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The path difference at a point on the sc...

The path difference at a point on the screen in Young's experiment is `5 lamda`. If the distance of that point from the central bright hand is 0.5 mm, then the band width is

A

`2.5 mm`

B

`1 mm`

C

`0.1` mm

D

10 mm

Text Solution

Verified by Experts

The correct Answer is:
C
Path diff. `5 lambda, x _(n)=0.5 mm beta = ?`
Path diff. `=(x_(n)d)/(D)=5lambda`
`:.(x_(n))/(5)=(D)/(d)lambda`
`(0.5)/(5)= beta`
`:. beta=0.1 mm`
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