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Question: In an interference experiment, third bright fringe is obtained on the screen with a light of 700 nm....

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

A

420 nm

B

500 nm

C

750 nm

D

630 nm

Answer

420 nm

Explanation

Solution

3Dλd\frac{\mathbf{3D\lambda}}{\mathbf{d}}= 5Dλ′d\frac{5D\lambda'}{d}

3l = 5l'

\ l' = 3λ5\frac{3\lambda}{5} = 3×7005\frac{3 \times 700}{5}nm = 420 nm