# The resolution of the eye is ultimately limited by the pupil diameter. What is the smallest...

## Question:

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 3.51 mm? Assume light with a wavelength of {eq}\lambda = 550 {/eq} 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.

## The angle of resolution:

In the question, we have to calculate the diameter of the spot, so we need to use the formula of the angle of resolution in which wavelength directly depends on the resolution, but the diameter is inverse as more the size of a spot less is the angle of resolution.

## Answer and Explanation: 1

Solution:

Given Data:

• Diameter of pupil, {eq}D = 3.51\;{\rm{mm}} {/eq}
• wavelength in vacuum, {eq}{\lambda _{vacuum}} = 550\;{\rm{nm}} {/eq}
• The distance from pupil to retina, {eq}L = 25.4\;{\rm{mm}} {/eq}

Calculation:

We know the formula for resolution is,

{eq}\sin {\theta _{\min }} = 1.22\dfrac{\lambda }{D} {/eq}

Where,

{eq}\lambda = {/eq} wavelength in vacuum

{eq}D = {/eq} Diameter of pupil

Substitute the values in above formulas,

Here, we use the wavelength for the vaccum thus,

{eq}\lambda = \dfrac{{{\lambda _{vacuum}}}}{n} {/eq}

Where,

{eq}n = {/eq} redfractive index

Substitute the values in above equation;

{eq}\begin{align*} \sin {\theta _{\min }} &= 1.22\dfrac{{550 \times {{10}^{ - 9}}}}{{3.51 \times {{10}^{ - 3}} \times 1.336}}\\ {\theta _{\min }} &= {\sin ^{ - 1}}\left( {1.91 \times {{10}^{ - 4}}} \right)\\ {\theta _{\min }} &= 8.19 \times {10^{ - 3}} \end{align*} {/eq}

The smallest diameter of spot is,

{eq}\begin{align*} 2y &= 2L\tan {\theta _{\min }}\\ 2y &= 2 \times 25.4 \times {10^{ - 3}} \times \tan \left( {8.19 \times {{10}^{ - 3}}} \right)\\ 2y &= 7.26\;{\rm{\mu m}} \end{align*} {/eq}

As given in question the size of spot is twice the distance from the main axis to the first minimum thus,{eq}2y = 7.26\;{\rm{\mu m}} {/eq}