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In an interference exeperiment, third br...

In an interference exeperiment, third bright fringe is obtained at a point on the screen with a light of 700 nm. What should be the wavelength of the light source in order to obtain 5th bright fringe at the same point?

A

630nm

B

500nm

C

420nm

D

750nm

Text Solution

Verified by Experts

The correct Answer is:
C

`n_(1)lamda_(1)=n_(2)lamda_(2)`
`3xx700=5xxlamda_(2)`
`implieslamda_(2)=(3xx700)/(5)=420` nm.
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