Question

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

In order to view the squares distinctly with the maximum possible magnifying power.

(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:

Answer 1

(a) Area of each square , $A=1m{m}^2$

Object distance $u=-9cm$

Focal length of the converging lens , $f=10cm$

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

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

By putting the values we get $v=-90cm$

Now magnification,

$m=\frac{u}{v}=\frac{-60}{-9}=10$

$\therefore$ Area of each square in the virtual image = ${10}^{2}A={10}^2\times 1=100m{m}^2$

(b) Magnifying power of the lens

$\frac{d}{|u|}=\frac{25}{9}=2.8$

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

The ,magnification magnitude is

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

The two quantities will be equal when the image is formed at the near point (25cm).