Question

In Young's double slit experiment, a glass plate of refractive index $1.5$ and thickness $5\times {10}^{-4}\text{cm}$ is kept in the path of one of the rays, then

• A. There will be no shift in the interference pattern.
• B. Fringe width will increase.
• C. Fringe width will decrease.
• D. Optical path of the ray will increase by $2.5\times {10}^{-4}\text{cm}$

Answer 1

The correct option is D Optical path of the ray will increase by $2.5\times {10}^{-4}\text{cm}$

Due to the presence of the glass plate of thickness $t=5\times {10}^{-4}\text{cm}$, the optical path of ray will become
$x_2=\mu t$

$\Rightarrow x_2=1.5\times 5\times {10}^{-4}=7.5\times {10}^{-4}\text{cm}$

Therefore, increase in optical path,

$\Delta x=x_2-x_1=\mu t-t$

$=7.5\times {10}^{-4}-5\times {10}^{-4}=2.5\times {10}^{-4}\text{cm}$

The whole fringe pattern shifts by $\frac{D}{d}(\mu -1)t$ but the fringe width remains the same.

Hence, option $\left(\text{D}\right)$ is correct.