Q. In an interference experiment, third bright fringes are 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) 630 nm
B) 500 nm
C) 420 nm
D) 750 nm
Explanation

(c) Here, $X_{3} = X_{5}$
$\frac{3D\lambda}{2d} = \frac{5D\lambda^{'}}{2d}$
$\Rightarrow 3\lambda = 5\lambda^{'}$ or $\frac{\lambda}{\lambda^{'}} = \frac{3}{5}$
$\lambda^{'} = \frac{3}{5} \times 700\ nm = 420\ nm$

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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