Question

A Fresnel biprism of angle $2^{\circ }$ is illuminated by light of wavelength $6280\mathring{A}$ from a source which is $0.10m$ away from it. What will be the width of the fringes (in mm) formed on a screen kept $0.9m$ away from the biprism?
(Take Refractive index of glass $=1.5,\pi =3.14$ )

Answer 1

Answer: 0.18

The deviation $\delta$ produced by each half of biprism $=(\mu -1)A$ where, $A$ is in radian.
Here,
$d=2(\mu -1)Ax$
$=2(1.5-1)\left(\frac{\pi }{180}\times 2^{\circ }\right)\times 0.1$
$d=\frac{\pi }{900}m$
$D=0.9+0.1=1m$
Now, $\beta =\frac{\lambda D}{d}$
$\beta =\frac{\left(6280\times 10^{-10}\right)\times 1}{\frac{\pi }{900}}$
$=1.8\times 10^{-4}m=0.18mm$