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

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:

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

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