5.OPTICS-SOLUTIONS-PART TEST-1

SOLUTIONS TO PART TEST-01 TOPIC : OPTICS 1. The ray just above x-axis shall after refraction at curved surface intersect x axis at (A) a distance x1 from O. Where 1    1 x1 R or x = R  15 cm 1   1 2 3 The critical angle for air-glass interface is  = sin–1 5 = 37°. The rays above the ray incident on curved surface at i = 37° shall suffer TIR and need not be considered. The ray incident as curved surface at  = 37° after refraction intersect curved surface at (B) a distance x2 from O. In  PBC 3  = tan53° = 4 5 or x = 1 x2  required width x = x 3 – x = 15  5 2 4 = 25 cm 1 2 2 4 4 2. Let O be the centre and A be the point source for ray incident at any point P thus for maximum angle of incidence i, 3. Given that OA i = sin–1 OP = sin–1 x R Number of waves in A = Number of waves in B + Number of waves in C 3d  d A B C  2d C A B C But  = f 3d d 2d Here C is constant (3 × 108 m/s) and frequency f does not depends on medium. 1 So,    3A = B + 2C 3(1.5) = B + 2(1.6) B = 4.5 – 3.2 = 1.3. 4. When object moves normal to the mirror, image velocity will be opposite to it. When object moves parallel to the mirror, image velocity will be in the same direction. → → 5. Vm =– Vom (normal to plane mirror)  → → → → V G  VmG = –( VoG  VmG ) → → → V G  VoG →  VOG  0  → = 2VmG VG  2V sin 6. Due to T.I.R., light is returned back and due to dispersion light splitted into different colors. 7. A = 90 –  (A angle of prism) for TIR i > c sin i > sin c sin(90 – ) > sin c cos  > sin c sin c = cos  > 6 5 3 2 4   < cos–1 4 = 37° . 5 5 8. Let the critical angle of interface between media 1 and 2 is C1 and between 1 and 3 is C2. n2 Then sin C1= n1 and sin C2= n3 n1 . From TIR at second interface 90 – C1 > C2 . taking sin of both side we get n cos C > sin C or  3 or n2  n2  n2 1 2 n1 1 3 2 9. For spherical surface using n2 – v n1 = u n2  n1 R  n – 2R 1 n  1  = R  n = 2n – 2  n = 2. 10. Put A =  and  =  A  min  sin  The relation  =  2  .  A  sin 2   sin A sin A / 2    2 sin A / 2.cos A / 2 sin A / 2  A = 2 = 90º 11. Error –  Error – II 12. For a real object a convex mirror always form a virtual image. 13. For M1 : 1/v + 1/u = 1/f ; 1/v = –1/20 + 1/30 ; 1/v = –10/600 ; v = –60 m1 = –2 for M2 : u = + 20. f = 10 1 1  V + 20 1 = 10  V = 20 20  m2 = – 20 = – 1  m = m1 × m2 = + 2 14. By mirror formula : 1  1  1 v  10 10  v = + 5 cm  m = + 1 ( m =  v ) 2 the image revolves in circle of radius u 1 cm. Image of a radius is erect  particle will revolve in the 2 same direction as the particle. The image will complete one revolution in the same time 2s. 2 velocity of image v =  r = 2 1  × 2 = 2 cm/s = 1.57 cm/s Ans. np 15. nrel = ns  ns 1.512 = 0.9 = 1.68 ns = Refractive index of surrounding. np = Refractive index of medium. 16. 1 =  1  1 P f u v 1 1 1 20 =  30  v 1 1 1 1 =  v 20 30 = 60 v = 60 cm As triangles OPC & ABC are similar 2  60 = x  x = 10 1 = 0.3 cm. 3 17. Focal length is minimum in case I, therefore power is maximum. ( P = 1/f) 18. By symmetry, r1 = r2 = 30° and by Snell's law, sin 45° = n sin 30° n = 19. The distance between nth bright fringe and (n + 1)th dark fringe is equal to half of fringe width =  = D 2 2d 20. The intensity at distance x-from central order bright on screen is  =  +  + 2  max =  cos    2 = 4 given  = 3/4   = 3  3 =  +  +2 . 3 = 2 (1 + cos ) cos  = 1/2  = /3, 2/3, –/3, ....  = 2/6, 2, –2/6, .... yd  cos    D    ,       y =   , ,..... D 6 6  d  6 6 where  = maximum intensity  = fringe width  Hence the required distance = 2 × 6 21. In  S PO :  = 3 = 0.20 mm  tan 2 = d/ 2 D As D > > d   is very small.  tan        d 2 2 2 2D  D  1  Fringe width = D    Ans. d  d  22. Angular width = tan  = y/D tan     = 10–3 (given) d  = y/D  y = D y  Dd  D n = d    180  No. of fringes within 0.12° will be 0.12  2 n = 360 103  [2.09]  The number of bright spots will be three. Hence (B).     2  4  1 2 23. max   1  =   = 9 min       max  min  9min  min  9  1  4 max  min 9min  min 9  1 5 24.   (Fringe width) Higher '' means wider fringes. At higher wavelength, same space on the screen is occupied by wider fringes, and hence lesser number of fringes. Also, n  = n   n  = n   4000  n  n = 40 1 1 2 2 1 1 2 2 6000 60 25. There are three and a half fringes from first maxima to fifth minima as shown.   + /2 = 7mm  = 2mm   = D d = 600 nm.   26. At path difference 6 , phase difference is 3  =  + O + 2  cos 3 = 3 O max = 4 (at central maxima) 3I0 So the required ratio is 4I0 = 0.75 27. For maxima at 'P' path difference at 'P' should be, P = , 2, 3, ....  P = yd/D  yd/D = , 2  y = d/2  d2   2D d d2 , 4D 6D , ..... Hence (C) is possible. 28. All rays originating from point source are incident normally on glass air interface, hence no ray suffers total internal reflection. 29. viG – vmG = –(voG – vmG) viG = – voG + 2vmG  1  30. Every point of the image will be at distance t 1   towards left of the object. The image will be parallel to   ‘AB’ and of the same size. SOLUTIONS TO PART TEST-02 TOPIC : OPTICS 1. Path difference at P x = AP – BP for maxima x = n n = x (n  +x)2 = x2 + 162 n22 x2 + 2nx = x2 + 162 162 – n22 x = 2n n = 1, 2, 3 are possible sol. 2. Using formula of spherical surface taking 'B' as object 2  1 = 2  1 1 (R being the radius of the curved surface)  = 2  (2R) R 2 3. 1 = 1 – 1  v f u 1  1  v 10 1  20  v = 20 cm 3 I = – 20 v 3 u × O = 20 1 × 1 = 3 ∴ The distance between tip of the object and image is S = = cm 1 4. 10 1 = v – 1 (15)  v = + 30 cm for small object (or length of image) v 2  30 2 dv = u2 du =  15   1 = 4 mm   5. Vi = v 2 u 2 V0  f v u  f = u  v 2    f 2  V = V – V = 1 2  V = 1     V as u increases v decreases rel 0 i  u  0   f  u   0 rel here u is negative so when it approach lens it will increases. 6. If mirror is turned, about an axis perpendicular to plane of mirror, then there will be no change in incident angle and reflected angle so angle between incident & reflected rays after rotation will be same as before. 1  3     1  1  7. 10 =  2 1 R R  .............(1)    1 2  7 1  3 / 2    1  1  when placed in medium of RI 5   =  7 / 5 1 R R  ................(2) F    1 2  Dividing (1) by (2) F = 70 cm 8. Due to refraction at water-air interface, the object is at a distance less than the focal length of the lense, thus, a virtual image is created inside water 9. Let the bubble B is at distance H from the face F1 of the cube. nc sin i = na sin r na h1 = nC H = 5 cm Similarly when looking from opposite face F2, nc sin i = na sin r na h2 = nC (12 – H) = 3 cm Solving H = 7.5 cm and nC = 1.5 2 10. Path difference at point O is d sin  = 0.5 mm corresponding phase difference,  =  2(0.5 103 ) × p = 5000  1010 = 2000  = 2 × 1000  O is a point correesponding – 1000th maxima. the point at 1 m below O corresponds to central maxima. So 4 m from O will be maxima position. 11. In case x no fringes are formed because light passes the slab normaly & in case y fringes are obtained. Suppose for point A ‘t’ is such that it satisfies the condition for bright interference. The same ‘t’ will be present throughout the line A A’ & therefore the line AA’ will be bright & a bright line will be seen. The same applies for dark lines.Hence fringes are straight line 12. Here path difference will be : x = [( –  ) t]/   = 2 w (2  1)t 1 2 =  / 1 (2  1)t 1 = 2 ( –  )t Hence (C) 13. Statement 1 is false because constructive interference can be obtained if phase difference between cohrent sources is 2, 4 , 6, etc. 14. In the situation of statement-1, the magnitude of image (v) and object (u) distance is same. The size of image = n1  v × size of object n2 u  v = u and n1  n  size of image  size of object. Hence statement-1 is false and statement-2 is true. 15. The lens can be assumed to be made of prisms of small but different angles. The lowest ray of beam is incident on prism of smaller angle while the topmost ray of beam is incident on prism of larger angle. Hence all rays of beam suffer different deviations. Therefore both statements are correct and statement-2 is the correct explanation. 