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(C) Let the concave mirror be $M_1$ and the convex mirror be $M_2$. For mirror $M_1$,the object $O$ is at $u = -15 \ cm$ and $f = -10 \ cm$.Using the

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$6$

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(C) Let the concave mirror be $M_1$ and the convex mirror be $M_2$. For mirror $M_1$,the object $O$ is at $u = -15 \ cm$ and $f = -10 \ cm$.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v} + \frac{1}{-15} = \frac{1}{-10} \Rightarrow \frac{1}{v} = \frac{1}{15} - \frac{1}{10} = \frac{2-3}{30} = -\frac{1}{30} \Rightarrow v = -30 \ cm$.This image $I_1$ acts as an object for mirror $M_2$. The distance of $I_1$ from $M_2$ is $u_2 = -(40 - 30) = -10 \ cm$.For mirror $M_2$,$f = +15 \ cm$ and $u = -10 \ cm$.Using the mirror formula $\frac{1}{v_2} + \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v_2} + \frac{1}{-10} = \frac{1}{15} \Rightarrow \frac{1}{v_2} = \frac{1}{15} + \frac{1}{10} = \frac{2+3}{30} = \frac{5}{30} = \frac{1}{6} \Rightarrow v_2 = +6 \ cm$.Thus,the final image is formed at a distance of $6 \ cm$ behind the convex mirror.

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