The smallest object separation a near-sighted person can resolve with her naked eye is 3.3 * 10^-5 m.Assume the diameter of her pupil is 3.0 mm,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).What is the index of refraction of the aqueous humor?

Question

The smallest object separation a near-sighted person can resolve with her naked eye is 3.3 * 10^-5 m.Assume the diameter of her pupil is 3.0 mm,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).What is the index of refraction of the aqueous humor?

Quizplus · Accepted Answer

The index of refraction of the aqueous humor can be calculated using the formula:n = s / (d * tan(theta))where s is the smallest object separation, d is the diameter of the pupil, and theta is the angle of resolution.Plugging in the given values, we get:n = (3.3 * 10^-5 m) / (3.0 mm * tan(theta))We can also use the small angle approximation to simplify the equation:tan(theta) = opposite / adjacent = s / (2 * near point)Plugging in the given values, we get:tan(theta) = (3.3 * 10^-5 m) / (2 * 0.20 m) = 8.25 * 10^-5Substituting this into the first equation, we get:n = (3.3 * 10^-5 m) / (3.0 mm * 8.25 * 10^-5) = 1.30Therefore, the index of refraction of the aqueous humor is 1.30.

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