An optical arrangement consists of two concave mirrors and and a convex lens with a common principal axis, as shown in the figure. The focal length of is 10 cm. The radii of curvature of and are 20 cm and 24 cm, respectively. The distance between a

Question

An optical arrangement consists of two concave mirrors $M_{1}$ and $M_{2}$ and a convex lens $L$ with a common principal axis, as shown in the figure. The focal length of $L$ is 10 cm. The radii of curvature of $M_{1}$ and $M_{2}$ are 20 cm and 24 cm, respectively. The distance between $L$ and $M_{2}$ is 20 cm. A point object $S$ is placed at the mid-point between $L$ and $M_{2}$ on the axis. When the distance between $L$ and $M_{1}$ is $\frac{n}{7} cm$ , one of the images coincides with $S$ . The value of $n$ is ______. [2023]

Smart Achievers · Accepted Answer

(1, 2, 3)

From reflection from mirror $M_{2}$ ,

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $\Rightarrow \frac{1}{v} + \frac{1}{(- 10)} = \frac{1}{(- 12)}$

$\Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{12}$ $∴$ $v = + 60 cm$

For refraction from lens $L$ ,

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ or, $\frac{1}{v} - \frac{1}{(- 80)} = \frac{1}{10}$

$∴$ $v = + \frac{80}{7}$ $({For image I}_{2})$

This image should be at the focus of $M_{1}$

$∴$ $\frac{20}{2} + \frac{80}{7} = \frac{n}{7}$ $\Rightarrow \frac{300}{14} = \frac{n}{7}$ $∴$ $n = 150$

If image $I_{2}$ is formed at the pole of mirror $M_{1}$ , then $\frac{n}{7} = \frac{80}{7}$

$∴$ $n = 80$

If image $I_{2}$ is formed at the centre of curvature of mirror $M_{1}$ ,

$∴$ $\frac{n}{7} = \frac{80}{7} + 20$

$∴$ $n = 220$

Solution

1 80

2 150

3 220

4 250

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