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In the Young's double slit experiment, t...

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

A

(a) `1.0cm`

B

(b) `1.5cm`

C

(c) `0.5cm`

D

(d) `2.0cm`

Text Solution

Verified by Experts

The correct Answer is:
C
`beta=(lambdaD)/(d)=(5000xx10^(-10)xx1)/(0.1xx10^-3)m=5xx10^-3m=0.5cm`
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