Q.An object is placed in front of a concave mirror of focal length 20 cm. The image formed is three times the size of the object. Calculate the two possible distances of the object from the mirror.

Step 1. Concave mirror, $f=-20$ cm (sign convention). Case (i), real inverted image: $m=-v/u=-3$, so $v=3u$.

Step 2. Substituting into the mirror equation $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$: $\dfrac{1}{3u}+\dfrac{1}{u}=\dfrac{1}{-20}$, i.e. $\dfrac{1+3}{3u}=\dfrac{4}{3u}=-\dfrac{1}{20}$, giving $u=-\dfrac{4\times20}{3}=-\dfrac{80}{3}$ cm.

Step 3. Case (ii), virtual erect image: $m=-v/u=+3$, so $v=-3u$. Substituting: $\dfrac{1}{-3u}+\dfrac{1}{u}=-\dfrac{1}{20}$, i.e. $\dfrac{-1+3}{3u}=\dfrac{2}{3u}=-\dfrac{1}{20}$, giving $u=-\dfrac{2\times20}{3}=-\dfrac{40}{3}$ cm.

Step 4. So there are genuinely two valid object positions producing a 3-times-size image: $u=-80/3$ cm (object beyond the focus, giving a real, inverted, enlarged image) and $u=-40/3$ cm (object between the focus and the mirror, giving a virtual, erect, enlarged image).

Solve the mirror equation twice, once for m = -3 (real image) and once for m = +3 (virtual image).

• Only finding one of the two valid solutions, forgetting a concave mirror can give either a real or a virtual enlarged image depending on object position.
• Sign errors substituting v = 3u or v = -3u into the mirror equation.