Question

A person unable to see object placed within 50cm from eyes.what is the defect?Find power of lens if near point is 25cm.

Answer 1

v= 25cm

u= -50cm

by lens formula

1/v - 1/ u = 1/f

1/25-1/50 = 1/f

1/50 = 1/f

100/50= 100/f = P

= P = 2 Dioptres

the defect is hypermetropia

Answer 2

Answer:

This defect is myopia and the power of the lens is -2 D.

Explanation:

Given that,

Distance of the object u = -50 cm

Distance of the image v = -25 cm

This defect is myopia because the person can not see the far object but close object can see clearly.

Using lens's formula

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

$\dfrac{1}{f}=\dfrac{1}{-25}-\dfrac{1}{50}$

$\dfrac{1}{f}=-\dfrac{1}{50}$

$f = -50$

The power of the lens is

$P =\dfrac{100}{-50}$

$P=-2\ D$

Hence, This defect is myopia and the power of the lens is -2 D.