Resolution

What is the minimum angular spread of a 633-nm-wavelength He-Ne laser beam that is originally 1.00 mm in diameter?

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Strategy
The diameter of the beam is the same as if it were coming through an aperture of that size, so D = 1.00 mm. You are given $λ$ , and you must solve for $θ$ .
Solution
$θ = \frac{(1.22) λ}{D} = \frac{(1.22) (633 nm)}{1.00 \times 10^{6} nm} = 7.72 \times 10^{−4} rad = 0.0442°$
17.7

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