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Physics Question on Wave optics

A parallel beam of monochromatic light of wavelength 600 nm600 \, \text{nm} passes through single slit of 0.4 mm0.4 \, \text{mm} width. Angular divergence corresponding to second order minima would be ______ ×10−3 rad \times 10^{-3} \, \text{rad}.

Answer

The angular position of the nn-th order minima is given by:
sin⁡θ=nλb.\sin \theta = \frac{n \lambda}{b}.
For the second-order minima (n=2n = 2):
sin⁡θ=2⋅600×10−90.4×10−3.\sin \theta = \frac{2 \cdot 600 \times 10^{-9}}{0.4 \times 10^{-3}}.
sin⁡θ=3×10−3.\sin \theta = 3 \times 10^{-3}.
The total divergence is:
Total divergence=2⋅3×10−3=6×10−3 rad.\text{Total divergence} = 2 \cdot 3 \times 10^{-3} = 6 \times 10^{-3} \, \text{rad}.
Final Answer:
6×10−3 rad6 \times 10^{-3} \, \text{rad}.

Explanation

Solution

The angular position of the nn-th order minima is given by:
sin⁡θ=nλb.\sin \theta = \frac{n \lambda}{b}.
For the second-order minima (n=2n = 2):
sin⁡θ=2⋅600×10−90.4×10−3.\sin \theta = \frac{2 \cdot 600 \times 10^{-9}}{0.4 \times 10^{-3}}.
sin⁡θ=3×10−3.\sin \theta = 3 \times 10^{-3}.
The total divergence is:
Total divergence=2⋅3×10−3=6×10−3 rad.\text{Total divergence} = 2 \cdot 3 \times 10^{-3} = 6 \times 10^{-3} \, \text{rad}.
Final Answer:
6×10−3 rad6 \times 10^{-3} \, \text{rad}.