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. 3 cm
B. 0.011 cm
C. 2 cm
D. 4 cm

Answer

B

Explanation

`a=0.3m,b=0.7m`
`therefore D=0.3++0.7=1m,mu=1.5`,
`lamda=6000xx10^(-10)m=6xx10^(-7)m`
angle of the prism `(alpha)=1^(0)=1xx(pi)/(180)=(3.14)/(180)`
and `X=(lamdaD)/(d)`
But in terms of `alpha and mu,d=2(a(mu-1)alpha`
`therefore X=(lamdaD)/(2a(mu-1)alpha)`
`=(6xx10^(-7)xx1)/(2xx0.3xx(1.5-1)xx(3.14)/(180))`
`=(180xx10^(-6))/(1.57)`
`=1.14xx10^(-4)m=1.14xx10^(-2)cm`
`=0.0114cm=0.011cm`