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. 60°
• B. 45°
• C. 15°
• D. 30°

Hint:

In diffraction, the first minimum occurs when the path difference equals the wavelength.

Answer 1

The Correct Option is A

{Understanding the diffraction formula} 
The condition for minima in diffraction is: \[ d \sin \theta = n \lambda \] For the first minimum: \[ \sin \theta = \frac{\lambda}{d} = \frac{2}{4} = \frac{1}{2} \] \[ \theta = 30^\circ \] Angular spread: \[ 2\theta = 60^\circ \] Thus, the correct answer is \( 60^\circ \). 
 

Answer 2

Step 1: Understanding the Concept
When a wave passes through a single slit, it undergoes diffraction. The angular position of the first minima on either side of the central maximum is given by: \[ a \sin \theta = \lambda \] where:
- \( a \) is the width of the slit,
- \( \lambda \) is the wavelength of the wave,
- \( \theta \) is the angle from the central axis to the first minimum.

Step 2: Substituting the Values
Given:
- Wavelength \( \lambda = 2.0 \, \text{cm} = 0.02 \, \text{m} \)
- Slit width \( a = 4.0 \, \text{cm} = 0.04 \, \text{m} \)
Substitute into the equation: \[ \sin \theta = \frac{\lambda}{a} = \frac{0.02}{0.04} = 0.5 \] \[ \theta = \sin^{-1}(0.5) = 30^\circ \]

Step 3: Calculating the Angular Spread
The angular spread of the central maximum includes the angle on both sides, so: 

\[ \text{Angular Spread} = 2\theta = 2 \times 30^\circ = 60^\circ \]

Step 4: Final Answer
\[ \boxed{\text{Option 1: 60°}} \]