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(C) Let $d$ be the distance between $L$ and $M_1$. The object $S$ is at a distance of $10 	ext{ cm}$ from $L$ (mid-point of $20 	ext{ cm}$). Since t

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$150$

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(C) Let $d$ be the distance between $L$ and $M_1$. The object $S$ is at a distance of $10 	ext{ cm}$ from $L$ (mid-point of $20 	ext{ cm}$). Since the focal length of $L$ is $10 	ext{ cm}$,the object $S$ is at the focus of the lens $L$. Light rays from $S$ pass through $L$ and become parallel to the principal axis. These parallel rays strike the concave mirror $M_1$. After reflection from $M_1$,they converge at the focus of $M_1$. The focal length of $M_1$ is $f_1 = R_1/2 = 20/2 = 10 	ext{ cm}$. So,the rays converge at a point $10 	ext{ cm}$ in front of $M_1$. This point acts as an object for the lens $L$. The distance of this point from $L$ is $u = -(d - 10) 	ext{ cm}$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$: $\frac{1}{v} - \frac{1}{-(d - 10)} = \frac{1}{10} \implies \frac{1}{v} = \frac{1}{10} - \frac{1}{d - 10} \quad (i)$. The rays emerging from $L$ strike the mirror $M_2$. For the final image to coincide with $S$,the rays must strike $M_2$ normally,meaning they must appear to come from the center of curvature of $M_2$. The radius of curvature of $M_2$ is $24 	ext{ cm}$. Since $S$ is $10 	ext{ cm}$ from $M_2$,the rays must converge at a point $10 	ext{ cm}$ from $M_2$ (which is $S$ itself). Thus,the image formed by $L$ must be at a distance $v = 10 	ext{ cm}$ from $L$ (towards $M_2$). Substituting $v = 10$ in $(i)$: $\frac{1}{10} = \frac{1}{10} - \frac{1}{d - 10} \implies \frac{1}{d - 10} = 0$,which is not possible. Re-evaluating: The rays after reflection from $M_1$ converge at $10 	ext{ cm}$ from $M_1$. Let this be $I_1$. $I_1$ is at distance $(d-10)$ from $L$. For the final image to be at $S$,the rays must strike $M_2$ such that they retrace their path. This happens if they are directed towards the center of curvature of $M_2$. The center of curvature of $M_2$ is at $24 	ext{ cm}$ from $M_2$,i.e.,$4 	ext{ cm}$ behind $L$. Using lens formula for $L$ with $u = -(d-10)$ and $v = +4$: $\frac{1}{4} - \frac{1}{-(d-10)} = \frac{1}{10} \implies \frac{1}{d-10} = \frac{1}{10} - \frac{1}{4} = \frac{2-5}{20} = -\frac{3}{20}$. This gives $d-10 = -20/3$,$d = 10 - 6.67 = 3.33$. Alternatively,if the rays from $M_1$ form an image at $S$ after passing through $L$ again: For $M_2$,the object is $S$ at $10 	ext{ cm}$. $f_2 = -12 	ext{ cm}$. $\frac{1}{v} + \frac{1}{-10} = \frac{1}{-12} \implies \frac{1}{v} = \frac{1}{10} - \frac{1}{12} = \frac{1}{60} \implies v = 60 	ext{ cm}$. This image acts as an object for $L$ at $u = -(20+60) = -80 	ext{ cm}$. $\frac{1}{v_L} - \frac{1}{-80} = \frac{1}{10} \implies \frac{1}{v_L} = \frac{1}{10} - \frac{1}{80} = \frac{7}{80} \implies v_L = 80/7$. This $v_L$ is the distance from $L$ where rays from $M_1$ must converge. Since $M_1$ has $f=10$,rays from $L$ (parallel) converge at $10 	ext{ cm}$ from $M_1$. So $d - 10 = 80/7 \implies d = 150/7$. Thus $n = 150$.

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