Q. In Young’s double slit experiment, the aperture screen distance is $2m$. The fringe width is $1\ mm$. Light of $600\ nm$ is used. If a thin plate of glass $(\mu = 1.5)$ of thickness $0.06\ mm$ is placed over one of the slits, then there will be a lateral displacement of the fringes by

A) $0\ cm$
B) $5\ cm$
C) $10\ cm$
D) $15\ cm$
Explanation

(b) Lateral displacement of fringes $= \frac{\beta}{\lambda}(\mu - 1)t$
$= \frac{1 \times 10^{- 3}}{600 \times 10^{- 9}}(1.5 - 1) \times 0.06 \times 10^{- 3} = 5\ cm$

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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