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:

For diffraction, the angle for minima is determined by the equation \( a \sin \theta = m \lambda \), where \( a \) is the slit width and \( m \) is the order of the minimum.

Answer 1

The equation for the first diffraction minimum is \( a \sin \theta = m \lambda \), with \( m = 1 \). Given \( \lambda = 600 \, \text{nm} = 6 \times 10^{-7} \, \text{m} \) and \( \theta = 30^\circ \), we substitute these values: \( a \sin 30^\circ = 1 \cdot 6 \times 10^{-7} \). Since \( \sin 30^\circ = \frac{1}{2} \), the equation becomes \( a \cdot \frac{1}{2} = 6 \times 10^{-7} \). Solving for \( a \) yields \( a = 1.2 \times 10^{-6} \, \text{m} \). Therefore, the slit width is \( 1.2 \times 10^{-6} \, \text{m} \).