Question

A concave mirror of focal length $15\, cm$ forms an image having twice the linear dimension of the object. The position of the object when the image is virtual will be:

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

Answer 1

Answer: (C)

7.5 cm

Answer 2

Focal length of concave mirror $f=-15\, cm$ magnification $=2$ (for virtual image) Linear magnification $m=\frac{\text { size of image }}{\text { size of object}}=-\frac{v}{u}$ $2=-\frac{v}{u}$ or $v=-2 u$ Now, from the relation $\frac{1}{f} =\frac{1}{v}+\frac{1}{u}$ or $-\frac{1}{15} =\frac{1}{u}-\frac{1}{2 u}$ $=\frac{1}{2 u}$ or $2 u=-15$ or $u=-7.5\, cm$