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(A) $1$. The image formed by the plane mirror $M_1$ is at a distance $x$ behind the mirror $M_1$. Since the distance between $M_1$ and $M_2$ is $y$,th

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$\frac{{{x^2} - {y^2}}}{{2y}}$

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(A) $1$. The image formed by the plane mirror $M_1$ is at a distance $x$ behind the mirror $M_1$. Since the distance between $M_1$ and $M_2$ is $y$,the image formed by $M_1$ is at a distance $(x + y)$ in front of the convex mirror $M_2$. Thus,the object distance for $M_2$ is $u = -(x + y)$.$2$. The image formed by the convex mirror $M_2$ is at the same location as the image formed by $M_1$. The image formed by $M_1$ is at a distance $x$ behind $M_1$,which is at a distance $(y - x)$ behind $M_2$ (or $(x - y)$ in front of $M_2$). Thus,the image distance for $M_2$ is $v = +(x - y)$.$3$. Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$,where $f$ is the focal length of the convex mirror:$\frac{1}{f} = \frac{1}{x - y} + \frac{1}{-(x + y)}$$\frac{1}{f} = \frac{1}{x - y} - \frac{1}{x + y}$$\frac{1}{f} = \frac{(x + y) - (x - y)}{(x - y)(x + y)}$$\frac{1}{f} = \frac{2y}{x^2 - y^2}$$f = \frac{x^2 - y^2}{2y}$

An object $O$ is placed in front of a small plane mirror $M_1$ and a large convex mirror $M_2$ of focal length $f$. The distance between $O$ and $M_1$ is $x$,and the distance between $M_1$ and $M_2$ is $y$. The images of $O$ formed by $M_1$ and $M_2$ coincide. The magnitude of $f$ is

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