Question

1. What is the optical power of the eye when viewing an object from a distance of 23.5 cm?

                                         

2.What is the far point for a person whose eyes have a relaxed power of 50.25 D?

3. What power lens is required to correct a myopic (nearsighted) eye with a far point of 6.50 m? Neglect the distance between the corrective lens and the eye.

4. What power lens is required to correct a hyperopic (farsighted) eye with a near point of 1.00 m? Assume the distance between the corrective lens and the eye is 1.50 cm.

5.What is the overall magnification of a microscope with an eyepiece magnification of 10 and objective magnification of 35?

6.A person has a prescription for -4.50 D corrective lenses. What type of vision problem does this person have? What is his/her far point?

Answer 1

1) Using the lens-to-retina distance of 2.00 cm and the equation P = 1/do + 1/di

determine the power at an object distance of 23.5 cm:

P = 1/0.235 + 1/0.02 = 54.26 D

2) Around 2.0 m

3) The distance of far point = x = 6.5 m

To view distant object, concave lens of focal length -6.5 m should be used.

f = -x = -6.5 m

Power of lens = P = 1/f = 1/-6.5 = -0.154 D

5) overall magnification = 10*35 = 350

6) The person is myopic (nearsighted)

f = 1/P = -1/4.5 = -0.22 m

Far point = x = -f = 0.22 m = 22 cm