In Fresnel's biprism `(mu=1.5)` experiment the distance between source and biprism is `0.3m` and that between biprism and screen is `0.7m` and angle of prism is `1^@`. The fringe width with light of wavelength `6000Å` will be

Options

A. (a) `3cm`
B. (b) `0.011cm`
C. (c) `2cm`
D. (d) `4cm`

Answer

B

Explanation

`beta=((a+b)lambda)/(2a(mu-1)alpha)`
where a=distance between source and biprism `=0.3m`
b=distance between biprism and screen `=0.7m`.
a=Angle of prism`=1^@`, `mu=1.5`, `lambda=6000xx10^(-10)m`
Hence, `beta=((0.3+0.7)xx6xx10^-7)/(2xx0.3(1.5-1)xx(1^@xxpi/180))`
`=1.14xx10^-4m=0.0114cm`.