Question

The far point of the eye decreases to 5 m and near point increases to 1m. Then, Find the power of the lens.

Answer 1

Far point of the eye is

Here the far point of the eye is reduced to 5m

We want the eye to see distant object clearly. Since far point is 5m, the image of distant object should be formed at 5m.

Image distance, v=-5m;

Object distance, u= -

1/f in m^{-1} is power

Power of lens used to correct the eye whose far point is 5m is P1

Near point of an eye is 25cm = 0.25m

It has increased to 1m

Thus to see an object at 25cm properly image should form at 1m.

u=-0.25m

v=-1m

So focal length f in meter should be

Power of lens used to correct the eye whose near point is 1m is P2

Thus power of lens to correct the eye whose far point is 5m and near point is 1m should be P

P= P1+P2