 λ 2 or n2 = n1 λ 1 12 × 600 = = 18 λ2 400 6. In the single slit diffraction pattern, the condition of diffraction minima is, Path difference = e sin θ = ± mλ , where m = 1, 2, 3, . . . . . etc. For the first diffraction minima on either side of central maxima m = 1 Path difference = ± λ Phase difference = 2π × λ Path difference = 2π × λ = 2π λ 7. The angular spread of central maxima in the single slit diffraction pattern is 2θ , where, λ Since, θ is very small e λ λ 2λ 2θ = − − = a a a sin θ = ± sin θ ≈ θ = ± λ e ∴ The wavelength of X-rays is of the order of 1Å i.e., 10 -10 m. Since wavelength of X-rays is very small compared to that of yellow light, therefore 2θ for X-rays is extremely small. Hence, the central maxima becomes narrower. 9. 10. Brewster’s law µ = tan ip d sin θ = n λ For the maximum number of possible interference maximas on the screen i.e., for the maximum value of n, (sin θ )max = 1 and nmax = d λ or ip = tan-1 (µ ) The condition for obtaining interference maximum on the screen is given d = 2λ , so nmax = 2 Hence, there must be five values of n equal to – 2, - 1, 0, 1, 2. 12. The angle subtended at the human eye θ= 1.22 λ y ≥ D d where λ is the wavelength of light with which the dots have been illuminated and d is the diameter of the pupil of eye. Thus, 13. D≤ d. y 1.2 λ 2 ≤ 3 × 10 −3 × 10 −3 1.2200 × 10 −9 m ≤ 30 m 6.1 ≤ 5m Using the law of Malus, the intensity of light transmitted is I = I0 cos2 θ For the unpolarised light the place of vibration is not confined in a particular direction and therefore the angle θ between the planes of transmission of polariser and analyser is not 1 2 fixed. The average value of cos θ = . Hence, the intensity of transmitted polarised light = 2 I0 and the intensity of untransmitted unpolarised light 2 = I0 − 14. I0 I0 = 2 2 sin θ I = I0 θ 2 In a single slit diffraction pattern, the intensity is given by For the central principal maxima θ = 0 15. 16. Re solving Power (RP ) ∝ 1 wavelength sin O I = I0 O = I0 2 RP at λ 2 RP at λ 1 = λ 2 5000 5 = = λ 1 4000 4 I Imax φ = cos 2 2 φ= 2π λ π . = λ 6 3 Phase difference = ∴ I π 3 = co s2 = I0 3 4 2π (path difference) λ