Power and focal length of human eye

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SUMMARY

The discussion focuses on calculating the power and focal length of the human eye lens, specifically for an eyeball length of 4cm. When focusing on an object at an infinite distance, the focal length equals the image distance, which is 4cm, resulting in a power of +25 diopters. For an object at the near point of 25cm, the thin lens formula (1/f = 1/d_ob + 1/d_im) is applied, yielding a focal length of 25cm and a power of +4 diopters. A long-sighted person with a near point of 2m requires corrective lenses with a power of -2.5 diopters to adjust their near point back to 25cm.

PREREQUISITES
  • Understanding of the thin lens formula (1/f = 1/d_ob + 1/d_im)
  • Knowledge of diopters and lens power calculations
  • Familiarity with the anatomy of the human eye, specifically the lens and retina
  • Concept of virtual images in optics
NEXT STEPS
  • Research the effects of lens curvature on focal length and power
  • Learn about the differences between contact lenses and spectacles
  • Explore the anatomy of the human eye in detail, focusing on vision correction
  • Investigate the physics of ray optics and image formation
USEFUL FOR

Students in optics, optometrists, and anyone interested in understanding vision correction and the physics of the human eye.

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I am extremely stuck on this question and do not know how to go about it. Could somebody help me?

1. (a) If the human eyeball is 4cm from lens to retina, what is the range of the power and focal length of the eye lens when focussing an object an infinite distance away and one at the normal near point 25cm away?

(b) A long-sighted person(whose eyeball length is still 4cm)has a near point of 2m. What is the power of the eye lens when focussing at the near point?
What power glasses lens would the person need to correct their near point back to 25cm?

Any help will greatly be appreciated!
 
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if the object distance is infinite, then the
focal length *is* the image distance.
(parallel rays converge to focus at focal length)
Did you somehow not realize that the image
is formed on the back of the eyeball, in focus?
For the 25[cm] object, use the focus equation:
("thin-lens") 1/f = 1/d_ob + 1/d_im .

Thin lenses held right in front of one another
have a "total power" that's *approximately*
the two powers multiplied together
(ray deflection angles add).

But another way to correct one's vision is to
have a lens that produces a virtual image
(of the object at 25[cm]) where the person sees
(that is, d_im = - 200[cm] from the eye).
This way shows that contact lenses are different
than spectacles (with d_im = - 199[cm])
 

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