An object $2\text{cm}$ tall is placed $5\text{cm}$ in front of a concave mirror whose focal length is $-10\text{cm}$. (i) Find the position of the image. (ii) State whether the image is real or virtual and whether it is upright or inverted. (iii) Calculate the magnification and the height of the image.

Step-by-step solution

1. Use the mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ with sign convention: object distance $u=-5\text{cm}$ (in front of the mirror) and focal length $f=-10\text{cm}$ (concave).
2. Rearrange to find $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{1}{10}-\left(-\frac{1}{5}\right)=-0.1+0.2=0.1$.
3. Thus $v=+10\text{cm}$. A positive image distance for mirrors indicates a virtual image formed $10\text{cm}$ behind the reflecting surface.
4. Since the image is virtual, it will be upright.
5. Magnification $m$ is given by $m=-\frac{v}{u}=-\frac{+10}{-5}=2$.
6. A positive magnification confirms the image is upright, and $|m|>1$ shows it is enlarged.
7. Height of the image $h_i=m\times h_o=2\times 2\text{cm}=4\text{cm}$.
8. Summarising: the image is $10\text{cm}$ behind the mirror, virtual, upright, with magnification $2$ and height $4\text{cm}$.

Concept tested: Concave mirror image formation

Published 25 August 2026 · How this solution was prepared and checked

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