COMEDK 2025 Afternoon Solved Paper

Concept:Single-slit diffraction: the first minimum occurs at $a \sin \theta = \lambda$; the central maximum spans from $-\theta$ to $+\theta$.Explanation:Given: frequency $f = 5 \times 10^9$ Hz, slit width $a = 3$ cm.Speed of light $c = 3 \times 10^8$ m/s.Wavelength $\displaystyle \lambda = \frac{c}{f} = \frac{3 \times 10^8}{5 \times 10^9} = 0.06$ m $= 6$ cm.For the first minimum: $a \sin \theta = \lambda$, so $\sin \theta = \frac{\lambda}{a} = \frac{6}{3} = 2$.Since $\sin \theta$ cannot exceed 1, no real minima exist within $\pm 90^\circ$.Thus, the small-angle approximation $\sin \theta \approx \theta$ (in radians) is used.Therefore, $\theta \approx \frac{\lambda}{a} = 2$ rad.The total angular width of the central maximum (from $-\theta$ to $+\theta$) is $2\theta = 2 \times 2 = 4$ rad.Answer:Option C (4 rad).