Solving Human Eye Problem Homework w/o Quotes

[Violagirl]

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

P_f = 1/x_f + 1/D

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

P_n = 1/x_n + 1/D

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

A = P_n - P_f

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.

 

[voko]

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.

 

[Violagirl]

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...

 

[voko]

Have you drawn a diagram as per #2?

 

[Violagirl]

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...

 

[voko]

See my go at it.

 

[Violagirl]

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