In Young's double slit experiment, the fringe width of the interference pattern produced on the screen is $2.4 \mu \mathrm{~m}$. If the experiment is carried out in another medium having refractive index 1.2 , the fringe width will be $\_\_\_\_$ $\mu \mathrm{m}$.
Select the correct option:
A
1.2
B
2
C
2.4
D
2.88
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Since Fringe width in a medium becomes
$$
\begin{aligned}
& \beta^{\prime}=\frac{\beta}{\mu} \quad\left(\text { Here } \beta=\frac{\lambda D}{d}\right) \\
& \beta^{\prime}=\frac{\beta}{\mu}=\frac{2.4}{1.2}=2 \mu \mathrm{~m}
\end{aligned}
$$
Hello 👋 Welcome to Competishun – India’s most trusted platform for JEE & NEET preparation. Need help with JEE / NEET courses, fees, batches, test series or free study material? Chat with us now 👇
