OPTICS-04-SUBJECTIVE SOLVED

SOLVED SUBJECTIVE PROBLEMS Problem 1. A 4 cm thick layer of water covers a 6 cm thick glass slab. A coin is placed at the bottom of the slab and is being observed from the air side along the normal to the surface. Find the apparent position of the coin from the surface. Solution: Using equation, the total apparent shift is 4cm h1 6cm h2 Air Water Glass s   1   1  h1 1     h2 1     1   2  4cm or s    1 6 1  1       4 / 3   3/ 2  = 3.0 cm 6cm Thus, h  h1  h 2  s  4  6  3 = 7.0 cm. Problem 2. How long will the light take in travelling a distance of 500 metre in water? Given that  for water is 4/3 and the velocity of light in vacuum is 31010 path. Solution: We know that   velocityof light in vaccum velocityof light in water cm/sec. Also calculate equivalent 4  31010 3 velocityof light in water or velocity of light in water  2.251010 cm/sec. Time taken = 500 100 2.251010  2.22 106 sec. Equivalent optical path =   distance travelled in water  4  500  666.64 m. 3 Problem 3. In Young’s experiment for interference of light two narrow slits 0.2 cm apart are illuminated by yellow light (  5896 Å). What would be the fringe width on a screen placed 1 m from the plane of slits. What will be the fringe width if the whole system is immersed in water (  4 / 3)? Solution: As here d  2 103 m,   5896 1010 m   D 1 5896 1010 and D = 1 m, 4 ()  d 2 103  2.948 10 m □ 0.3 mm Now if the system is immersed in water, ()w  w  v [as v  f and f = constt.]   c or ()  ( / ) as   c  w  v   ()w  (0.3)  (3/ 4)  0.225 mm. Problem 4. A beam of light consisting of two wavelengths 6500 Å and 5200 Å is used to obtain interfer- ence fringes in a Young’s double slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maxima for the wavelength 6500 Å. (b) What is the least distance from the central maxima where the bright fringes due to both the wavelengths coincide? The distance between the slits is 2 mm and the distance between the plane of the slits and screen is 120 cm. Solution: (a) According to theory of interference, position y of a point on the screen is given by y  D (x) d and as for 3rd maximum x  3 y  D (3)  120 (3  6500 108 )cm  0.117 cm d 0.2 also as   (D / d), y  3, i.e.,   (y / 3)  0.039 cm. (b) If n is the least number of fringes of 1 ( 6500 Å), which are coincident with (n + 1) of smaller wavelength 2 ( 5200 Å), y  n  (n  1) , i.e., n 1    1 n  2 or n  2 1  2  5200  5200  4 6500  5200 1300 So y  4  4  0.039  0.156 cm. Problem 5. In a modified Young’s double slit experiment, a monochromatic uniform and parallel beam of light of wavelength 6000 Å and intensity (10 / ) W / m2 is incident normally on two circular apertures A and B of radii 0.001 m and 0.002 m respectively. A perfectly transparent film of thickness 2000 Å and refractive index 1.5 for wavelength 6000 Å is placed in front of aperture A. Calculate the power (in watts) received at the focal spot F of the lens. The lens is symmetrically placed with respect to the aperture. Assume that 10% of the power re- ceived by each aperture goes in the original direction and is bought to the focal spot. Solution: As intensity and power are defined as I  E , St P  E , and P  IS So power received at A and B is respectively, P  10  (0.001)2  105 W A  and P  10  (0.002)2  4 105 W B  and as only 10% of the incident power passes, P  10 105  106 W and P  10  4 105  4 106 W A 100 B 100 Now as due to introduction of film the path difference produced x  ( 1)t  (1.5 1)  2000  1000 Å So,   2 (x)  2 1000     But as in interference, I  I1  I2  2 6000 3 F cos and if S is the area of focal sport, P  IS  IAS  IBS  2S( IAIB ) cos i.e., P  PA  PB  2 cos(/ 3) or, P  106 [1  4  2( 1 4 )  (1/ 2)]  7106 watt. Problem 6. In the figure shown, for an angle of incidence i at the top surface, what is the minimum refractive index needed for total internal reflection at the vertical face? Solution: Applying Snell’s law at the top surface sin r  sini …(i) For total internal reflection the vertical face sin c  1 Using geometry,   90o  r  sin(90  r)  1 or cosr  1 …(ii) On squaring and adding equation (i) and (ii), we get 2 sin2 r  2 cos2 r  1  sin2i or   . Problem 7. A point source of light is placed at the bottom of a tank containing a liquid (refractive index =  ) upto a depth h. A bright circular spot is seen on the surface of the liquid. Find the radius of this bright spot. Solution: Rays coming out of the source and incident at an angle greater than c will be reflected back into the liquid there- S fore, the corresponding region on the surface will appear dark. As is obvious from the figure, the radius of the bright spot is given by R  h tan   hsin c R  Since c cos or sin   1 c   R  h . Problem 8. The cross-section of the glass prism has the form of an isosceles triangle. One of the equal faces is silvered. A ray of light incident normally on the other equal face and after being reflected twice, emerges through the base of prism along the normal. Find the angle of the prism. Solution: From the figure,   90o  A Also and Thus, i  90o    A   90o  2i  90o  2A   90o    2A   r  2A …(i) From geometry, A      180o . or A  180  36o . 