Question

A certain myopic person has a far point of $150\text{cm}.$ What power a corrective lens will allow him to see distant objects clearly? If he is able to read a book at $25\text{cm},$ while wearing the glasses, find his near point.

• A. $-0.67\text{D},21.43\text{cm}$
• B. $-0.50\text{D},25\text{cm}$
• C. $+1.5\text{D},20\text{cm}$
• D. $+2\text{D},21.43\text{cm}$

Answer 1

The correct option is A $-0.67\text{D},21.43\text{cm}$

Here, the distance of the far point, $x=150\text{cm}$

The defect can be corrected by using concave lens of focal length,

$f=-x=-150\text{cm}=-1.5\text{m}$

So, the power of the lens is given by,
$P=\frac{1}{f}=\frac{1}{-1.5}=-0.67\text{D}$

So, for an object placed at $25\text{cm}$,

$u=-25\text{cm};f=-150\text{cm}$

From the lens equation, we have

$v=\frac{uf}{u+f}=\frac{(-25)\times (-150)}{(-25)+(-150)}$

$\Rightarrow v=-21.43\text{cm}$

Therefore, the near point will be at a distance of $21.43\text{cm}$ i.e. less than $25\text{cm}$.

Hence, option $\left(A\right)$ is correct.