First slide

In YDSE one of the slits is covered with a thin transparent sheet

Question

If a transparent medium of refractive index $μ = 1.5$ 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

Moderate

A

5 cm
B

2.5 cm
C

0.25 cm
D

0.1 cm

Solution

Shift in the fringe pattern $x = \frac{(μ - 1) t . D}{d}$
$= \frac{(1 .5 - 1) \times 2 .5 \times 10^{- 5} \times 100 \times 10^{- 2}}{0 .5 \times 10^{- 3}} = 2 .5 cm .$

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