Question

HELLO !

A person with a defective eye vision is unable to se the objects nearer than 1.5m .He wants to read a book at a distance of 30 .Find the focal length and the power of the lens he needs in his spectacles.

Answer 1

This person suffers from the defect of hypermetropia.

For him u = -30cm, v = -1.5 m

= -150cm

Therefore, focal length of corrective lens to be used by him is

1/f = 1/v- 1/u = 1/-150 - 1/-30 = 4/150 = 37.5cm

The positive sign shows that the lens needed is a convex lens of focal length 37.5 cm.

Hence, power of lens neededP =1/f = 100/37.5 = 2.67D.

Mark it brainliest ♥

Answer 2

Answer:

Explanation:

The condition of then person is $\boxed{Hypermetropic}$

Here ,

u = -30cn

v = -1.5m = -150cm

Focal length of lens of the spectacles in order to be corrected is :-

1/f = 1/v-1/u

=> 1/f = 1/-150 -1/-30

=> f = 150/4

=>f = 37.5

As focal length is 37.5cm it shows he needs a CONVEX LENS of focal length 37.5

Therefore ,The power of the lens is 100/37.5 = 2.67D