A) zero
B) \[\infty \]
C) \[\frac{1}{2}\]
D) 2
Correct Answer: C
Solution :
: For a mirror, \[\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\] For convex mirror\[f\]is positive For virtual image\[v\]is positive \[u\]is always negative as per convention \[\therefore \]\[\frac{1}{v}-\frac{1}{f}=\frac{1}{f}\]or\[\frac{1}{v}=\frac{1}{f}+\frac{1}{f}=\frac{2}{f}\]or\[v=\frac{f}{2}\] \[\therefore \]\[\frac{v}{u}=\frac{f}{f\times 2}=\frac{1}{2}\] \[\therefore \]Magnification\[=\frac{1}{2}\]
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