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In an interference experiment, third bri...

In an interference experiment, 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

500 m

B

630 nm

C

750 nm

D

420 nm

Text Solution

Verified by Experts

The correct Answer is:
D
`n_(1) lambda_(1)=n_(2)lambda_(2)`
`implies 3 xx 700=5 xx lambda_(2)`
`implies lambda_(2)=420 nm`
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