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

In a single-slit diffraction experiment, the angular position of the first minimum is given by the equation: \[ a \sin \theta = m \lambda \] where: - \( a \) is the width of the slit,
- \( \theta \) is the angle of the first minimum,
- \( \lambda \) is the wavelength of the light,
- \( m \) is the order of the minimum (for the first minimum, \( m = 1 \)).
Given: - \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \),
- \( \theta = 30^\circ \),
- \( m = 1 \).
Substituting the given values into the equation: \[ a \sin(30^\circ) = (1) \times (600 \times 10^{-9}) \] \[ a \times \frac{1}{2} = 600 \times 10^{-9} \] \[ a = 1200 \times 10^{-9} \, \text{m} = 1.2 \, \text{mm} \] Thus, the width of the slit is \( 1.2 \, \text{mm} \).