What is the angular resolution of the eye at a wavelength of $500 \mathrm{~nm}$ for a pupil diameter of $2.8 \mathrm{~mm}$ ?
A) $3.3 \times 10^{-5} \mathrm{rad}$
B) $2.2 \times 10^{-4} \mathrm{rad}$
C) $4.5 \times 10^{-5} \mathrm{rad}$
D) $8.9 \times 10^{-5} \mathrm{rad}$
E) $1.8 \times 10^{-4} \mathrm{rad}$

Question

Quizplus · Accepted Answer

The angular resolution $\theta$ of an optical system can be estimated using the formula $\theta \approx \frac{\lambda}{D}$, where $\lambda$ is the wavelength of light and $D$ is the diameter of the aperture (in this case, the pupil diameter). Substituting $\lambda = 500 \times 10^{-9} \, \mathrm{m}$ (or $500 \, \mathrm{nm}$) and $D = 2.8 \times 10^{-3} \, \mathrm{m}$ (or $2.8 \, \mathrm{mm}$) into the formula gives $\theta \approx \frac{500 \times 10^{-9}}{2.8 \times 10^{-3}} = 1.7857 \times 10^{-4} \, \mathrm{rad}$, which is closest to the option $B) 2.2 \times 10^{-4} \mathrm{rad}$.

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