The photosensitive cells (rods and cones) in the retina are most densely packed in the fovea- the part of the retina used to see straight ahead. In the fovea, the cells are all cones spaced about \(1 \mu \mathrm{m}\) apart. Would our vision have much better resolution if they were closer together? To answer this question, assume two light sources are just far enough apart to be resolvable according to Rayleigh's criterion. Assume an average pupil diameter of \(5 \mathrm{mm}\) and an eye diameter of \(25 \mathrm{mm}\). Also assume that the index of refraction of the vitreous fluid in the eye is \(1 ;\) in other words, treat the pupil as a circular aperture with air on both sides. What is the spacing of the cones if the centers of the diffraction maxima fall on two nonadjacent cones with a single intervening cone? (There must be an intervening dark cone in order to resolve the two sources; if two adjacent cones are stimulated, the brain assumes a single source.)

Step by step solution

01

Write down the given information

We are given the following information: - The average pupil diameter is \(5 \mathrm{mm}\) - The eye diameter is \(25 \mathrm{mm}\) - The index of refraction of the vitreous fluid in the eye is \(1\) - The cones are spaced about \(1 \mu \mathrm{m}\) apart

02

Apply Rayleigh's criterion to find angular separation

According to Rayleigh's criterion, the minimum angular separation between two light sources that can be resolved is: \(\theta = 1.22 \frac{\lambda}{D}\) where: - \(\theta\) is the angular separation - \(\lambda\) is the wavelength of light - \(D\) is the diameter of the aperture (the pupil diameter)

03

Assume the wavelength of light

To calculate the angular separation, we need to assume a wavelength of light. We will assume the wavelength of yellow light, which is approximately \(550 \mathrm{nm}\).

04

Calculate the angular separation

Using the formula for Rayleigh's criterion and the given information: \(\theta = 1.22 \frac{550 \times 10^{-9} \mathrm{m}}{5 \times 10^{-3} \mathrm{m}}\) \(\theta = 1.34 \times 10^{-4} \mathrm{rad}\)

05

Calculate the linear separation on the retina

To find the linear separation on the retina, we can use the formula: \(L = \theta \times D\) where \(D = 25 \mathrm{mm}\) is the eye diameter, and \(L\) is the linear separation on the retina. \(L = 1.34 \times 10^{-4} \mathrm{rad} \times 25 \times 10^{-3} \mathrm{m}\) \(L = 3.35 \times 10^{-6} \mathrm{m}\)

06

Calculate the number of intervening cones

We are asked to find the spacing of the cones if the centers of the diffraction maxima fall on two nonadjacent cones with a single intervening cone. So, we need to find if the linear separation \(L\) is enough for this condition. Since the cones are spaced about \(1 \mu \mathrm{m}\) apart, for two nonadjacent cones with a single intervening cone, we would need a separation of \(2 \times 1 \mu \mathrm{m} = 2 \times 10^{-6} \mathrm{m}\).

07

Compare the calculated linear separation with the required separation

We found that the linear separation on the retina for two resolvable light sources is \(3.35 \times 10^{-6} \mathrm{m}\). This is greater than the required separation of \(2 \times 10^{-6} \mathrm{m}\) for two nonadjacent cones with a single intervening cone. This means that our vision would not have much better resolution if the cones were closer together since the current spacing is sufficient according to Rayleigh's criterion.

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