Two transparent slabs having equal thickness but different refractive indices $μ_{1}$ and \ $m u_{2}$ 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 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 $P_{0}$ which is equidistant from th eslits ?

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Solution

The change in path difference due to the two slabs is $(μ_{1} - μ_{2})$ t (as in problem N o. 16) for having a minimum at p, the path
Difference should change by $\frac{1}{2}$
So, $\frac{λ}{2} = (μ_{1} - μ_{2}) t \Rightarrow t = \frac{λ}{2 [μ_{1} - μ_{2}]}$

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The change in path difference due to the two slabs is (μ1−μ2) t (as in problem N o. 16) for having a minimum at p, the path Difference should change by 12 So, λ2=(μ1−μ2)t⇒t=λ2[μ1−μ2]

The change in path difference due to the two slabs is $(μ_{1} - μ_{2})$ t (as in problem N o. 16) for having a minimum at p, the path
Difference should change by $\frac{1}{2}$
So, $\frac{λ}{2} = (μ_{1} - μ_{2}) t \Rightarrow t = \frac{λ}{2 [μ_{1} - μ_{2}]}$