Question

Use the mirror equation to deduce that an object placed between f and 2f of a concave mirror produces a real image beyond 2f. [3 MARKS]

Answer 1

Each point: 1 Mark

We know
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ or

$\frac{1}{v}=\frac{1}{f}–\frac{1}{u}$

f < 0 (concave mirror); u < 0 (object on the left)

For 2f < u < f implies $\frac{1}{2f}$ > $\frac{1}{u}$ > $\frac{1}{f}$ or $\frac{-1}{2f}$ < $\frac{-1}{u}$ < $\frac{-1}{f}$

or $\frac{1}{f}$ – $\frac{1}{2f}$ < $\frac{1}{f}$ – $\frac{1}{u}$ < 0 or $\frac{1}{2f}$ < $\frac{1}{v}$ < 0

which means v < 0 (image on the left; real) the image lies beyond 2f. The image is real because v is negative.