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Resolving Power of the Eye

  1. Apr 23, 2017 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    θrad=1.22λ/D

    tanθ=Ym/L


    3. 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
     
  2. jcsd
  3. Apr 23, 2017 #2

    gneill

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    Staff: Mentor

    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)?
     
  4. Apr 23, 2017 #3
    It causes the light to bend in the eye. I forgot to mention that in my post, sorry! λfilmvac/n which turns out to be 412nm or 0.412μm
     
  5. Apr 23, 2017 #4

    gneill

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    Staff: Mentor

    So it alters the wavelength. What value of wavelength did you use when you calculated the value for θrad?
     
  6. Apr 23, 2017 #5
    I used λfilm but I wrote 0.42μm instead of 0.412μm which was messing up the calculations.Thank you.
     
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