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.(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, $A=1m{m}^2$

$u=-9,f=10$

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow v=-90cm$

Magnification, $m=v/u=-90/(-9)=10$

Area of each square is ${10}^2\times A=100m{m}^2$

(b)

The magnifying power of the lens is

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

(c)

No, magnification of an image by a lens and angular magnification (or magnifying power) of an optical instrument are two separate things. The latter is the ratio of the angular size of the object (which is equal to the angular size of the image even if the image is magnified) to the angular size of the object if placed at the near point (25 cm). Thus, magnification magnitude is $\left|\right(v/u\left)\right|$ and magnifying power is $\left(25/\left|u\right|\right)$. Only when the image is located at the near point $\left|v\right|=25cm$, are the two quantities equal.