Question

A person with defective eye vision is unable to see the objects nearer than 1.8m. He wants to read a book at a distance of 32 cm . The power of lens he needed is ​

Answer 1

Convex Lens

Let the focus of the lens be $f$.

The power of the lens must be such that the person will be able to see things at a distance of $32$ cm, for which the virtual image has to be formed at a distance of $1.8$ m.

So, $u=-32$ cm, $v=-180$ cm

Using the formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, we get

$\quad \frac{1}{-180}-\frac{1}{-32}=\frac{1}{f}$

$\Rightarrow \frac{1}{f}=\frac{37}{1440}$

$\Rightarrow f=38.92$

$\therefore$ power of the lens is

$\quad P=\frac{100}{f}$

$\Rightarrow P=\frac{100}{38.92}$

$\Rightarrow P=2.57\:m^{-1}$

Answer: Therefore that person will need a convex lens of power $2.57\:m^{-1}$.

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