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An object is placed at a distance of 60cm from a convex mirror where the magnification produced is
1/2 where should the object be placed to get a magnification of 1/3.​

Answer 1

Given :-

When object is placed at 60cm magnification= 1/2

To find :-

where should the object place to get magnification 1/3

Solution :-

We have,

• object distance (u)= (-60cm,)

• Maginfication (m) = 1/2

Now find the image distance (v) using magnification,

$\implies\sf m= \dfrac{-v}{u}\\ \\ \implies\sf \dfrac{1}{2}= \dfrac{-v}{-60}\\ \\ \implies\sf -60= -2v\\ \\ \implies\sf \cancel{\dfrac{-60}{-2}}= v\\ \\ \implies\sf 30= v\\ \\ \bullet \sf Image \ distance = 30 cm$

Now by using mirror formula, find the focal length ,

$\boxed{\sf \dfrac{1}{f}= \dfrac{1}{v}+\dfrac{1}{u}}$

$\implies\sf \dfrac{1}{f}=\dfrac{1}{30} +\dfrac{1}{-60}\\ \\ \implies\sf \dfrac{1}{f}= \dfrac{1}{30}-\dfrac{1}{60}\\ \\ \implies\sf \dfrac{1}{f}= \dfrac{2-1}{60}\\ \\ \implies\sf \dfrac{1}{f} = \dfrac{1}{60}\\ \\ \implies\sf f= 60cm$

●Now we have magnification= 1/3

$\implies\sf m= \dfrac{-v}{u}\\ \\ \implies\sf \dfrac{1}{3}= \dfrac{-v}{u}\\ \\ \implies\sf u= -3v\\ \\ \implies\sf v= \dfrac{u}{-3}$

Focal length remains same !

Again using mirror formula ,

$\implies\sf \dfrac{1}{f}= \dfrac{1}{v}+\dfrac{1}{u}\\ \\ \implies\sf \dfrac{1}{60}= \dfrac{1}{\dfrac{u}{-3}}+\dfrac{1}{u} \\ \\ \implies\sf \dfrac{1}{60}= \dfrac{-3}{u}+\dfrac{1}{u}\\ \\ \implies\sf \dfrac{1}{60}= \dfrac{-3+1}{u}\\ \\ \implies\sf \dfrac{u}{60}= -2\\ \\ \implies\sf u= -2\times 60 \\ \\ \implies\sf u= -120cm$

$\underline{\bigstar{\sf\ Object \ distance \ (u) = -120cm}}$

● Hence, object is placed at 120 cm behind the mirror to get magnification 1/3