16. From symmetry the ray shall not suffer TIR at second interface, because the angle of incidence at first interface equals to angle of emergence at second interface. Hence statement 1 is false 17. If phase difference at point P is zero then n d sin = n d sin'  ' = 37° y 3 and as tan' = D  y = – 4 m It is negative because upper path in medium n2 is longer than lower path in the same medium. 18. Path lengths in medium 2 are equal for point O. Therefore path difference = d sin (0.3) 4    n1  0.3mm, n2   3   0.36 mm 10 9 p    dsin 4 3 10 9 2 ,   = 2 p    1  4 / 3  = 10   4  4 / 3 1   3   = 0 + 0 + 20 cos  2  = 0 0.3   2  10/ 9   3   10/ 9     19. As we go up from point O, path difference will increase. At O, phase difference is 3 + 3 and when it becomes 4, there will be maximum. Extra path difference created in medium 2 must lead to 2 phase 3 difference. 2 2 a . d sin1. n2 = 3 20. Using values sin = 3 25 1  1  1 3 y  tan  = = D  y = 300 cm = 150 cm  v u Here v = 2.5 (Distance of Retina as position of image is fixed) u = – x 1  1  1  2.5 x For f min : x is minimum 1 min  1  1 2.5 25 21. For f max : x is maximum 1 max  1  1 2.5  22. For near sighted man lens should make the image of the object with in 100 cm range For lens u = –  v = – 100 1 lens 23. (A)  1  1  100   Image (Erect, (B) enlarged) F1 (C) (D) 24. By using ( – 1)t = n , we can find value of n, that is order of the fringe produced at P, if that particular strip has been placed over any of the slit. If two strips are used in conjuction (over each other), path difference due to each is added to get net path difference created. If two strips are used over different slits, their path differences are subtracted to get net path difference. Now, n1 = (1  1) t1  = 5 n2 = 4.5 and n3 = 0.5 For (A), x = (B –1)tB = (1.5 – 1) 2.5 = n 4B order of the fringe is 4.5 i.e. 5th dark. 3B for (B), x = |(A – 1)tA – (B – 1)tB | 2B x = (1.5 – 1)5 – (2.5 – 1)1.5 = n 1B net order is 5 – 0.5 = 4.5 i.e. fifth dark. CB 1B for (C), x = |(A – 1)tA – [(B – 1)tB + (C – 1)tC]| net order is 5 – (0.5 + 4.5) = 0 2B i.e. it is central bright again at P. 3B for (D), x = |(A – 1)tA + (B – 1)tB – (C – 1)tC | net order is (5 + 0.5) – (4.5) = 1 i.e. first bright SOLUTIONS TO PART TEST-03 TOPIC : OPTICS 1. Given NED = 30º  BED = 120º BCDE is cyclic quadrilateral  BCD = 60º The line CE will be angle bisector of BCD  BE = a tan 30º =  now tan i = BE = a 3 = AB a / 2  sin i = now by snell’s law 1 × sin i = n sin r  = n × 1  n = 2 2. For normal incidence path difference between ray 1 and ray 2 is 2  t  For minimum thickness increment 2  t = 2  (t – t1) =  41 9.610–7 = 4  1.2 = 2 × 10–7 m 3. In absence of film or for =0 intensity is maximum at screen. As the value of  is increased, intensity shall decrease and then increase alternately. Hence the correct variation is 4. Acceleration of block AB = 3mg 3m  m = 3 g ; acceleration of block CD = 4 2mg 2g 2m  m = 3 Acceleration of image in mirror AB ( ViG = –VoG + 2VmG) = 2 acceleration of mirror 2.   3g  3 =   = g   2 2. 2 g  4g Acceleration of image in mirror CD =   =  3  4g    3g   17g  Acceleration of the two image w.r.t. each other = 3   = .  2  5. Let be the angle of emergence from the first prism be ‘e’ Snell's law on surface AB 4 1 × sin 53º = 3 r = 37º  A = r1 + r2 sin r r2 = 67º – 37º = 30º 4 Snell's law on AC 3 sin 30º = 1 × sin e 2 e = sin–1 3 Then for net deviation to be double, the incident ray on side A B of second prism should make angles i or e with normal. Hence the angle between the given then will be 2e or i + e. 6. Dispersion will not occur for a light of single wave length  = 4000Å. 7. As AB is common and O1B = B  O1BA and BA are congruent By symmetry A is perpendicular to O1 to O2 and  O1BA =  BA  BA = 45° and  BAC = 90° 8. From the figure : Amplitude = x =  – cos = (1 – cos) =   2 (By binomial expansion) Ans. 9. For convex mirror |m| < 1 for any real object Now, Vimage = – m2 Vobject  | Vimage | < | Vobject | always. 