A point source is situated at a distance x

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(C) For a concave mirror,the mirror formula is given by $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$.Using sign convention,$u = -x$ and $v = -v_{i}$ (whe

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Vedclass · Answer

(C) For a concave mirror,the mirror formula is given by $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$.Using sign convention,$u = -x$ and $v = -v_{i}$ (where $v_{i}$ is the image distance,and we are interested in $|v_{i}|$).Substituting these,we get $\frac{1}{f} = -\frac{1}{x} - \frac{1}{v_{i}}$,which simplifies to $\frac{1}{v_{i}} = -(\frac{1}{f} + \frac{1}{x}) = -(\frac{x+f}{xf})$.Thus,$|v_{i}| = \frac{xf}{f-x}$.As $x 	o 0$,$|v_{i}| 	o 0$.As $x 	o f$,$|v_{i}| 	o \infty$.For $x > f$,the image becomes real and inverted. The mirror formula is $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$. With $u = -x$,we have $\frac{1}{v} = \frac{1}{f} - \frac{1}{x} = \frac{x-f}{xf}$,so $v = \frac{xf}{x-f}$.Thus,$|v| = \frac{xf}{x-f}$.As $x 	o f^{+}$,$|v| 	o \infty$.As $x 	o \infty$,$|v| 	o f$.Comparing these behaviors,the graph in option $C$ correctly shows the asymptotic behavior at $x = f$ and the limits as $x 	o 0$ and $x 	o \infty$.

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