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 :

Solution

Physics sir · 3 steps

Find where the first dark fringe sits, then double it: the central maximum spans both sides.

1. First minimum. In single slit diffraction the first dark fringe on each side is at a sin θ = λ. The central maximum lies between these two minima.

   sin θ = λ/a = (2.0 cm)/(4.0 cm) = 0.5

2. No small-angle shortcut. Here λ/a is large, so sin θ ≈ θ does not hold. Use the exact value: sin θ = 0.5 gives θ = 30°. The 1.5 m screen distance does not affect an angle.
3. So. The central maximum runs from -30° to +30°, so its angular spread is 2θ = 60°, option (a).

The idea to keep. Angular width of the central maximum is 2θ, with sin θ = λ/a. When λ is comparable to a, use the exact sine, not the small-angle approximation.

Answer from the NTA final answer key. Solution by PrepWiser.