Q. If a transparent medium of refractive index $\mu = 1.5$ and thickness $t = 2.5 \times 10^{- 5}m$ is inserted in front of one of the slits of Young’s Double Slit experiment, how much will be the shift in the interference pattern? The distance between the slits is $0.5\ mm$ and that between slits and screen is $100\ cm$
A) $5\ cm$
B) $2.5\ cm$
C) $0.25\ cm$
D) $0.1\ cm$
Explanation
(b) Shift in the fringe pattern $x = \frac{(\mu - 1)t.D}{d}$
$= \frac{(1.5 - 1) \times 2.5 \times 10^{- 5} \times 100 \times 10^{- 2}}{0.5 \times 10^{- 3}} = 2.5\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.