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

The angular spread of the central maxima in a single-slit diffraction pattern is calculated using the formula for the angular width of the central maximum:

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

where:

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

Given values:

• Wavelength, \(\lambda = 2.0\) cm = 0.02\) m.
• Slit width, \(a = 4.0\) cm = 0.04\) m.

Substituting these values into the formula yields:

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

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

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

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

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

The result is \(60\degree\).