Question

A man with normal near point $\left(25\text{cm}\right)$ reads a book with small print using a magnifying thin convex lens of focal length $5\text{cm}$. What is the closest and farthest distance at which he can read the book when viewing through the magnifying glass?

• A. $4.17\text{cm},5\text{cm}$
• B. $5\text{cm},7.4\text{cm}$
• C. $17\text{cm},8\text{cm}$
• D. $5\text{cm},6.3\text{cm}$

Answer 1

The correct option is A $4.17\text{cm},5\text{cm}$

As for normal eye, far point and near point are $\infty$ and $25\text{cm}$ respectively.

So, for magnifier,
${\text{v}}_{\text{max}}=-\infty$
${\text{v}}_{\text{min}}=-25\text{cm}.$

Using lens formula,

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

$\Rightarrow \text{u}=\frac{f}{\left(\frac{f}{\text{v}}\right)-1}$

$\Rightarrow {\text{u}}_{\text{min}}=\frac{f}{\left(\frac{f}{{\text{v}}_{\text{min}}}\right)-1}$

$\Rightarrow {\text{u}}_{\text{min}}=\frac{5}{\left(\frac{-5}{25}\right)-1}=-\frac{25}{6}$

$\Rightarrow {\text{u}}_{\text{min}}=-4.17\text{cm}$

Similarly,
${\text{u}}_{\text{max}}=\frac{f}{\left(\frac{f}{{\text{v}}_{\text{max}}}\right)-1}$


$\Rightarrow {\text{u}}_{\text{max}}=\frac{5}{\left(\frac{-5}{\infty }\right)-1}=-\frac{5}{1}$

$\Rightarrow {\text{u}}_{\text{max}}=-5\text{cm}$

So, the closest and farthest distance of the book from the magnifier (or eye) for clear viewing are $4.17\text{cm}$ and $5\text{cm}$ respectively.

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