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

The correct option is B

$2.5cm$

Transparent medium: A medium through which light can pass easily.

Refractive index: It is the measure that determines how the light will bend or refract when moving from one medium to another.

Step 1: Given

Refractive index, $\mu =1.5$

Thickness, $t=2.5\times {10}^{-5}m$

The distance between the slit and the screen is, $D=100cm=100\times {10}^{-2}m$

The distance between the slits is, $d=0.5mm=0.5\times {10}^{-3}m$

Step 2: Determine the shift in the interference pattern

The shift in the interference pattern is given by,

$$\begin{gathered} x=\frac{(\mu -1)tD}{d} \\ =\frac{(1.5-1)\times 2.5\times {10}^{-5}\times 100\times {10}^{-2}}{0.5\times {10}^{-3}} \\ =2.5cm \end{gathered}$$

Hence, option (B) is the correct option.