Question

A man with defective eyes cannot see clearly objects at a distance of more than $60cm$ from his eyes. The power of the lens to be used will be

• A. $-1.66D$
• B. $-60D$
• C. $+60D$
• D. $\frac{+1}{1.66}D$

Answer 1

The correct option is A

$-1.66D$

Step 1- Given data

Distance at which eye can clearly see the object is $d=60cm$

Step 2- Formulae and concept

Since the person can not see the distance more than $60cm$ , means that the person is suffering from myopia or near sightedness. So distance of $60cm$ will be considered as a new far point.

The relation between image distance $v$ , object distance $u$ and the focal length $f$ of the lens is given by $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$.

The power is calculated as reciprocal of focal length and it is expressed as $P=\frac{1}{f}$

Step 3- Calculations

The far point of a healthy eye is at infinity and it is considered object distance i.e., $u=-\infty$

The new far point of a myopic eye is taken as image distance i.e., $v=-60cm$

The focal length is calculated as $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{-60cm}-\frac{1}{-\infty },orf=-60cm$

The power is calculated as $P=\frac{1}{f}=\frac{100}{-60cm}=-1.66D$

Thus, the power of the lens to be used will be $-1.66D$

Hence, option (A) is correct.