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