Question

Interference Fringes are observed with a biprism of refracting angle 2° and refractive index 1.5 on a screen 100m away from it. If the distance between the source and the biprism is 20m and the frienge width is 0.10 mm what is the wavelength of light?

Answer 1

This question is based on Young's double slit experiment.

first of all find deviation , $\delta=(\mu-1)A$
$\delta=(1.5-1)2^{\circ}=1^{\circ}=\frac{\pi}{180}rad$
we know, $d=2asin\delta\approx2a\delta$
where a is distance between source and biprism. e.g., a = 20m
so, d = 2 × π/180 × 20 = 2π/9 m

now, use formula of fringe width, $\beta=\frac{\lambda D}{d}$

here, $\beta=0.1mm=0.1\times10^{-3}$
D = 100m , d = 2π/9 m
so, 0.1 × 10^-3 = λ × 100/(2π/9)
=> 10^-4 × 2π/9 × 10^-2 = λ
=> λ = 2π/9 × 10^-6 m
=> λ = 0.697 × 10^-6 = 697 × 10^-9 m
=> λ = 697 nm