1. ## Blurred Circle Problem

Hello,

This is my first thread on here.

My classmate and I are stumped by a blurred circle question in our exercise pack (2nd year students in 3 year Dispensing Optics course).

The questions wants the blurred circle size but only gives three pieces of information.

The patient's eye is an axially myopic.
A point object is viewed at 6 metres distance.
And the pupil diameter is 6mm.

We've inferred the axial myopia means the axial length of the eye is greater than +22.22mm.
And figure we need to fill out the rest of the formula -
Blurred Image size = Pupil Size x (Dioptric Power ÷ Refractive Error).

We just don't know what steps we need to take from there or even how the 6m distance helps at all in order to get the Dioptric Power or Refractive Error to fill in the rest of the formula.

Any Dispensers who see this, your help would mean a lot to us.

2. 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.

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