What is the far point of a person whose eyes have a relaxed power of 51.1 d?

Explanation:

The power of the eye is given as

P = 1/di + 1/do

P is power of the eye

di is distance between eye lens and image

do is distance between eye lens and image

Given that,

P = 51.1 d

Note: the power unit "d" is called dioptre and it means per meter

If the person is seeing clearly, the distance between the eye lense and the image (di) must be equal to the distance between the eye lenses and the retina which is 2cm

Thus, di = 2cm = 0.02m

P = 1/di + 1/do

1/do = P — 1/di

1/do = 51.1 — 1/0.02

1/do = 51.1—50

1/do = 1.1m

taking reciprocal

do = 1/1.1

do = 0.9091m

Then, do = 90.91cm

The far point the person can see is 90.91cm

it has to be b, about 97 percent.