A student is not capable of focusing on objects closer than 0.48m from the eyes

Question

A student is not capable of focusing on objects closer than 0.48m from the eyes and no farther than 1.02m from them. To improve the vision, separate prescriptions are needed, one to correct for myopia, and the other to correct for hyperopia.

a) What is in meters the focal length of the lens needed to to enable the student to see objects that are 0.20m away from the eye? Express your answer in meters.

b) What is in diopters the power of that lens?

c) What is in meters the focal length of the lens needed to correct for the myopia? Express your answer in meters.

What if we want to design a bifocal for the student. The top portion will allow him to see far away objects. The lens then will be the lens you calculated in part c. To be able to see objects that are 0.20m away from the eye, we need to glue another lens at the bottom portion of this lens.

d) What power lens should we glue to the lens you have determined in part c, to enable the student to see objects that are 0.20m away from the eye?Express your answer in diopters.

Explanation / Answer

(a) From lens formula 1/f = 1/v - 1/u

f = focal length of lens,      v = shortest distance = 0.48 m,    u= object distance = 0.20 m

1/f = 1/0.48 - 1/0.20

f = - 0.342 m          ( since focal length is -ve so lens used is concave lens)

(b) power P = 100 / f ( c.m)

P = 100 / -34.2 c.m = -2.92 dioptre

(c) Let x = distance of far point of myopic eye = 1.02 m   - given

f = Focal length of concave lens,            for concave lens u = infinite

From lens formula , 1/f = 1/v - 1/u =   1/-x - 0                          ( v = -x,   u = infinite)

f = -x = -1.02 m

Hence focal length of concave lens put in front of myopic eye = distance of far point of eye.

* For a bifocal lens the upper part of the lens help to see far off object( as given) so it will be concave.its power is

P = 100/ f(c.m) = 100 / - 102 cm = - 0.98 D

For bottom part of lens to see nearby objects:

1/f = 1/v - 1/u                           ( v = -0.48 m, u = -0.20 m)

1/f = 1/-0.48 +1/0.20

f = 0.342m= 34.2 cm

P = 100/34.2 = +2.92 D   ( hence a convex lens of power 2.92D should be used at bottom part)

(d) From lens formula 1/f = 1/v -1/u                       ( v = - 1.02 m,    u = object distance = -0.20 m)

1/f = 1/-1.02 + 1/0.20

f = - 0.248 m = -24.8 cm

P = 100 / f = - 4.03 D                   ( lens used is concave)

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