The smallest object separation a near-sighted person can resolve with her naked eye is 4.1 * 10^-5 m.Assume her eye has a near point of 20 cm and a far point of 2.0 m,and she is using blue-green light ($\lambda$ = 520 nm).The index of refraction of the aqueous humor is 1.30.What is the diameter of her pupil?

Question

The smallest object separation a near-sighted person can resolve with her naked eye is 4.1 * 10^-5 m.Assume her eye has a near point of 20 cm and a far point of 2.0 m,and she is using blue-green light ($\lambda$ = 520 nm).The index of refraction of the aqueous humor is 1.30.What is the diameter of her pupil?

Quizplus · Accepted Answer

The diameter of the pupil can be calculated using the Rayleigh criterion formula for the minimum resolvable angle, $\theta = 1.22 \frac{\lambda}{D}$, where $\lambda$ is the wavelength of light and $D$ is the diameter of the aperture (in this case, the pupil). Given the smallest object separation (resolution) and the distance to the object, we can find the angle $\theta$ and then solve for $D$. The calculation involves understanding that the resolution given is the smallest detail that can be resolved by the eye, and using the given parameters, including the wavelength of blue-green light (520 nm = $520 \times 10^{-9}$ m) and the index of refraction of the aqueous humor, we can find the diameter of the pupil required to achieve this resolution.

The Smallest Object Separation a Near-Sighted Person Can Resolve with Her