# Q 9.24  (a)  ) At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?(b) What is the magnification in this case? (c) Is the magnification equal to the magnifying power in this case?

P Pankaj Sanodiya

a)

maximum magnifying is possible when our image distance will be equal to minimum vision point that is,

$v=-25$

$f = 10cm$     (Given)

Now according to the lens formula

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

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

$\frac{1}{u}=\frac{1}{-25}-\frac{1}{10}$

$\frac{1}{u}=-\frac{7}{50}$

$u=-\frac{50}{7}=-7.14cm$

Hence required object distance for viewing squares distinctly is 7.14 cm away from the lens.

b)

Magnification of the lens:

$m = \left | \frac{v}{u} \right |=\frac{25}{50}*7= 3.5$

c)

Magnifying power

$M = \frac{d}{u} =\frac{25}{50}*7= 3.5$

Since the image is forming at near point ( d = 25 cm ), both magnifying power and magnification are same.

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