JEE Main 2023 — Wave Optics Question with Solution

In Young's double-slit experiment, the distance between the slits and the screen, $L=1\text{ m}$, the wavelength of the light, $\lambda =600\text{ nm}=600\times 10^{-9}\text{ m}$, and the position of the 5th bright fringe from the central maximum, $y=5\text{ cm}=0.05\text{ m}$.

The distance between the central maximum and the $n$th bright fringe is given by the formula:

$$y_n=\frac{n\lambda L}{d}$$

where $d$ is the distance between the two slits.

We can rearrange this formula to solve for $d$:

$$d=\frac{n\lambda L}{y_n}$$

Substituting the values given in the question, we get:

$$d=\frac{5\times 600\times 10^{-9}\times 1\times 100}{5}$$

$$d=6\times 10^{-5}\text{ m}$$

$$d=60\mu \text{m}$$

Therefore, the separation between the slits is $60\mu \text{m}$.