In Young's double slit experiment, the fringe width is ` 1 xx 10^(-4) m` if the distance between the slit and screen is doubled and the distance between the two slit is reduced to half and wavelength is changed from `6.4 xx 10^(7) m` to `4.0 xx 10^(-7)m` , the value of new fringe width will be

Options

A. `0.10` mm
B. `0.15` mm
C. `0.20` mm
D. `0.25` mm

Answer

D

Explanation

Distance between the centres of adjacent fringes = Fringe width
Fringe width, `beta=(lambdaD)/(d)` …(i)
where `lambda` is the wavelength of light used, D is the distance of the screen from the slits, d is the distance between the slits.
`:.beta.=(lambda.D.)/(d,)`
Divide (ii) by (i), we get
`(beta.)/(beta)=((lambda.)/lambda)((D.)/(D))((d)/(d.))`
Substituting the given values, we get
`(beta.)/(beta)=((4.0xx10^(-7))/(6.4xx10^(-7)))((2D)/(D))((d)/((d//2)))=2.5`
`beta.=2.5beta=2.5xx0.10mm=0.25mm`