Q. 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 5$^{th}$ bright fringe at the same point?

A) 500 nm
B) 630 nm
C) 750 nm
D) 420 nm
Explanation

(d) $n_{1}\ \lambda_{1} = n_{2}\ \lambda_{2}$
$\Rightarrow 3 \times 700 = 5 \times \lambda_{2}$
$\Rightarrow \ \lambda_{2} = 420\ nm$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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