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(C) For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Using the sign convention,$u = -x$,where $x$ is the distance

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(C) For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Using the sign convention,$u = -x$,where $x$ is the distance of the object from the mirror $(x > 0)$.Thus,$-\frac{1}{v} - \frac{1}{x} = -\frac{1}{f} \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{x} = \frac{x-f}{fx} \implies v = \frac{fx}{x-f}$.The linear magnification $m$ is given by $m = -\frac{v}{u} = -\frac{fx/(x-f)}{-x} = \frac{f}{x-f}$.The magnitude of linear magnification is $|m| = \left| \frac{f}{x-f} ight|$.As the object moves away from the focal point $(x = f)$ towards infinity $(x 	o \infty)$:$1$. At $x = f$,$|m| 	o \infty$.$2$. At $x = 2f$,$|m| = |f / (2f - f)| = 1$.$3$. As $x 	o \infty$,$|m| 	o 0$.Therefore,the graph must show $|m| 	o \infty$ at $x = f$,$|m| = 1$ at $x = 2f$,and $|m| 	o 0$ as $x 	o \infty$. This corresponds to the graph shown in option $C$.

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If the distance between an object and its two times magnified virtual image produced by a curved mirror is $15 \,cm$, the focal length of the mirror must be:

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Assertion : $A$ point object is placed at a distance of $26 \ cm$ from a convex mirror of focal length $26 \ cm$. The image will not form at infinity.
Reason : For the above-given system,the equation $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ gives $v = \infty$.

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