Question

when an object is placed at a distance of 60 cm from a convex spherical mirror ,the magnification produced is 1/2 .Where should the object be placed to get magnification of 1/3​

Answer 1

Answer -

-120 cm

Given:

The distance of the object is (u₁) = -60 cm

Magnification m₁ = ¹/₂

To be found:

The distance of the object (u₂) =?

Now,

Magnification (m₂) = ¹/₃

So,

∴ m₁ = $\boxed{\frac{-v_{1}}{u_1}}$

$\implies v_{1} = -m_{1}u_{1} = \frac{-1}{2} \times (-60)$

$v_{1}= 30 cm$

By using mirror formula,

$\frac{1}{f_{1}} = \frac{1}{v_{1}}+\frac{1}{u_{1}} = \frac{1}{v_{1}} + \frac{1}{(-60)} = \frac{1}{30} -\frac{1}{60} = \boxed{\frac{1}{60}}$

$\implies f_{1} = 60cm$

Now,

$m_{2} = \frac{-v_{2}}{u_{2}} \implies v_{2} = \frac{-u_{2}}{3}$

Since,

$f_{1} =f_{2}$

$\therefore \frac{1}{f_{1}} = \frac{1}{v_{2}} + \frac{1}{u_{2}}$

$= \frac{1}{u_{2}} = \frac{1}{f_{1}} - \frac{1}{v_{2}}$

$\frac{1}{u_{2}} = \frac{1}{60} + \frac{3}{u_{2}}$

$\implies \frac{1}{u_{2}}-\frac{3}{u_{2}} = \frac{1}{60}$

$\implies \frac{-2}{u_{2}} = \frac{1}{60}$

$\implies \boxed{u_{2}=-120cm}$