Power and focal length of human eye

[123Sub-Zero]

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!

 

[lightgrav]

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])

 

[quicknote]

I can provide an explanation for the concepts of power and focal length of the human eye. The power of the eye refers to its ability to bend light, which is necessary for clear vision. This is controlled by the eye's lens, which can adjust its shape to focus on objects at different distances. The focal length of the eye is the distance between the lens and the retina, where the light is focused to form an image.

(a) In the scenario given, the range of power and focal length of the eye lens can be calculated using the thin lens equation: 1/f = 1/u + 1/v, where f is the focal length, u is the object distance, and v is the image distance. When focusing on an object at an infinite distance, the object distance (u) is essentially 0, and the image distance (v) is equal to the focal length (f). Therefore, the power of the eye lens in this case would be 0 diopters (D) and the focal length would be equal to the distance between the lens and retina, which is 4cm.

When focusing on an object at the normal near point of 25cm, the object distance (u) is 25cm and the image distance (v) is also 25cm. Using the same equation, we can calculate the power of the eye lens to be 4D and the focal length to be 2cm.

(b) For a long-sighted person with an eyeball length of 4cm and a near point of 2m, the power of the eye lens when focusing at the near point can be calculated as 1/0.02 - 1/2 = 0.49D. This means that the eye lens is not strong enough to focus on objects at a close distance like 2m.

To correct this, the person would need glasses with a lens power of -2.04D, which is the difference between the power of their eye lens (0.49D) and the desired power for clear vision at 25cm (2.53D).

I hope this explanation helps in understanding the concept of power and focal length of the human eye. If you have any further questions, please feel free to ask.