In a biprism experiment, fifth dark fringe is obtained at a point. A thin transparent film of refractive index '$\mu$' is placed in one of the interfering paths. Now $7^{th}$ bright fringe is obtained at the same point. If '$\lambda$' is the wavelength of light used, the thickness of film is equal to

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

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

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:
$\Delta x_2 - \Delta x_1 = (\mu - 1)t$

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

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

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

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