JEE Main 2026 Apr 2 Shift 1, Physics Q21: Diffraction

The angular width of the central maximum in single-slit diffraction is ${\theta }_{cm}=\frac{2\lambda }{d}$.

${\theta }_{cm}=\frac{2\times 628\times 10^{-9}}{0.2\times 10^{-3}}=6.28\times 10^{-3}$ rad

Converting to degrees by multiplying by $180/\pi$:

${\theta }_{cm}=6.28\times 10^{-3}\times \frac{180}{\pi }\approx 0.36^{\circ }=36\times 10^{-2}$ degrees

$\therefore \alpha =36$