# Q 9.23  A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 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.

P Pankaj Sanodiya

Given,

Object distance u = -9cm

Focal length of convex lens = 10cm

According to the lens formula

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

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

$\\\frac{1}{v}=\frac{1}{10}-\frac{1}{9}\\\Rightarrow v=-90\ cm$

a) Magnification

$m=\frac{v}{u}=\frac{-90}{9}=10 \ cm$

The area of each square in the virtual image

$=10\times10\times1=100mm^2=1cm^2$

b) Magnifying power

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

c) No,

$magnification = \frac{v}{u}$

$magnifying\ power = \frac{d}{|u|}$.

Both the quantities will be equal only when image is located at the near point |v| = 25 cm

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