Question

Two transparent slabs having equal thickness but different refractive indices µ1 and µ2 are pasted side by side to form a composite slab. This slab is placed just after the double slit in a Young's experiment so that the light from one slit goes through one material and the light from the other slit goes through the other material. What should be the minimum thickness of the slab so that there is a minimum at the point P0 which is equidistant from the slits?

Answer 1

Minimum Thickness is  $\frac {\lambda}{2(\mu_1 - \mu_2)}$

Explanation:

We know that,

• As per the Young's Experiment,
• The Phase Difference is equal to $\frac {2\pi}{\lambda}$ times Path Difference.

Or

$Phase\ Difference = \frac {2\pi}{\lambda} \times Path\ Difference$

• The change in path difference due to the two slabs is $(\mu_1 - \mu_2)t$

• For having a minimum at $P_0\ ,$

• the path difference should change by $\frac {\lambda}{2}.$

So,  

$\frac {\lambda}{2}$

= $(\mu_1 - \mu_2)t$

So, t = $\frac {\lambda}{2(\mu_1 - \mu_2)}$