10. Now m Cair = Cmed 3  108  = 2  108 = 1.5 R As f =  – 1 (Plano convex) where R = Radius of curvature i.e. the radius of that sphere of which the given lens is the part.  n  OAC  R2 = 32 + (R – t)2 R2 = 32 + R2 – 2Rt + t2 R = 32  t2 = 2t 9  0.09 2  0.3 ( t = 3 mm) 9.09 R = 0.6 90.9 = 6 = 15.16  R = 15.2 cm (approx)  f = 15.16 1.5  1 = 30.32 cm or f = 30.4 cm (approx) 11. v For first & second position u = 1 , u O v = 2  O v 2 1 u2 = 2 = 4.84  v  2.2 and v + u = 96  v = 66 , u = 30 u O = v  2.2  11  A is True 2 u 5 Distance between two position of lens = v - u = 36 cm  B is True Focal length of lens f = uv u  v  66  30 66  30 = 20.63  C is False Distance of lens from shorter image = u = 30 cm  D is True 12 b 13. by cauchy's formula µ = a + 2 λ : a & b are constant of medium. The index of refraction for light at the red end of the visible spectrum is lesser than at the violet end.Hence statement -2 is false 14. Draw an incident ray on the mirror and trace the corresponding reflected ray. If a point object moves along this ray, its image will always lies on the traced reflected ray. Hence when a point object moves near the principal axis of a fixed spherical mirror along a straight line, then its image formed by the spherical mirror also moves along a straight line.Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. 15 - 17 sin c = µair  µliquid The critical angle is C = 45° Hence all rays from S which are incident on interface at an angle of incidence greater than 45° are reflected back (and appear to come from S1) and fall on screen between OP. Light also directly falls on OP from S.  Interference pattern is formed in region OP. Since waves from S do not under go phase change during reflection, We can take S and S1 to be coherent sources in same phase. Hence zero order bright is formed at y = 0. 15. No interference is observed above y = 0. Therefore y coordinate of second order bright below y = 0 is     5000 2 10   D   10  1 y = – 2 = – 2    = 2   2  = –500 m.  2d   2  103         16. The region in which inteference pattern can be observed is OP. OP = AP  1 metre 17. If the screen is brought nearer than 1 mm, no rays from O shall under go TIR and Hence overlapping of incident and reflected beams from O shall not occur on screen.  Minimum distance D = 1 mm. 18. By snell's law n = sin i sin r b and n = a + 2 So at inclined face angular incidence is same for all ray and refraction is from denser to rarer. So emergent angle will be less for red. 19. Lets find out the radius of curvature of equi. convex lens. 1  ( – 1)  1 – 1   1   3   2      f R – R  – 1   10 2 R  R = 10 cm.       Now for lens : 1 – 1  1 V – 20 10 1  1 V 20  for surface of tube (of R = 10 cm.) 2 – 1  2 – 1  4 / 3 – 1  4 / 3 – 1  V = + 80 cm. V u R V  20 – 10 (b) Now for mirrors. As the object for the mirrors is at 20 cm so the image will be at 20 cm only  u = – 2 f  v = – 2f also.  magnification = m = y y0  –v u y –  – 20      y = + (1 mm) – (1mm)  – 20  I so the final images are like. so the distance between the images is 4 mm. (c) Now, these I2 and I4 behave as the 2 sources for fringe pattern.   = D vD = fd = (c / )D fd    3  108  0.8 =  4    1015 (4 10–3 ) = 60 m.     20. Velocity and acceleration of central maximum = velocity and acceleration of screen  it does not move to the left or right on the screen.] vscreen = 0 + gt = 100 m/s  → = 100 ˆj m/s (in vector form) and → s = 10 ˆj Ans. 4 21. Solutions : Given, vf = 10 m/s () vi + vf = 13 vi = 13  10 = 3 m/s Refraction from surface AB (object I1) 3 / 2  1 = (3 / 2)1 3 1 1 2 9  =  =  v = ; v 6 1 2v 2 6 6 2 dv n1 v 2 du By diff. of the image formula, dt = n u 2 dt  3 = 1 3 / 2 (9 / 2)2  (6)2 v0  v0 = 8 m/s. Ans. 22. R = 10cm Applying snell’s law sin 3 sin r = 4  r = 53° sin( r  ) By sine law in  ABC 10 sin(  r) = (10  x) 10  x ; 10 = 4 5 (sinr cos cosr sin) 4 5  4  4   5 3 3   5  ; 10 + x = 200 7  x = 200  70 7 = 130 7