In the shown figure M1 and M2 are two concave mirrors of same focal length 10 cm. AB and CD are their principal axis respectively. A point object O is kept on the line AB at a distance 15 cm from M1. The distance between the mirrors is 20 cm. Considering two successive reflections first on M1 and then on M2. The distance of final image from the line AB is

For reflection through M1u = –15 cmf = –10 cm1v=−110+115⇒v=−30cmFor reflection through M2u = 10 cmf = –10 cm⇒v=−cm (from mirror formula)Now magnification, m = =−vu=510=12So, finally formed image will be at the distance of 32 cm from the line AB

In the shown figure M1 and M2 are two concave mirrors of same focal length 10 cm. AB and CD are their principal axis respectively. A point object O is kept on the line AB at a distance 15 cm from M1. The distance between the mirrors is 20 cm. Considering two successive reflections first on M1 and then on M2. The distance of final image from the line AB is

For reflection through M1u = –15 cmf = –10 cm1v=−110+115⇒v=−30cmFor reflection through M2u = 10 cmf = –10 cm⇒v=−cm (from mirror formula)Now magnification, m = =−vu=510=12So, finally formed image will be at the distance of 32 cm from the line AB

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In the shown figure M₁ and M₂are two concave mirrors of same focal length 10 cm. AB and CD are their principal axis respectively. A point object O is kept on the line AB at a distance 15 cm from M₁. The distance between the mirrors is 20 cm. Considering two successive reflections first on M₁ and then on M_2.The distance of final image from the line AB is

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1.5 cm

4.5 cm

3 cm

1 cm

answer is B.

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Detailed Solution

For reflection through M₁
u = –15 cm
f = –10 cm
$\frac{1}{v} = - \frac{1}{10} + \frac{1}{15} \Rightarrow v = - 30 cm$
For reflection through M₂
u = 10 cm
f = –10 cm
$\Rightarrow v = - cm$ (from mirror formula)
Now magnification, m = $= - \frac{v}{u} = \frac{5}{10} = \frac{1}{2}$
So, finally formed image will be at the distance of $\frac{3}{2}$ cm from the line AB