Focal length of her contact lenses

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SUMMARY

To determine the focal length of Anne's contact lenses, we start with her current vision limitation of 2.4 m and the distance from her eye to the retina, which is 1.8 cm. Using the lens formula \(1/p + 1/q = 1/f\), we need to calculate the focal length required for her to see objects as close as 24.0 cm. The focal length of the contact lens must compensate for the biological lens of her eye, effectively summing the focal lengths of both lenses due to their close proximity.

PREREQUISITES
  • Understanding of the lens formula \(1/p + 1/q = 1/f\)
  • Knowledge of focal lengths and their significance in optics
  • Basic principles of ray tracing in optics
  • Familiarity with the anatomy of the human eye, particularly the retina
NEXT STEPS
  • Calculate the focal length of the biological lens in the eye for distant vision
  • Determine the required focal length for viewing objects at 24.0 cm
  • Explore ray tracing techniques for optical systems
  • Study the differences between contact lenses and eyeglasses in terms of focal length calculations
USEFUL FOR

Students studying optics, optometrists, and anyone interested in the physics of vision correction through lenses.

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Homework Statement



Anne is farsighted; the nearest object she can see clearly without corrective lenses is 2.4 m away. It is 1.8 cm from the lens of her eye to the retina.

What should the focal length of her contact lenses be so that she can see clearly objects as close as 24.0 cm from her eye?

Homework Equations



1/p+1/q=1/f. I keep getting 1.67 and this is wrong and i am stuck

The Attempt at a Solution

 
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What units do they want the answer in?
 
You'll need to find a few things:
First, find the focal length of the biological lens in the viewer's eye when he/she is working at his/her limit of vision.
Then find the required focal length the viewer should have to view the closer object.

Then consider: how do these two differ?
The contact lens is placed right up next to the lens in the viewers eye, so you can consider the net focal length required to be the sum of the contact lens and the viewer's lens in his/her eye. It's much easier than the case of case of eyeglasses... which have a separation between the two lenses (the eyeglass lens and the eye lens itself).

There's another way to approach the problem also.
First, where would the viewers eye place the image of the object that is too close?
Second, make THAT image the object for a second lens (at the position of the eye)... and you know where you WANT to place the second image of this... right at the back of the eye.
Drawing out scaled ray-tracing for this technique might help.
This second approach is, in fact, the only way to approach the eyeglasses problem, in which case you should then know where the glasses will sit in relation to the person's eye.

Both ways should work out to the same result. DO as Kurdt says, watch your units (and be consistent in them!). It's also easy to goof up all the math, since you are working with reciprocals of the distances and focal lengths.
 

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