Question

A parallel beam of monochromatic light of wavelength \(600 \, \text{nm}\) passes through single slit of \(0.4 \, \text{mm}\) width. Angular divergence corresponding to second order minima would be ______ \( \times 10^{-3} \, \text{rad}\).

Answer 1

Correct Answer: 6

The angular position of the $n$-th order minima is given by:
\[\sin \theta = \frac{n \lambda}{b}.\]
For the second-order minima ($n = 2$):
\[\sin \theta = \frac{2 \cdot 600 \times 10^{-9}}{0.4 \times 10^{-3}}.\]
\[\sin \theta = 3 \times 10^{-3}.\]
The total divergence is:
\[\text{Total divergence} = 2 \cdot 3 \times 10^{-3} = 6 \times 10^{-3} \, \text{rad}.\]
Final Answer:
$6 \times 10^{-3} \, \text{rad}$.