The resolution of the eye is ultimately limited by the pupil diameter. What is the smallest diameter spot the eye can produce on the retina if the pupil diameter is 3.93 mm? Assume light with a wavelength of λ = 550 nm. (Note: The distance from the pupil to the retina is 25.4 mm. In addition, the space between the pupil and the retina is filled with a fluid whose index of refraction is n = 1.336.) Hint: The size of the spot is twice the distance from the main axis to the first minimum.

D = 3,246 10⁻³ mm

Explanation:

We can treat this exercise as a diffraction problem with a circular opening. The expression that describes diffraction is

d sin θ = m λ

In our case d is the width of the pupil 3.93 mm and we will use m = 1

sin θ = λ / d

Since these angles are very small, you can approximate the sin θ=θ

θ = lam / d

This expression is exact for a linear slit, for the case of circular slits must be multiplied by a constant

θ = 1.22 λ / d

Now let's analyze the wavelength, as it is within a medium with refractive index its value changes

λ ₙ = λ ₀ / n

Let's replace

θ = 1.22 λ / (n d)

Let's calculate

θ = 1.22 550 10⁻⁹ / (1,336 3.93 10⁻³)

θ = 1,278 10⁻⁴ rad

This is the smallest angle that can be in the retina, let's use trigonometry to find the distance (y), we must use half the angle for the right triangle

Let's not forget that all angles are in radians

tan θ/2 = y / L

y = L tan θ / 2

y = 25.4 tan (1.278 10⁻⁴/2) = 25.4 0.639 10⁻⁴

y = 1.623 10⁻³ mm

The spot size twice this value

D = 2y

D = 3,246 10⁻³ mm