Question

The focal length of a lens suggested to a person with Hypermetropia is $100cm$. Find the distance of near point and power of the lens.

Answer 1

Given: The focal length of a lens suggested to a person with Hypermetropia is 100cm.

To find the distance of near point and power of the lens.

Solution:

Let the distance of near point be ’$d$’ and focal length be '$f=100cm$' then

$f=\frac{25d}{d-25}cm⟹100(d-25)=25d⟹100d-25d=2500⟹d=\frac{2500}{75}⟹d=\frac{100}{3}=33.33cm$

Hence the distance of near point will be 33.33cm

We know,

Power, $P=\frac{1}{f}D$, $f$ is in metres

$⟹P=\frac{1}{100\times {10}^{-2}}⟹P=1D$

is the power of the lens.