Question

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. $30\degree$
• B. $15\degree$
• C. $60\degree$
• D. $45\degree$

Answer 1

Answer: (C)

$60\degree$

Answer 2

Given: - Wavelength of the microwave: $\lambda = 2.0 \, \text{cm} = 0.02 \, \text{m}$ - Width of the slit: $a = 4.0 \, \text{cm} = 0.04 \, \text{m}$ - Distance from the slit to the screen is irrelevant for angular spread calculation.

Step 1: Calculating the Angular Spread of the Central Maximum

The angular position $\theta$ of the first minimum in a single-slit diffraction pattern is given by: $a \sin \theta = m \lambda$ where: - $m = 1$ for the first minimum. Substituting the given values: $0.04 \sin \theta = 1 \times 0.02$ $\sin \theta = \frac{0.02}{0.04}$ $\sin \theta = 0.5$ Thus: $\theta = \sin^{-1}(0.5) = 30^\circ$

Step 2: Angular Spread of the Central Maximum

The angular spread of the central maximum is given by $2\theta$: $\text{Angular spread} = 2 \times 30^\circ = 60^\circ$

Conclusion:

The angular spread of the central maxima of the diffraction pattern is $60^\circ$.