A) 45 cm
B) 30 cm
C) 7.5 cm
D) 22.5 cm
Correct Answer: C
Solution :
Focal length of concave mirror \[f=-15cm\] magnification =2 (for virtual image) Linear magnification \[m=\frac{size\,of\,image}{size\,of\,object}=-\frac{\upsilon }{u}\] \[2=-\frac{\upsilon }{u}\] or \[\upsilon =-2u\] Now from the relation \[\frac{1}{f}=\frac{1}{\upsilon }+\frac{1}{u}\] or \[-\frac{1}{15}=\frac{1}{u}-\frac{1}{2u}=\frac{1}{2u}\] or \[2u=-15\] or \[u=-7.5cm\]You need to login to perform this action.
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