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                 \[x=\frac{(1.5-1)\times 6\times {{10}^{-5}}\times 2}{1\times {{10}^{-3}}}\]                 \[=0.5\times 12\times {{10}^{-2}}\]                 = 0.06m                 = 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