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

Answer: The defect is hypermetropia.

Power = -2Dioptre

Explanation:

v = -25cm

u = -50cm

by using lens formula

1/f=1/v-1/u

1/f=-1/25-(-1/50)

1/f=-1/25+1/50

  f=-50cm

P=100/f(in cm)

 =100/-50

  -2Dioptre

In myopia we use u= infinity hence zero always.....

 

Answer 2

Answer:

The defect is hypermetropia.

The power of the lens is  $+2 D$.

Explanation:

When a person is not to see nearby objects clearly, the defect is called hypermetropia.

Sign Convention:

All distances which are measured along the direction of incidence of light are taken as positive while those which measured along the direction opposite to that of incidence of light are taken as negative.

Given:

The near point of the person is 25 cm, so let the object is placed at distance 25 cm from his eyes.

The person is unable to see distances less than 50 cm from his eyes, so image is formed at distance 50 cm from his eyes.

Therefore,

• Distance of object from the eyes, $u = -25 cm$.

• Distance of the image from the eyes, $v = -50 cm$.

Using the lens formula,

$\dfrac 1f = \dfrac 1v - \dfrac 1u\\= \dfrac {1}{-50} - \dfrac {1}{-25}\\=\dfrac{-1}{50}+\dfrac{1}{25}\\=\dfrac{-25+50}{50\times25}\\=\dfrac{25}{1250}\\=\dfrac{1}{50}\\\Rightarrow f = 50\ cm = 0.5\ m.$

It is the focal length of the eye of the person.

Therefore, the power of the lens is given by

$P = \dfrac 1f = \dfrac {1}{0.5} = +2 D$.