A concave mirrorr of focal length `15 cm` forms an image having twice the linear dimensions of the object. The position of the object when the image is virtual will be

Options

A. 45 cm
B. 30 cm
C. `7.5 cm`
D. `22.5 cm`

Answer

C

Explanation

Focal length of concave mirror `f = - 15 cm`
magnification `=2`(for virtual image)
Linear magnification `m=("size of image")/("size of object") =-(upsilon)/(u)`
`2=-(upsilon)/(u)` or `upsilon=-2u`
Now, from the relation
`(1)/(f)=(1)/(upsilon)+(1)/(u)`
Or `(1)/(15)=(1)/(u)-(1)/(2u)=(1)/(2u)`
or `2u =- 15` or `u = -7.5 cm`.