**Question:**

A card sheet divided into squares each of size $1 \mathrm{~mm}^{2}$ is being viewed at a distance of $9 \mathrm{~cm}$ through a magnifying glass (a converging lens of focal length $9 \mathrm{~cm}$ ) held close to the eye.

(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?

(b) What is the angular magnification (magnifying power) of the lens?

(c) Is the magnification in (a) equal to the magnifying power in (b)?

Explain.

**Solution:**

Note :

Here we took focal Length as 10 cm because if we take it as 9 cm then image distance will be zero ,which does not make any sense.

(a) Area of each square, *A* = 1 mm2

Object distance, *u* = −9 cm

Focal length of a converging lens, *f *= 9 cm

For image distance *v*, the lens formula can be written as:

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

$\frac{1}{10}=\frac{1}{v}+\frac{1}{9}$

$\frac{1}{v}=-\frac{1}{90}$

$\therefore v=-90 \mathrm{~cm}$

Magnification, $m=\frac{v}{u}$

$=\frac{-90}{-9}=10$

∴Area of each square in the virtual image = (10)2*A*

= 102 × 1 = 100 mm2

$=1 \mathrm{~cm}^{2}$

(b) Magnifying power of the lens $=\frac{d}{|u|}=\frac{25}{9}=2.8$

(c) The magnification in (a) is not the same as the magnifying power in (b).

The magnification magnitude is $\left(\left|\frac{v}{u}\right|\right)$ and the magnifying power is $\left(\frac{d}{|u|}\right)$.

The magnification magnitude is $\left(\left|\frac{v}{u}\right|\right)$ and the magnifying power is $\left(\frac{d}{|u|}\right)$.

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