Question

The least distance of vision of a long-sighted person is $60cm$. By using a spectacle lens, this distance is reduced to $12cm$. The power of the lens is:

• A. $+5.0D$
• B. $+\frac{20}{3}D$
• C. $-\frac{10}{3}D$
• D. $+2.0D$

Answer 1

The correct option is B $+\frac{20}{3}D$

Given that the least distance of vision is $60cm$, which is reduced using a lens to $12cm$. This implies that the image of an object at $12cm$ is formed at $60cm$ from the lens.

Hence, $v=-60cm$ , $u=-12cm$
$\therefore$ by using the lens' formula $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
We have $\frac{1}{f}=\frac{1}{-60}-\frac{1}{-12}$
$\Rightarrow \frac{1}{f}=\frac{1}{15}{cm}^{-1}$ or $\frac{100}{15}{m}^{-1}$
So, the power of lens, $P=\frac{1}{f}=\frac{100}{15}=+\frac{20}{3}D$