5 Problem 9. For the optical arrangement shown in the figure. =1 =1.33 1cm O C 20cm 40cm Solution: According to Cartesian sign convention u = - 40 cm, R = - 20 cm   1, 2  1.33 Applying equation for refraction through spherical surface, we get 2  1  2  1 v u R 1.33  v 1 40  1.33 1 20 After solving, v = - 32 cm. The magnification is m  h2  1v h1 h2   1(32) 1 1.33(40) 2u or h2  0.6 cm The positive sign shows that the image is erect. Problem 10. A lens has a power of +5 diopter in air. What will be its power if completely immersed in water ? Given   3 ;   4 g 2 w 3 Solution: Let fa and fw be the focal lengths of the lens in air water respectively, then Pa  a and P  w fw fa = 0.2 m = 20 cm Using lensmaker’s formula P  1  ( 1)  1  1  a f g R R  ...(i) a  1 2  1   g 1  1  1  f    R R  w  w   1 2  P  w  (   )  1  1   w f g w R R  ...(ii) w  1 2  Dividing equation (ii) by equation (i), we get, Pw Pa or Pw  (g  w )  1 (g 1) 3  Pa  5 D 3 3 Problem 11. A thin plano-convex lens of focal length f is split into two halves. One of the halves is shifted along the optical axis. The separation between object and image plane is 1.8 m. O The magnification of the image formed by one of the half- lenses is 2. Find the focal-length of the lens and separation between the two halves. 1.8m Solution: This is a modified displacement method problem. Here a  1.8m and a  d  2 a  d 1 Solving we get d = 0.6 m a2  d2  f  4a  0.4 m . Problem 12. In an interference arrangement similar to Young’s double-slit experiment, slits S1 and S2 are illuminated with coherent microwave sources each of frequency 1 MHz. The sources are synchronized to have zero phase difference. The slits are separated by distance d = 150.0 m. The intensity I() is measured as a function of  , where  is defined as shown in figure. If I0 is maximum intensity, calculate I() for (a)   0o , (b)   30o and (c)   90o . Solution: For microwaves, as c  f ,   c  3108  and as f 106 x  dsin    2 (dsin )   300 m 2 (150 sin )  sin  300 So, I     I1 I2 2( I1I2 ) cos with I1  I2 and   sin  reduces to I  2I [1  cos(sin )]  4I cos2 (sin / 2) and as IR will be max. when cos [((sin ) / 2]  max  1 2 So that, and hence, So (a) If   0 (b) If   30o (c) I   90o (IR )max  4I1  I0 (given) I  I cos2 [(sin ) / 2] I  I cos2 (0)  I I  I cos2 ( / 4)  (I / 2) I  I cos2 ( / 2)  0 …(i) Problem 13. What is the effect on the interference fringes in a YDSE due to each of the following operations. (a) the screen is moved away from the plane of the slits (b) the (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength (c) the separation between the two slits is increased (d) the monochromatic source is replaced by source of white light (e) the whole experimental setup is placed in a medium of refractive index  Solution : (a) Angular separation ( ) of the fringes remains constant. But the linear separation or d fringe width increase in proportion to the distance (D) from the screen. (b) As  decreases, fringe width (  ) decreases (c) As d increases, fringe width (  1 ) d decreases (d) The interference pattern due to different component colours of white light overlap (in- coherently). The central bright fringes of different colours are at the same position on either side of the central white fringe is blue; the farthest is red. (e) Since in a medium the wavelength of light is  '   , therefore the fringe width is given by    ' D  D . d d Thus, fringe width decreased by a factor  . Problem 14. Light from a source consists of two wavelength 1  6500A and 2  5200A . If D = 2m and d = 6.5 mm find the minimum value of y ( 0) lengths coincide. where the maxima of both the wave- Solution: Let n1 th maxima of 1 coincides with n2 th maxima of 2 Then or y  y 1 2 n11D  n2 2D or n1  1  5200  4 d d n2 2 6500 5 Thus, fourth maxima of 1 coincides with fifth maxima of 2 The minimum value of y ( 0) is given by 4 D 4(0.65106 )(2) y  1   0.8 mm . d 6.5103 Problem 15. In a YDSE,   60 nm; D = 2m;   6 mm. Find the positions of a point lying between third maxima and third minima where the intensity is three-fourth of the maximum intensity on the screen. Solution: Using equation I  4I cos2  0 2 here I  3 (4I )  3I 4 0 0  cos   3 2 2 Thus, or Since   n   2 6   2n   3   2 p  2  d.yn     D   2 dyn    2n    or y D 3   n  1  D n  6  d   For the point lying between third minima and third maxima, n = 3 and y   3  1  D 3  6  d or y3    17 D 6 d Putting   0.6 106 m; D  2m;d  6 mm 17 (0.6 106 )(2) y3  6 6103  5.67 mm .