In the Young's double slit experiment, the spacing between two slits is `0.1mm`. If the screen is kept at a distance of `1.0m` from the slits and the wavelength of ligth is `5000Å`, then the fringe width is

Options

A. 1.0 cm
B. 1.5 cm
C. 0.5 cm
D. 2.0 cm

Answer

C

Explanation

d= 0.1 mm, D= 1m, `lambda= 5000Å`
`beta= (lambdaD)/(d)= (5000 xx 10^(-10) xx 1)/(0.1 xx 10^(-3))= 5 xx 10^(-3) m= 0.5 cm`