Light of wavelength 500 nm is incident on a single slit of width 0.2 mm. The angular position of the first diffraction minimum is approximately:
A$2.5 \times 10^{-3}$ rad
B$2.5 \times 10^{-4}$ rad
C$5.0 \times 10^{-4}$ rad
D$2.5 \times 10^{-6}$ rad
Answer & Solution
Correct answer: A. $2.5 \times 10^{-3}$ rad
$\sin\theta \approx \theta = \lambda/a = (500\times10^{-9})/(0.2\times10^{-3}) = 2.5 \times 10^{-3}$ rad.
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