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In single-slit diffraction with $\lambda = 500$ nm and slit width $a = 0.1$ mm, the angular position of the first minimum is approximately:
A$0.5$ rad, ignoring the slit width entirely here
B$0$ rad, since the central maximum has no edge value
C$0.05$ rad, by simple ratio without the lambda value
D$5\times 10^{-3}$ rad, since $\sin\theta \approx \lambda/a$
Answer & Solution
Correct answer: D. $5\times 10^{-3}$ rad, since $\sin\theta \approx \lambda/a$
$\sin\theta \approx \lambda/a = 500\times 10^{-9}/10^{-4} = 5\times 10^{-3}$ rad.
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