Question

A person can not view objects beyond 30 cm.

Find the power of the lens to rectify his defect.​

Answer 1

Answer:

The near point of a normal eye is equal to 25 centimeters.

According to the question the person is not able to see the objects clearly when they are placed at a distance less than 30 cm from the eye.

Hence, the person is not able to see the objects placed at the near point of his or her eye.

Thus,

the person is suffering from the defect of vision called as hypermetropia.

HYPERMETROPIA :-

In this type of defect of vision the person is not able to see the nearby objects clearly but is able to see far objects clearly. In this type of defect of vision the near point of a person goes far.

CAUSES OF HYPERMETROPIA :-

☸️ The eye lens becomes thin that is the focal length of eye increases.

☸️The eyeball contracts.

CORRECTION OF HYPERMETROPIA :-

HYPERMETROPIA can be corrected using a convex lens of appropriate focal length.

POWER OF LENS :-

$p = \frac{1}{f}$

$p = \frac{1}{0.3}$

$p = \frac{10}{3}$

p = +3.33

So,

The power of lens will be +3.33 diopter.

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