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:

The diffraction formula \( a \sin \theta = m \lambda \) is useful for calculating the width of a slit when the wavelength and diffraction angle are known.

Answer 1

The condition for the first minimum in a single-slit diffraction pattern is given by: \[ a \sin \theta = m\lambda \] For the first minimum, \( m = 1 \), so: \[ a \sin 30^\circ = (1)(600 \times 10^{-9} \text{ m}) \] Since \( \sin 30^\circ = 0.5 \), we get: \[ a \times 0.5 = 600 \times 10^{-9} \] Solving for \( a \): \[ a = \frac{600 \times 10^{-9}}{0.5} = 1.2 \times 10^{-6} \text{ m} \] Thus, the width of the slit is \( 1.2 \times 10^{-6} \) m.