Magnification is the ratio between height of the image to height of the object i.e., $m = \frac{height of image}{height of object} = \frac{- v}{u}$
i)If ‘m’ is $+ ve$ the image is virtual and erect.

Ii) If ‘m’ is $- ve$ the image is real and inverted.

iii)If $m > 1$ , image is bigger than the object.

iv) $m < 1$ image is smaller than the object.

v)If $m = 1$ image size is equal to object size.

vi)Magnification can also be written as $m = \frac{- image distance}{object distance}$

A 1 cm object is placed perpendicular to the principal axis of a convex mirror of focal length 7.5 cm. Then the object should be placed infront of the mirror at a distance to get the size of the image as 0.6 cm is

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7.5 cm

3 cm

10 cm

5 cm

answer is B.

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Detailed Solution

Given that for a convex mirror

f = + 7.5 cm

size of object = 1 cm

size of image = 0.6 cm

we get $m = \frac{size of image}{size of object} = \frac{0.6}{1} = \frac{3}{5}$

(Since we are using convex mirror m should be +ve because it always form virtual images)

$m = \frac{+ 3}{5} = \frac{- v}{u}$

$\Rightarrow v = \frac{- 3 u}{5}$ , u = ?, f = + 7.5 cm

We know $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

$\frac{1}{7 .5} = \frac{1}{u} + \frac{1}{(- 3 u / 5)}$

$\Rightarrow \frac{2}{15} = \frac{1}{u} - \frac{5}{3 u} = \frac{3 - 5}{3 u}$

$\Rightarrow \frac{2}{15} = \frac{- 2}{3 u}$

$∴ 3 u = - 15$

$u = - 5 cm$

So object should be placed infront of convex mirror at a distance of 5 cm.

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