Optics problem (myopia, glasses)

Click For Summary

Homework Help Overview

The discussion revolves around an optics problem involving a person with myopia standing in front of a mirror while wearing corrective glasses. The focus is on whether the individual can see themselves clearly, considering the specifications of the glasses and the distances involved.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between object distance, image distance, and the effects of corrective lenses on vision. Questions arise regarding the correctness of calculated distances and the implications of myopia on vision clarity.

Discussion Status

Participants are actively questioning the assumptions made about object and image distances, with some suggesting that the original poster's calculations may not align with standard interpretations. There is a recognition of the complexity involved in combining the effects of the corrective lenses and the eye's natural lens.

Contextual Notes

There is a mention of the original poster's teacher providing a different value for the object distance, leading to confusion. The discussion also touches on the implications of the glasses correcting vision to infinity and the role of the eye's accommodation in focusing on near objects.

tomkoolen
Messages
39
Reaction score
1
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?
 
Physics news on Phys.org
tomkoolen said:
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?

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

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.
 
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?
 
tomkoolen said:
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?

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.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K