Tiêu đề 29. Wave optics med Ans.pdf ✅

1. (a): The intensity decreases in proportion for the distance squared. 2. (a): Wave front is the locus of all points, where the particles of the medium vibrate with the sake phase 3. (c): Converging spherical 4. (b): Huygens 5. (c) : They will remain spherical, with the same curvature but sing of curvature reversed 6. (a): Light propagates rectilinearly due to wave nature. 7. (d): The Huygen’s construction of wave front does not explain the phenomena of origin of spectra. 8. (a): Here =600nm =600.1nm−600nm=0.1nm As   = c Vs 3 1 1 8 1 50 10 50 3 10 600 0.1 − − − =  = =      = ms kms s nm nm v c s 9. (b): Wavelength of H line =6563Å = m 10 6563 10−  Red shift observed in star is, Å m 10 ( ' ) 15 15 10−  − = =  And speed of light, 8 1 3 10 − c =  ms Let the velocity of the star with which it is receding away from the earth be v  Red shift relation  −=  c v ' ('−)   = c v 5 1 10 8 10 6.86 10 6563 10 3 10 15 10 − − − =      = ms 10. (b): the relation between ,c and  are =c For small changes in  and   −  =   As =589.6−589.0 = +0.6nm There fore , using Doppler shift c v − radial =  −  =   Or 5 1 1 3.06 10 306 589.0 0.6 − − =+  =      v + c ms kms radial Therefore , the galaxy is moving away from us 11. (b): Velocity of star in line of light of 8 4 1 3 10 7.5 10 100 0.025 −  =−   =−   −  v= c ms 12. (a): using =  c v Here 1 3 1 30 30 10 − − v= kms =  ms c 3 10 ms 0.58Å 8 1 =  = − Å v c 0.58 5800 30 10 3 10 . 3 8  =   = = 13. (b): g v v v v c   = = Å v g 4000 1.5 6000 = =    = 14. (d): Resulting amplitude = + + 2 1 2 cos 2 2 2 A A1 A A A Here A1 = A2 =1cm,= 3rad 2 2 ( 1) 0 1 1 2 1 1 cos3 2 2 = +  − = A= + +     15. (d): If source are coherent IR = I 1 + I 2 + 2 I 1 I 2 cos I I I 2I cos0 4I 0 = + +  = If sources are incoherent, 2 2 4 2 0 1 2 I I I I I I R = + = = = 16. (d): Hyperbola 17. (a): Fringe width, d D = Here, D = 1 m and d= m 3 1 10−  m mm nm m 5 10 0.5 1 10 1 5 10 500 5 10 4 3 7 7 =  =    = = =  − − − − 18. (a): Angular width of fringe rad rad 1800 3.14 180 0.1 0.1 =  =  =  d m 4 9 3.44 10 3.14/1800 600 10 − − =   =    =
19. (d): Here , 9 25 min max = I I Or 9 25 2 1 2 1 2 =         − + A A A A 4 3 5 1 1 2 1 2 1 2 1 =  = − + A A A A A A  width ratio of two slits 16:1 1 16 2 2 2 1 2 1 = = = A A d d 20. (b): Constructive interference occurs when the path difference ( ) S1 p − S2 p is an integral multiple of  Or S1 p−S2 p=n Where (tgk¡) n = 0,1,2,3..... 21. (c): Fringe width d D = ....... (i) Where D is the distance between the screen and slit and d is the distance between two slits. When a film of thickness t and refractive index  is placed over one of the slit, the fringe pattern is shifted by distance S and is given by d tD S ( −1) = ...... (ii) Given : S =20 ...... (iii) From equation (i), (ii) and (iii) we get, ( −1)t =20 Or cm cm t 3 8 2.5 10 20 20 5000 10 ( 1) − −    =   − =  −1=0.4or =1.4 22. (a): Fringe width d D = According to the question and d d D D ' 10 2 ' = = d D d D d D  =  =   = 20 1 10 ( / 2) ' ' ' 20 '   = (using (i)) 23. (d): A laser beam is used for locating distant objects because it has small angular spread 24. (b): Here, m 4 1 2.4 10−  =  1=6400Å,2 =4000Å 8 5 6400 4000 1 2 1 2 = =   =    Or m 4 4 2 1 2.4 10 1.5 10 8 5 8 5 − − =   =   =  Decreases in fringe width m 4 4 1 2 (2.4 1.5) 10 0.9 10 − − = − = −  =  25. (d): The resultant intensity 2 cos2 0  I = I Here, 0 I is the maximum intensity and 3  = 2 0 2 0 2 0 2 3 6 cos 3 2 cos         =   =        I =I I I 0 4 3 I = I 26. (c): Width ratio, 1 81 2 1 2 1 = =   I I  Amplitude ratio, 9 :1 1 81 2 1 2 1 = = = I I A A 27. (a): A low flying aircraft reflects the T.V. signal. The slight shaking on the T.V. screen may be due to interference between the direct signal and the reflected signal 28. (c): Fringe width , d D = Also, mv h = Here h is Planck’s constant. This wavelength is inversely proportional to the velocity. Hence , the fringe width increases with decreases in electron speed 29. (c): Fringe width d D = Where  is the wavelength of light, D is the distance between screen and the slits and d is the distance between two slits D. d   = (As  and d are constant) Or D d   = Substituting the given values , we get m Å m m m 6 10 6000 (5 10 ) (3 10 )(10 ) 7 2 5 3 =  =    = − − − −
