question_answer

In Youngs double slit experiment, the aperture screen distance is 2 m. The slit width is 1 mm. Light of 600 nm is used. If a thin plate of glass \[(\mu =1.5)\] of thickness 0.06 mm is placed over one of the slits, then there will be a lateral displacement of the fringes by:

A)  zero                                     

B)         6 cm

C)         10 cm                                  

D)         15 cm

Correct Answer: B

Solution :

When a thin glass plate of thickness t is placed over one of the slits, then lateral displacement is given by                                    \[x=\frac{(\mu -1)tD}{d}\] Given, \[\mu =1.5\,,t=0.06\,mm=6\times {{10}^{-5}}m\] \[D=2m,\]           \[d=1\,mm=1\times {{10}^{-3}}\,m\] Putting the values in the above relation, we get \[x=\frac{(1.5-1)\times 6\times {{10}^{-5}}\times 2}{1\times {{10}^{-3}}}\] \[=0.5\times 12\times {{10}^{-2}}\] \[=0.06\,m\] \[=6\,cm\] Note: Lateral displacement is independent of \[\lambda \]i.e., if white light is used. Then shift of red colour fringe = shift of violet colour fringe