# Homework Help: Optics problem (myopia, glasses)

1. Dec 9, 2012

### tomkoolen

A person with myopia is standing 50 cm in front of a mirror with glasses on (-1.75 dpt). Will he be able to see himself clearly?

1/f = 1/v + 1/b

I get that b = 26 cm. However my teacher says it should be 100. Is this correct?

2. Dec 9, 2012

### rude man

Trick question! Remember that his glasses correct to give a sharp image at infinity, whatever the strength of the glasses happen to be.

3. Dec 9, 2012

### haruspex

If b stands for the object distance, 100cm is clearly correct. If you disagree, by what argument? If b is something else, what is it?

4. Dec 9, 2012

### rude man

Come to think of it, neither the OP nor his teacher, nor I for that matter, answered the question, which is a bit vague to begin with.

Presumably, his spectacles enable him to see clearly at infinity. If he still enjoys power of accommodation, he can shorten the focal length of his eye lens to focus an object at 1m to the focal point on his retina. An older person would not necessarily be able to do that.

A myopic person has image distance (distance betw. his eye lens and retina) too short so he needs a diverging lens to focus infinity on his retina. Trying to use 1/f = 1/p + 1/q is tricky because the separation of his eye lens and his spectacle lens consitutes a "thick" lens with undefined image distance q, depending on how he wears his glasses.

5. Dec 10, 2012

### tomkoolen

Alright thank you very much! I now get:

-1.75 = 1/1 + 1/b (you guys call it q)
then b does not equal 1, however the teacher says it does. Could you explain this?

6. Dec 10, 2012

### rude man

This formula is meaningless. Your b is a negative number!

You are ignorig the eye lens. q is the effective image distance. The net effect of the eye lens plus the negative spectacle lens must still be a positive focal length, otherwise you don't get a real image on the retina.

As I've said before, the -1.75 diopter number has no significance by itself. It's just the correction the viewer needs to get a sharp image at infinity. Then, to see an object at a distance of p = 1m he needs to squeeze his eye lens down, thereby decreasing its focal length, so that the divergent rays from the virtual object in the mirror are again in sharp focus on his retina.

It can be shown that, if you stack two lenses of focal lengths f1 and f2 close together that the effective focal length f of the combined lens is given by 1/f = 1/f1 + 1/f2.
So if f1 = focal length of eye lens (about 1 cm?) and 1/f2 = -1.75 m-1, the right formula for focusing on a virtual object at p = 1m becomes 1/f = 1/p + 1/q where f and q ~ 0.01m, p = 1m, and q represents the (reduced) q of a myoptic eye. f is now the combination spectacle lens and his squeezed-down eye lens.

In any case, it should be obvious that p = 1m; it's just twice the distance from the eye to the mirror.