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:

Options

A. Double
B. Half
C. Four times
D. One-fourth

Answer

C

Explanation

Fringe width `beta= (lambdaD)/(d)`
Now `d.=(d)/(2)` and `D. = 2D`
So, `beta.= (lambda(2D))/(d//2)= (4lambdaD)/(d)`
`beta.= 4beta`