Question

An object of 5.0 cm size is placed at a distance of 20.0 cm from a converging mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed to get the sharp image? Also calculate the size of the image.

Answer 1

Concave mirror is a converging mirror.
Distance of the object from the mirror 'u' = -20 cm

Height of the object 'h_o' = 5 cm

Focal length of the mirror 'f' = -15 cm

We have to find the distance of the image 'v' and the height of the image 'h_i'.

Using the mirror formula, we get

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \frac{1}{-15}=\frac{1}{v}+\frac{1}{-20}=\frac{1}{v}-\frac{1}{20} \\ \mathrm{or}\frac{1}{v}=\frac{1}{20}-\frac{1}{15} \\ \mathrm{or}\frac{1}{v}=\frac{3-4}{60}=\frac{1}{60} \\ \mathrm{or}v=-60\mathrm{cm} \\ \mathrm{Thus},\mathrm{the}\mathrm{image}\mathrm{is}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}60\mathrm{cm}\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{the}\mathrm{mirror}. \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\mathrm{using}\mathrm{the}\mathrm{magnification}\mathrm{formula},\mathrm{we}\mathrm{get} \\ m=\frac{-v}{u}=\frac{h_i}{h_o} \\ \mathrm{or}\frac{h_i}{5}=\frac{-(-60)}{-20} \\ \mathrm{or}\frac{{\mathrm{h}}_{\mathrm{i}}}{5}=-3 \\ \mathrm{or}{\mathrm{h}}_{\mathrm{i}}=-3\times 5=-15\mathrm{cm} \\ \mathrm{Therefore},\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}15\mathrm{cm},\mathrm{and}\mathrm{the}\mathrm{negative}\mathrm{sign}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{formed}\mathrm{below}\mathrm{the}\mathrm{principal}\mathrm{axis}. \end{gathered}$$