Question

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

Answer 1

Answer: refer the material.

Explanation:

Answer 2

Answer:

The Distance of the  near point is 33.33m and power of the lens is 1 Dioptre.

Explanation:

In this question,

We have been asked

Focal length of a lens suggested to a person with Hypermetropia is 100cm.

We know that,

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

Here v = -25cm

        f = -100cm

Putting the values,we get

$-\frac{1}{100}\ =\ -\frac{1}{25}\ -\ \frac{1}{u}$

$\frac{1}{u}\ =\ -\frac{1}{25}\ +\ \frac{1}{100}$

$\frac{1}{u}\ =\ \frac{-4\ +\ 1}{100}$

$\frac{1}{u}\ =\ \frac{-3}{100}$

$u\ =\ -33.33cm$

Therefore,distance of near point is 33.33cm

Power of the lens is given by = $\frac{1}{f}$

100cm = 1 metre.

Power = $\frac{1}{1}$

Power = 1 Dioptre