Two spherical mirrors $M_{1}$ and $M_{2}$ , one convex and other concave having same radius of curvature R are arranged coaxially at a distance 2 R (consider their pole separation to be 2 R ). A bead of radius a is placed at the pole of the convex mirror as shown. The ratio of the sizes of the first three images of the bead is

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3: 11: 41

$\frac{1}{3} : \frac{1}{11} : \frac{1}{41}$

$1 : \frac{1}{2} : \frac{1}{3}$

1: 2: 3

answer is C.

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Detailed Solution

The first image is formed due to the reflection from concave mirror $M_{2}$ .

$\Rightarrow \frac{1}{v_{1}} + \frac{1}{(- 2 R)} = \frac{2}{- R} \Rightarrow \frac{1}{v_{1}} = \frac{1}{2 R} - \frac{4}{2 R} \Rightarrow \frac{1}{v_{1}} = \frac{- 3}{2 R} \Rightarrow v_{1} = \frac{- 2 R}{3} \Rightarrow m_{1} = - (\frac{\frac{- 2 R}{3}}{2 R}) = - \frac{1}{3} \Rightarrow Radius of image circle = \frac{1}{3} a, (in magnitude).$

Now the second image is formed by convex mirror $M_{1}$ .

The second image will be formed because the image formed by the first acts as object for it.

$\Rightarrow Object distance = 2 R - \frac{2 R}{3} = \frac{4 R}{3} \Rightarrow \frac{1}{v_{2}} + \frac{1}{(- \frac{4 R}{3})} = \frac{1}{(\frac{R}{2})} \Rightarrow \frac{1}{v_{2}} = \frac{2}{R} + \frac{3}{4 R} \Rightarrow v_{2} = \frac{4 R}{11} \Rightarrow m_{2} = \frac{- v_{2}}{u_{2}} = \frac{- \frac{4 R}{11}}{- \frac{4 R}{3}} = \frac{3}{11} \Rightarrow m_{2} = \frac{3}{11}$