In Young's double slit experiment, the distance between the two slits is `0.1 mm` and the wavelength of light used is `4 xx 10^(-7) m`. If the width of the fringe on the screen is 4 mm , the distance between screen and slit is

Options

A. `0.1` mm
B. `1` cm
C. `0.1` cm
D. `1` m

Answer

D

Explanation

Fringe width, `beta=(lambdaD)/(d), D=(betad)/(lambda)`
`D=(4xx10^(-3)xx0.1xx10^(-3))/(4xx10^(-7))=1 m`