JEE Main 2025 Apr 2 Shift 1, Physics Q23: Diffraction

For single-slit diffraction, the $n^{\text{th}}$ minimum is at angle ${\theta }_n$ with $a\sin {\theta }_n=n\lambda$.

For small angles, $\sin {\theta }_n\approx {\theta }_n\approx \frac{n\lambda }{a}$.

The angular separation between the $2^{\text{nd}}$ minimum on one side and the $3^{\text{rd}}$ minimum on the other side is the sum of their angles:

$$\theta ={\theta }_2+{\theta }_3=\frac{2\lambda }{a}+\frac{3\lambda }{a}=\frac{5\lambda }{a}$$

Set $\theta =30°=\frac{\pi }{6}\text{rad}$ and $\lambda =628\text{nm}$:

$$\frac{\pi }{6}=\frac{5\times 628\times 10^{-9}}{a}$$

$$a=\frac{6\times 5\times 628\times 10^{-9}}{\pi }=\frac{6\times 5\times 628\times 10^{-9}}{3.14159}$$

Using $\pi \approx 3.14$ and $628/\pi \approx 200$:

$$a\approx 6\times 5\times 200\times 10^{-9}=6000\text{nm}=6\mu \text{m}$$

$\therefore$ Slit width $=6\mu \text{m}$.