The nearpoint of an eye is 151 cm. A corrective lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. Part 1) What should be the focal length of this lens?
Answer in units of cm.
Part 2) What is the power of the needed corrective lens?Answer in units of diopters.

a) 0.3 m

b) 3.3 diopters

Explanation:

Given that

Object distance is 26cm from the eye.

Image distance is 105 cm

From the above, we will use the formula

1/f = 1/v + 1/u

where

f is the focal length of the lens,

v is the image distance,

u is the object distance.

Since the image is a virtual image, we will give our v a negative sign. So, on calculating, we have

1/f = 1/25 - 1/151

1/f = 0.0333

f = 1/0.0333

f = 30.03 cm

f = 0.3 m

b) The power of the needed corrective lens is the reciprocal of the focal length in metres;

1/0.3 = 3.3 diopters

This is usually in diopters

continuous conducting path for current, power supply