Question

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 $0.5mm$ and that between slits and screen is $100cm$

• A. $5cm$
• B. $2.5cm$
• C. $0.25cm$
• D. $0.1cm$

Answer 1

Answer: (B)

$2.5cm$

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