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In young's double slit experiment if the...

In young's double slit experiment if the seperation between coherent sources is halved and the distance of the screen from coherent sources is doubled, then the fringe width becomes:

A

Half

B

Four times

C

One-fourth

D

Double

Text Solution

Verified by Experts

The correct Answer is:
B
In young double slit experiment
Fringe width `beta = (lambda D)/(d)`
Now, `d.. = (d)/(2)` and `D. = 2D`
So, `beta. = (lambda(2D))/(d//2) = (4 lambda D)/(d)`
`beta. = 4beta`
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