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Human Eye problem

  • Thread starter Violagirl
  • Start date
  • #1
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Homework Statement



A) Under optimum conditions, the smallest black dot that can be seen subtends an angle of 2.3 x 10-6 rad. If a dot is viewed at a distant of 0.25 m, the near point of a normal adult, what is the smallest diameter it can have and still be seen? B) The maximum resolution is obtained when the image falls on the fovea centralis. At 10° away from this region, the acuity is 10 times poorer. What is the minimum size spot that can be seen at that angle under these conditions?

Homework Equations



Pf = 1/xf + 1/D

Pf = person's far point, 1/xf = distance a person sees an object, D = image distance

Pn = 1/xn + 1/D

Pn person's near point, 1/xn - distance a person sees an object, D = image distance

A = Pn - Pf

A = power of accommodation

sin θ = 1.22 (λ/d)-power of acuity equation

λ = wavelength, d = diameter


The Attempt at a Solution



I've puzzled over this problem. I'm not sure what the relevant equation would be unless I'm missing one. Any help is appreciated.
 

Answers and Replies

  • #2
6,054
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You are given a distance to the dot. The dot is essentially a small circle perpendicular to that distance. Its diameter and the distance to the eye form a right triangle. Your are also given an angle in this triangle.
 
  • #3
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I guess I feel like I'm missing something. I'm not given a wavelength of which this object is being viewed. I otherwise say I could algebraically solve for d that way from the acuity equation. Otherwise I'm not really sure what else to consider...
 
  • #4
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Have you drawn a diagram as per #2?
 
  • #5
114
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Attached is the diagram I drew out for it. Am I thinking about it right? I honestly have no idea how to set it up as we did not cover this topic much in class...
 

Attachments

  • #6
6,054
390
See my go at it.
 

Attachments

  • #7
114
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Oh! Ok so I took .25 m * sin (2.3 * 10^-7 ) and got d = 5.75 * 10^-7 m. Thank you!
 

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