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

The Correct Option is C

To determine the angular spread of the central maxima in a single-slit diffraction pattern, we use the formula for the angular width of the central maximum:

\(\theta = 2 \times \sin^{-1} \left( \frac{\lambda}{a} \right)\)

where:

• \(\lambda\) is the wavelength of the incident wave.
• \(a\) is the width of the slit.

Given:

• Wavelength, \(\lambda = 2.0\) cm = 0.02\).
• Width of the slit, \(a = 4.0\) cm = 0.04\).

Substitute these values into the formula:

\(\theta = 2 \times \sin^{-1} \left( \frac{0.02}{0.04} \right)\)

\(\theta = 2 \times \sin^{-1} (0.5)\)

Since \(\sin^{-1} (0.5) = 30\degree\),

\(\theta = 2 \times 30\degree = 60\degree\)

Thus, the angular spread of the central maxima is \(60\degree\).

Therefore, the correct answer is \(60\degree\).

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 \( 60^\circ \).