Question

A biprism is placed 5 cm from the slit illuminated by sodium light of wavelength
5890 A. The width of fringes obtained on a screen 75 cm from the biprism is 9.424X
10 cm. What is the distance between the two coherent sources?​

Answer 1

Given that,

Distance of bi-prism from the slit = 5 cm

Wavelength $\lambda=5890\ \AA$

Width of fringes $\beta=9.424\times10^{-2}\ cm$

Distance of screen = 75 m

We need to calculate the total distance

$D=5+75=80\ cm$

We need to calculate the distance between the two coherent sources

Using formula of width of fringes

$\beta=\dfrac{D\lambda}{d}$

$d=\dfrac{5890\times10^{-10}\times80}{9.424\times10^{-2}}$

$d = 0.0005\ m$

$d=0.05\ cm$

Hence, The distance between the two coherent sources is 0.05 cm.