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Question

An object is placed at a distance of 30 cm. from a concave mirror. The magnification produced is $${1}/{2}$$. where should the object be placed to get the magnification of 1/3.


Solution

$$U = -30$$
$$m = \dfrac{1}{2}$$
we know that 
$$m = \dfrac{-V}{U}$$
so V = 15
$$f = 10$$
for $$m = \dfrac{1}{3}$$


$$ \dfrac{-V}{U} = \dfrac{1}{3}$$
$$V = \dfrac{-V}{3}$$

$$ \dfrac{1}{f} = \dfrac{1}{V} + \dfrac{1}{U}$$

$$\dfrac{1}{10} = \dfrac{3}{-U} + \dfrac{1}{U}$$

U = -40
Therefore for magnification to be $$\dfrac{1}{3}$$ the object must be placed 40cm upon in front of the mirror.


Physics

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