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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

630 mm

B

500 mm

C

420 mm

D

750 mm

Text Solution

Verified by Experts

The correct Answer is:
C

`lamda_(1)=700nm,n_(1)=3n_(2)=5,lamda=?`
For same path diff.
`n_(1)lamda_(1)=n_(2)lamda_(2)`
`lamda_(2)=(n_(1)lamda_(1))/(n_(2))=(3xx700)/(5)=420nm`
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