Question

An old person is able to see an object nearest to 1 m.What should be the power of lens required so that he can see an object placed at nearest distance of distinct vision?

• A. 3D
• B. 4D
• C. 1/2D
• D. 1/4D

Answer 1

The correct option is A 3D

Hypermetropia is a condition of the eye where the person is not able to see things clearly when nearer to the eye.

The normal near point of the eye is =$25cm$

For corrective lens to be used ,

Focal length,= $f$

For corrective lens object distance =The normal near point of the eye = $u=-25cm$ and image distance = defected near point of man = $v=-100cm$

Lens formula:

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

$\frac{1}{f}=\frac{1}{-100}-\frac{1}{-25}=\frac{1}{25}-\frac{1}{100}=\frac{3}{100}$

$f=33.33cm=0.33m$

Power, $P=\frac{1}{f}=\frac{1}{0.333}=3D$

Hence, Power of lens is $3D$