# Optics and Contact Lenses

## Homework Statement

Without her contact lenses, a student can focus an object located 0.80 m to infinity from her eyes. The power lens, measured in diopters, is the reciprocal of the focal length, measured in meters. Assuming that the image distance in the eye is 0.02 m, what lens power does she require for reading a text located 0.25 m from her eyes.

## Homework Equations

(1/f)=(1/di)+(1/do)
D=(1/f), where f is expressed in meters

## The Attempt at a Solution

Well, the student's near point is 0.80 m, and the object is at 0.25 m from her eyes.
Thus, (1/f)=(1/-0.8)+(1/0.25) and D=2.75
But how does the image distance in the eye come into play? I don't really know what it means by image distance in the eye. Can someone help?
Thanks.

Related Introductory Physics Homework Help News on Phys.org
This student can focus properly on an object at 0.80 m from her eyes.
This means that she is able to focus the image of this object on her retina which is about 2 cm behind the "crystalline lens". This information can help you calculate the focal length of this "crystalline lens" and then proceed with the question.

The "crystalline lens" has a variable focal length. Muscles on the eye can change the shape of the lens and thereby modify its focal length. Here, when reading at 80cm, the "crystalline lens" was at is shorter focal length.

sorry but i still don't really get it
how does the 0.02 m affect the equation?

1/x + 1/x' = 1/f

with x=0.02m and x'=0.8m

so the focal length (f) of the crystaline lens is appromatively 0.02 m

But you are right, maybe there is no need to know this focal length.

With additional contact lenses (f') you get for a closer distance (x"):

1/x + 1/x" = 1/f + 1/f' = 1/x + 1/x' + 1/f'

and therefore

1/f' = 1/x" - 1/x'

this does not depend on x indeed