Question

Two transparent slabs having equal thickness but different refractive indices µ₁ and µ₂are 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

Given:-

Refractive index of the two slabs are µ₁ and µ₂.

Thickness of both the plates is t.

When both the strips are fitted, the optical path changes by

\[∆ x = \left( \mu_1 - 1 \right)  t -   \left( \mu_2 - 1 \right)  t\]

\[= \left( \mu_1 - \mu_2 \right)  t\]

For minimum at P₀, the path difference should be \[\frac{\lambda}{2}.\]

i.e. \[∆ x = \frac{\lambda}{2}\]

So, \[\frac{\lambda}{2} = \left( \mu_1 - \mu_2 \right)t\]

\[ \Rightarrow t = \frac{\lambda}{2\left( \mu_1 - \mu_2 \right)}\]

Therefore, minimum at point P₀ is \[\frac{\lambda}{2\left( \mu_1 - \mu_2 \right)}.\]