A human eye has a fixed distance deye = 2.40 cm between its lens and retina. The

Question

A human eye has a fixed distance deye = 2.40 cm between its lens and retina. The eye muscles adjust focus by changing the focal length feye of the lens.

a) Calculate the range of feye for an eye with good vision, able to focus from 25.0 cm away to infinity.

b) An imperfect eye can only focus on objects closer than 86.0 cm away. Is it nearsighted or farsighted? What is the power in diopters of a corrective lens placed 2.00 cm in front of the eye?

c) Consider an eye under water (see Sec 25-2), which causes its focal length to become about feye = 72.0 cm. What focal length corrective lens placed 2.00 cm in front of the eye will bring objects at infinity into focus? Hint: The image formed by the first lens becomes the object for the second lens.

Explanation / Answer

Part A)

Apply 1/f = 1/p + 1/q

For the near vision...

1/f = 1/25 + 1/2.4

f = 2.19 cm

For the far vision

1/f = 1/infinity + 1/2.4

f = 2.4 cm

Thus the range is 2.19 cm to 2.4 cm

Part B)

The eye is farsighted

We need to take objects at 25 cm and focus them at 86 cm in front of the eye.

That is 23 cm in front of the lens for the obejct and 84 cm in front of the lens for the VIRTUAL image

1/f = 1/23 + 1/-84

f = 31.7 cm

D = 1/f (in meters)

D = 1/.317

D = 3.16 Diopters

Part C)

For the eye 1/f = 1/p + 1/q

1/72 = 1/p + 1/2.4

p = -2.48

So the lens must focus an image 2.48 cm behind the eye (4.48 cm behind the lens)

1/f = 1/infinity + 1/4.48

f = 4.48 cm

D = 1/.0448

D = 22.3 Diopters

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