Question

A concave mirror produces a real image 1 cm tall of an object 2.5 mm tall placed 5 cm from the mirror. Find the position of the image and the focal length of the mirror.

Answer 1

Distance of the object from the mirror (u) = -5 cm
Height of the image (h_i)​ = -1 cm
Height of the object (h_o)​ = 2.5 mm = 0.25 cm
We have to find the image distance (v) and the focal length of the mirror (f).
Using the magnification formula, we get
$$\begin{gathered} m=\frac{h_i}{h_o}=\frac{-v}{u} \\ \mathrm{or}\frac{-1}{0.25}=\frac{-\mathrm{v}}{-5} \\ \mathrm{or}\mathrm{v}=\frac{-5}{0.25}=-20\mathrm{cm} \end{gathered}$$
Thus, the distance of the image is -20 cm.
It means that the image will form 20 cm in front of the mirror.
Now, using the mirror formula, we get
$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}$

$$\begin{gathered} \frac{1}{f}=\frac{1}{-20}+\frac{1}{-5} \\ \mathrm{or}\frac{1}{f}=-\frac{1}{20}-\frac{1}{5} \\ \mathrm{or}\frac{1}{f}=-\frac{1}{20}-\frac{4}{20}=\frac{-5}{20}=\frac{-1}{4} \\ f=-4\mathrm{cm} \end{gathered}$$

Thus, the focal length of the mirror is 4 cm.