Question

In biprism experiment the fringes are observed in the focal plane of the eyepiece at a distance of 1.2 m from slit. The distance between the central bright band and the ${20}^{th}$ bright band is 0.4 cm. When convex lens is interposed between the biprism and eyepiece, at a distance of 90 cm from the eyepiece, the distance between two magnified virtual images is found to be 0.9 cm. Find the wavelength of light used.

Answer 1

Data : D = 1.2 cm
20X = 0.4 cm
or X = $\frac{0.4}{20}$
= 2 x ${10}^{-2}$ cm = 2 x ${10}^{-4}$ cm
v = 90 cm = 0.9 cm
u = 1.2 - 0.9 = 0.3 cm
$d_1$ = 0.9 cm = 0.9 x ${10}^{-2}$ m
$\lambda$ = ?