In biprism experiment the maximum intensity is ' $\mathrm{I}_0$ '. If the path difference between the two interfering waves is ' $\lambda / 4$ ' then intensity at the point on the screen is

$$ \left[\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right] $$

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 $7^{\text {th }}$ bright fringe is obtained at the same point. If ' $\lambda$ ' is the wavelength of light used, the thickness of film is equal to