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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 $100 \,cm$

Wave Optics

Solution:

Shift in the fringe pattern $x=\frac{(\mu-1) t . D}{d}$
$=\frac{(1.5-1) \times 2.5 \times 10^{-5} \times 100 \times 10^{-2}}{0.5 \times 10^{-3}}=2.5 \,cm$