Physics - Wave Optics Question with Solution | TestHub

The path difference for the $n^{th}$ dark fringe is given by $\Delta x=(2n-1)\frac{\lambda }{2}$.

For the $5^{th}$ dark fringe ($n=5$), the initial path difference at the point is:
$\Delta x_1=(2(5)-1)\frac{\lambda }{2}=4.5\lambda$

When a thin transparent film of thickness $t$ and refractive index $\mu$ is introduced in one of the interfering paths, the optical path changes by $(\mu -1)t$.

The new fringe at the same point is the $7^{th}$ bright fringe. The path difference for the $n^{th}$ bright fringe is $\Delta x=n\lambda$.

For the $7^{th}$ bright fringe ($n=7$), the new path difference is:
$\Delta x_2=7\lambda$

The change in path difference is equal to the additional optical path introduced by the film:
$\Delta x_2-\Delta x_1=(\mu -1)t$

$7\lambda -4.5\lambda =(\mu -1)t$

$2.5\lambda =(\mu -1)t$

$t=\frac{2.5\lambda }{\mu -1}$

Answer: $\frac{2.5\lambda }{(\mu -1)}$