Question

A plate of thickness $t$ is made up of a material of refractive index $\mu$. It is placed in front of one of the slits, in a double-slit experiment. What should be the minimum thickness $t$ of the plate, which will make the intensity at the center of the fringe pattern zero ?

• A. $(\mu -1)\frac{\lambda }{2}$
• B. $(\mu -1)\lambda$
• C. $\frac{\lambda }{2(\mu -1)}$
• D. $\frac{\lambda }{(\mu -1)}$

Answer 1

The correct option is C $\frac{\lambda }{2(\mu -1)}$

Intensity at the center will be zero if the path difference, at that point, is equal to $\frac{(2n-1)\lambda }{2}$.

i.e. $(\mu -1)t=\frac{(2n-1)\lambda }{2}$

$\Rightarrow t=\frac{(2n-1)\lambda }{2(\mu -1)}$

For minimum thickness, $n=1$.
$\therefore t=\frac{\lambda }{2(\mu -1)}$

$\begin{array}{c}\text{Why this question?}\\ \text{Note :- When a slab of thickness}t\text{and refractive index}\mu \text{is placed in front of one of the slits, the path difference at center is}(\mu -1)t.\end{array}$