Question

In a Young's double slit experiment, let $A$ and $B$ be the two slits. A thin film of thickness $t$ and refractive index $\mu$ is placed in front of $A$. Let $\beta =$fringe width. The central maximum will shift

• A. towards $A$
• B. towards $B$
• C. by $t(\mu -1)\frac{\beta }{\lambda }$
• D. by $\mu t\frac{\beta }{\lambda }$

Answer 1

Answer: (A), (C)

towards $A$
by $t(\mu -1)\frac{\beta }{\lambda }$

Let $d=$distance between the slits,
$\lambda =$ wavelength of light,
$D=$ distance from the slits to the screen.
From a point $P$ on the screen at a distance $x$ from the centre of the screen, path difference$=\Delta =x\frac{d}{D}$.
Path difference introduced due to film$=t(\mu -1)$.
For central maximum at $P$, $x\frac{d}{D}=t(\mu -1)$
or $x=t(\mu -1)\frac{D}{d}$
Now, $\beta =\frac{\lambda D}{d}$
or $\frac{D}{d}=\frac{\beta }{\lambda }$
$\therefore x=t(\mu -1)\frac{\beta }{\lambda }$