Question

A Young's double slit experiment is being performed by illuminating the two slits with a monochromatic source of light.
When a glass plate of thickness $24\mu \text{m}$ is placed in front of one of the slits, the entire pattern shifts through a distance of $5\text{mm}$. What should be the thickness of the glass plate made of the same material so that the pattern shifts by $15\text{mm}$?

• A. $8\mu \text{m}$
• B. $48\mu \text{m}$
• C. $72\mu \text{m}$
• D. $12\mu \text{m}$

Answer 1

The correct option is C $72\mu \text{m}$

When a glass plate is placed in front of one of the slits, the shift $\left(S\right)$ in the interference pattern is given by,

$S=\frac{D}{d}(\mu -1)t$

As all other parameters are constant,
$\Rightarrow \frac{S_1}{S_2}=\frac{t_1}{t_2}$

$\frac{5\times {10}^{-3}}{15\times {10}^{-3}}=\frac{24\times {10}^{-6}}{t_2}$

$\therefore t_2=72\times {10}^{-6}\text{m}=72\mu m$

Hence, $\left(D\right)$ is the correct answer.