30. (d): Here, =3 , =2 , =60 A1 A A2 A The resultant amplitude at a point is = + + 2 1 2 cos 2 2 2 R A1 A A A (3 ) + (2 ) + 23  2  cos60 2 2 A A A A 9 4 6 19 2 2 2 = A + A + A = A 31. (a): (1) (2) or (3) 32. (b): the colours of a thin oil film are due to interference 33. (b): Fringe width in first case d D 1 1   = ........ (i) Fringe width in second case . 2 2 2 d D  = ...... (ii) Divided equation (ii) by (i) 1 1 2 2 1 2 1 2 1 2 . . 2 1 . 2 1 / / 2     =   =   =    or D d D d mm mm Å Å 0.8 0.5 6000 7500 2 1 =   = 34. (a): The path difference at the centre is zero. The path difference at the th 4 bright fringe =4 Hence m 10 6000 10− =   Path difference 10 7 4 6000 10 24 10 − − =   =  m 6 2.4 10− =  35. (b): Let th n fringe of 2500 Å coincide with th (n − 2) fringe of 3500Å 3500(n − 2)=2500n 1000n=7000,or n= 7  th 7 order fringe of st 1 source will coincide with th 5 order fringe of nd 2 source 36. (c): Given : 5 10 m,D 1 m,d 1mm 7 =  = = − Distance of th n bright fringes from the centre d nD = Where n = 1,2,3...... So the distance of th 5 bright fringes d D = 5 Distance of th n dark fringe from the centre d D n        = − 2 1 Where n= 1,2,3...... rd 3 dark fringe d D d D  =        = − 2 5 2 1 3 Distance between them d D d D  =        = − 2 5 2 5 5 12.5 10 m 1.25mm 2 1 10 5 1 5 10 4 3 7 =  =      = − − − 37. (b): 2 2 min max ( ) ( ) A B A B I I − + = 2 2 ( ) ( ) 1 9 A B A B − + = Or or A B A B A B A B − = + − + = 3 3 1 3 Or 2 =4 =2  = 2 B A A B or A B 1 4 2 2 2 1  = = B A I I 38. (c): Further from the centre than the first maxima for green light 39. (d): When a bright fringe is formed opposite to one of the slits, 2 d x = Path difference D d D d d D xd 2 2 2 = =  = If it is th n order bright fringe Path difference D d n 2 2  = or  = D d n 2 2 40. (a): We know that d m D ym  = Therefore for wavelength , 1 d D y 1 1 10 = And for wavelength , 2 2 1 2 2 1 2 2 , 5    =  = y y d D y 41. (c): Here, 2 2  t = t 141nm 4 1.33 750 4 min  =    = 42. (b): here a mm m 3 2 2 10− = =  nm m m 9 7 500 500 10 5 10 − − = =  =  D = 1m
The distance between the first minima on either side on a screen is 3 7 2 10 2 2 5 10 1 − −     =  = a D 5 10 m 0.5 10 m 0.5mm 4 3 =  =  = − − 43. (c): Conservation of energy holds good and energy is redistributed 44. (b) For destructive interference the path difference should be an odd multiple of 2  45. (b) Angular position of first dark fringe 2d 2d (2 1 1) 1  =   =  − 2730 10 rad 2 0.1 10 5460 10 6 3 10 − − − =     = 0.16o 180 2730 106 =  =   46. (a) The condition for bright fringes is, path difference,  = dsinbright = N Where N = 0,1,2 The angular position of the bright fringes is         = − d N sin 1 bright 47. (a) = =  =  b a b a I I 2 2 2 1 Fringe visibility is given by 2 2 2 2 max min max min (a b) (a b) (a b) (a b) I I I I V + + − + − − = + − = 1 2 B 1 b a 2(a / b) 2(a b ) 4ab 2 2 2 2  + =         + = + = 48. (d) n11 = n22 4000Å 24 16 6000Å n n 1 2 1 2 =   =  = 49. (c) Angular fringe width d   =  = d Here, d = 0.01 mm 10 m −5 = radians 180 1o   = = 10 1.74 10 m 180 −5 −7  =    = 0.174 10 m 0.174 m 6 =  =  − 50. (a) Fringe width d D  = and    =  = D d 51. (c) Fringe width d D  = 7.9 68.5 10 5.4 10 1 2 2 1 2 1 =   =   =    52. (a) Width of slit, x n D d  = Here, n = 1, D = 1, x= 2.5mm 2.5 10 m −3 =  500nm 500 10 m −9  = =  2 10 m 0.2mm 2.5 10 1 500 10 1 d 4 3 9 =  =      = − − − 53. (d) When red light is replaced by blue light the diffraction bands become narrow and crowded. 54. (b) The position of nth minima in the diffreaction pattern is d nD xn  = d 2D d D x x (3 1) 3 1  =   − = − Or 2 10 m 3 10 2 0.50 6000 10 x x 2D d 4 3 10 3 1 − − − =      = −  = 55. 56. (b) The angular width of the central maximum is 2/ a where a is the width of the slit. If the value of a is doubled the angular width of the central maximum decrease to half its earlier value this implies that the central maximum becomes much sharper. Furthermore if a sis doubled the intensity of the central maximum becomes four times thus the central maximum becomes four times thus the central maximum becomes much sharper and brighter. 56. (a) Here, 6000Å 6000 10 m 6 10 m −10 −7  = =  =  a 0.3mm 0.3 10 m 3 10 m −3 −4 = =  =  For first minima asin =  where a is the slit width