Blurred Circle Problem

[JonathanJT]

[drk]

Well, the 6M target distance just means that light enters the eye parallel, so no initial vergence.

I don't remember how to do this...there's an "entrance pupil" and stuff and I don't remember.

If I were to guess, it's a simple trigonometric proportion. There has to be a "cone" of light of a certain diameter that leaves the pupil (I imagine you'd ignore crystallline lens power) that focuses to a point in the inside of the eye, and a new cone begins again. The size of the retinal blur circle is, of course, the cross section of that cone.


Obviously (and as we all know) the larger the pupil "cone" the larger the blur circle "cone".
And we know that the position of the focal point has a great bearing on the size of the blur circle on the retina.
And we know that the position of the focal point is the "dioptric power" of the eye. I assume they give you an average emmetropic power, and that's what you'd use.
But you'd need to know axial length, which they're going to say is longer than the average emmetropic length.
Are there any "default" assumptions that the course uses?

That's all I got.