Question

In the ideal double – slit experiment, when a glass – plate $(\mu =1.5)$ of thickness t is introduced in the path of one of the interfering beams $(wave–length\lambda )$, the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass – plate is

• A. $2\lambda$
• B. $\frac{2\lambda }{3}$
• C. $\frac{\lambda }{3}$
• D. $\lambda$

Answer 1

The correct option is A $2\lambda$

The pattern shifts such that nth bright fringe occupies the position of central maximum.
$\therefore$ shift = $\frac{Dn\lambda }{d}$ $\therefore$ $\frac{n\lambda D}{d}=\frac{D(\mu -1)}{d}t$
$n\lambda =(\mu -1)t$ or $t=\frac{n\lambda }{(\mu -1)}$
For minimum value of t, n = 1
$\therefore$ $t=\frac{\lambda }{(1.5-1)}=2\lambda$