What Is the Smallest Spot Diameter the Human Eye Can Detect?

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Homework Help Overview

The problem involves determining the smallest diameter spot that the human eye can detect, given specific parameters such as pupil diameter, light wavelength, and the index of refraction of the fluid in the eye. The context is rooted in optics and human vision.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations related to the resolving angle and the impact of the index of refraction on light behavior within the eye. There are questions about the effects of using different wavelengths in calculations.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the index of refraction and its effect on wavelength. There is acknowledgment of potential errors in calculations, but no consensus has been reached on the correct approach or final answer.

Contextual Notes

Participants are navigating through the implications of the index of refraction and its effect on wavelength, which may influence the calculations. There is also a note about the specific values used in the calculations, highlighting a potential source of error.

aDabOfRanch
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Homework Statement


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 2.59 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.

Homework Equations


θrad=1.22λ/D

tanθ=Ym/L

The Attempt at a Solution


Found the resolving angle using θrad=1.22λ/D where D is the diameter of the pupil, and the angle turned out to be 1.98E-4 radians. Then I used the angle to find the distance from the main axis to the first minima by plugging in θ into tanθ=Ym/L solving for Y1 I got 5.03
Using the hint provided I doubled 5.03 to get 10.06μm however the answer is 9.85μm
 
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Hi aDabOfRanch,

Welcome to Physics Forums!

What do you think might be the effect of the index of refraction of the vitreous humor (fluid filling the eye)?
 
gneill said:
Hi aDabOfRanch,

Welcome to Physics Forums!

What do you think might be the effect of the index of refraction of the vitreous humor (fluid filling the eye)?
It causes the light to bend in the eye. I forgot to mention that in my post, sorry! λfilm=λvac/n which turns out to be 412nm or 0.412μm
 
aDabOfRanch said:
It causes the light to bend in the eye. I forgot to mention that in my post, sorry! λfilm=λvac/n which turns out to be 412nm or 0.412μm
So it alters the wavelength. What value of wavelength did you use when you calculated the value for θrad?
 
gneill said:
So it alters the wavelength. What value of wavelength did you use when you calculated the value for θrad?

I used λfilm but I wrote 0.42μm instead of 0.412μm which was messing up the calculations.Thank you.
 

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