Focal length of convex mirror $M_{1}$ is 20 $c m$ and that of concave mirror $M_{2}$ is 30 $c m$ . Find position of second image $I_{2}$ . Take first reflection from $M_{1}$ .

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$v_{1} = - 10 c m a n d v_{2} = - 120 c m$

$v_{1} = - 10 c m a n d v_{2} = + 120 c m$

$v_{1} = + 10 c m a n d v_{2} = + 120 c m$

$v_{1} = + 10 c m a n d v_{2} = - 120 c m$

answer is B.

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Detailed Solution

For $M_{1}$ ,

$u_{1} = - 20 c m f_{1} = + 20 c m$

Using the mirror formula, $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ , we have

$\frac{1}{v_{1}} + \frac{1}{- 20} = \frac{1}{+ 20}$

Solving, we get $v_{1} = + 10 c m$

For $M_{2}$ ,

$u_{2} = - 40 c m (P I_{1} = 40 c m) f_{2} = - 30 c m$

Using the mirror formula, $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ , we have

$\frac{1}{v_{2}} + \frac{1}{- 40} = \frac{1}{- 30} ∴ v_{2} = P_{2} I_{2} = - 120 c m$

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