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

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