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In Young's double slit experiement when ...

In Young's double slit experiement when wavelength used is `6000Å` and the screen is `40cm` from the slits, the fringes are `0.012cm` wide. What is the distance between the slits?

A

`0.24 cm`

B

`0.2 cm`

C

`2.4 cm`

D

`0.024 cm`

Text Solution

Verified by Experts

The correct Answer is:
B
Fringe width `beta=(Dlambda)/d`
`d=(Dlambda)/beta=(6000xx10^(-10)xx(40xx10^(-2)))/(0.012xx10^(-2))=0.2cm`
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