Question

A bright object 50 mm high stands on the axis of a concave mirror of focal length 100 mm and at a distance of 300 mm from the concave mirror. How big will the image be?

Answer 1

Distance of the object from the mirror 'u' = -300 mm

Height of the object 'h_o' = 50 mm

Focal length of the mirror 'f' = -100 mm

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

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \frac{1}{-100}=\frac{1}{v}+\frac{1}{-300} \\ -\frac{1}{100}=\frac{1}{v}-\frac{1}{300} \\ \frac{1}{v}=\frac{1}{300}-\frac{1}{100} \\ \frac{1}{v}=\frac{1}{300}-\frac{1}{100} \\ \frac{1}{v}=\frac{-2}{300} \\ v=-150\mathrm{mm} \\ \mathrm{Thus},\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}150\mathrm{mm}\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{the}\mathrm{mirror}. \\ \mathrm{Now},\mathrm{using}\mathrm{the}\mathrm{magnification}\mathrm{formula},\mathrm{we}\mathrm{get} \\ m=\frac{h_i}{h_o}=\frac{-v}{u} \\ \frac{h_i}{50}=\frac{-(-150)}{-300}=-\frac{1}{2} \\ {h}_i=-\frac{50}{2}=-25\mathrm{mm} \\ \mathrm{Thus},\mathrm{the}\mathrm{size}\mathrm{of}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}25\mathrm{mm}. \end{gathered}$$