A small aperture is illuminated with a parallel beam of light of wavelength λ= 628 nm. The emergent beam has an angular divergence of 20. The size of the aperture is

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$18 μ m$

$36 μ m$

$9 μ m$

$27 μ m$

answer is D.

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Detailed Solution

$2\theta = {2^0} \Rightarrow \theta = {1^0} = \frac{\pi }{{180}}rad\,\,;a\sin\theta = \lambda \$

$as{\text{ }}\theta {\text{ }}is{\text{ }}small \Rightarrow a\theta = \lambda \$

$\large a = \frac{\lambda }{\theta } = \frac{{628\times{{10}^{ - 9}}}}{{3.14}} \times 180 = 36\,\mu m\$

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