 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2017/paper-2/section-a/2/solution

For an object near the front focal plane, the objective's numerical aperture in air is

NA = sin θ ≃ \frac{D}{2 f} = \frac{0.005}{2 ( 0.020 )} = 0.125.

Δ x ≃ \frac{0.61 λ}{NA} = \frac{1.22 λ f}{D} = 2.93 \times 1 0^{- 6} m ≃ 2.9 μ m .

The factor

1.22

comes from the first zero of the Airy disk. Using the exact

sin (arctan (D / (2 f)))

changes the result by less than

1%

. A coarser

λ f / D

estimate gives

2.4 μ m

; the chosen criterion must be stated. This estimates the aperture limit, before aberrations and detector sampling.

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