1. The Human Eye; Corrective Lenses
Figure Diagram of a human eye.

2. The Human Eye; Corrective Lenses
The Human Eye Resembles a camera in its basic functioning, with an adjustable lens, the iris, &the retina.
Figure Diagram of a human eye.

3. The lens makes small adjustments to focus at different distances.
In the eye, most of the refraction is done at the surface of the cornea.
The lens makes small adjustments to focus at different distances.
Figure Accommodation by a normal eye: (a) lens relaxed, focused at infinity; (b) lens thickened, focused on a nearby object.

4. Some Vision Terminology
Near Point  The closest distance at which eye can focus clearly. For a normal eye, this is about 25 cm.
Far Point  The farthest distance at which object can be seen clearly. For a normal eye, this is close to infinity.
Nearsightedness  Myopia: The far point is too close for the eye to function properly.
Farsightedness  Hyperopia: The near point is too far away for the eye to function properly.

5. Nearsightedness Can be corrected with a diverging lens.
Figure 33-27a. Correcting eye defects with lenses: (a) a nearsighted eye, which cannot focus clearly on distant objects, can be corrected by use of a diverging lens

6. Farsightedness Can be corrected with a converging lens.
Figure 33-27b. (b) a farsighted eye, which cannot focus clearly on nearby objects, can be corrected by use of a converging lens.

7. Example: Farsighted eye. Sue is farsighted with a near point of 100 cm
Example: Farsighted eye. Sue is farsighted with a near point of 100 cm. What lens power should her reading glasses must have so that she can read a newspaper at a distance of 25 cm? Assume the lens is very close to the eye.
Solution: Using do = 25 cm and di = -100 cm gives f = 0.33 m. The lens power is +3.0 D.

8. Example Nearsighted eye.
A nearsighted person has near and far points of 12 cm and 17 cm, respectively (a) What lens power is needed for this person to see distant objects clearly? (b) What then will be the near point? Assume that the lens is 2.0 cm from the eye (typical for glasses).
Solution: The lens should put a distant object at the far point of the eye. Using do = ∞ and di = -15 cm (since the eye is 2 cm from the lens) gives f = m. The lens power is -6.7 D.

9. Vision is blurry under water because light rays are bent much less than they would be if entering the eye from air. This can be avoided by wearing goggles.
Figure (a) Under water, we see a blurry image because light rays are bent much less than in air. (b) If we wear goggles, we again have an air–cornea interface and can see clearly.

10. Magnifying Glass
A simple Magnifying Glass is a converging lens It allows us to focus on objects closer than the near point, so that they make a larger, and therefore clearer, image on the retina.
Figure 33-33a. Leaf viewed (a) through a magnifying glass. The eye is focused at its near point

11. If the eye is focused at the Near Point this can be written:
See figures above! The Power of a magnifying glass is defined by its Angular Magnification:   
In the figures, N is the Near Point distance & f the Focal Length If the eye is relaxed this becomes (assuming small angles!):
If the eye is focused at the Near Point this can be written:

12. Example : A jeweler’s “loupe.”
An 8-cm focal-length converging lens is used as a “jeweler’s loupe,” which is a magnifying glass Estimate: (a) The magnification when the eye is relaxed, (b) The magnification if the eye is focused at its near point N = 25 cm.
Solution: a. With the relaxed eye focused at infinity, M = N/f, which is about a factor of 3.
b. With the eye focused at its near point, M = 1 + N/f, or about a factor of 4.