A microwave of wavelength 2.0 cm falls normally on a slit of width 4.0 cm. The angular spread of the central maxima of the diffraction pattern obtained on a screen 1.5 m away from the slit, will be:

A microwave of wavelength $$\lambda = 2.0$$ cm falls normally on a slit of width $$a = 4.0$$ cm, and we need the angular spread of the central maximum.

In single‐slit diffraction the first minimum occurs where $$a\sin\theta = \lambda$$, so that $$\sin\theta = \frac{\lambda}{a} = \frac{2.0}{4.0} = 0.5$$ and hence $$\theta = 30°$$.

The central maximum extends from $$-\theta$$ to $$+\theta$$ on either side, giving a total angular spread of $$2\theta = 2 \times 30° = 60°$$.

The correct answer is Option C) 60 degrees.