A man with normal near point (25 cm) reads a book with small print using a magnifying glass:
Question:

A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.

(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?

(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?

Solution:

(a) Focal length of the magnifying glass, f = 5 cm

Least distance of distance vision, d = 25 cm

Closest object distance = u

Image distance, v = −d = −25 cm

According to the lens formula, we have:

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

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

$=\frac{1}{-25}-\frac{1}{5}=\frac{-5-1}{25}=\frac{-6}{25}$

$\therefore u=-\frac{25}{6}=-4.167 \mathrm{~cm}$

Hence, the closest distance at which the person can read the book is 4.167 cm.

For the object at the farthest distant $\left(u^{\prime}\right)$, the image distance $\left(v^{\prime}\right)=\infty$

According to the lens formula, we have:

$\frac{1}{f}=\frac{1}{v^{\prime}}-\frac{1}{u^{\prime}}$

$\frac{1}{u^{\prime}}=\frac{1}{\infty}-\frac{1}{5}=-\frac{1}{5}$

$\therefore u^{\prime}=-5 \mathrm{~cm}$

Hence, the farthest distance at which the person can read the book is 5 cm.

(b) Maximum angular magnification is given by the relation:

$\alpha_{\max }=\frac{d}{|u|}$

$=\frac{25}{\frac{25}{6}}=6$

Minimum angular magnification is given by the relation:

$\alpha_{\min }=\frac{d}{\left|u^{\prime}\right|}$

$=\frac{25}{5}=5$