Optics Question 1138

Question: 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

[KCET 2003]

Options:

A) 500 nm

B) 630 nm

C) 750 nm

D) 420 nm

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Answer:

Correct Answer: D

Solution:

$ n _{1}{\lambda _{1}}=n _{2}{\lambda _{2}}\Rightarrow 3\times 700=5\times {\lambda _{2}}\Rightarrow {\lambda _{2}}=420\ nm $



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