Question

In a diffraction experiment, the slit is illuminated by light of wavelength 600 nm. The first minimum of the pattern falls at \( \theta = 30^\circ \). Calculate the width of the slit.

Hint:

In single-slit diffraction, the angular position of the first minimum is given by \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the angle, and \( m \) is the order of the minimum.

Answer 1

1. Diffraction Condition for Minima:

The condition for the first minimum in a single-slit diffraction pattern is described by:

\[ a \sin \theta = m \lambda \]

Where:

• \( a \) represents the slit width.
• \( \theta \) is the diffraction angle for a minimum (with \( m = 1 \) for the first minimum).
• \( \lambda \) denotes the wavelength of the light.
• \( m \) indicates the order of the minimum (set to \( 1 \) for the first minimum).

2. Provided Data:

• Wavelength \( \lambda \): \( 600 \, \text{nm} \) which is \( 600 \times 10^{-9} \, \text{m} \).
• Angle of first minimum \( \theta \): \( 30^\circ \).
• Order of the minimum: \( m = 1 \) for the first minimum.

3. Value Substitution:

Substituting the given data into the formula for the first minimum yields:

\[ a \sin 30^\circ = 1 \times 600 \times 10^{-9} \]

Knowing that \( \sin 30^\circ = \frac{1}{2} \), the equation simplifies to:

\[ a \times \frac{1}{2} = 600 \times 10^{-9} \]

Solving for \( a \):

\[ a = \frac{600 \times 10^{-9}}{\frac{1}{2}} = 1.2 \times 10^{-6} \, \text{m} = 1.2 \, \mu\text{m} \]

4. Result:

• The determined width of the slit is \( a = 1.2 \, \mu\text{m} \).