Question

A large concave mirror has a radius of curvature of 1.5 m. A person stands 10 m in front of the mirror. Where is the person's image?

Answer 1

Given,
The radius of curvature of the mirror 'R' = -1.5m
Focal length of the mirror $\text{'}\mathrm{f}\text{'}=\frac{1}{\mathrm{R}}=\frac{1}{-1.5}=-0.75\mathrm{m}$
Distance of the object 'u' =-10 m
We have to find the distance of the image 'v'.
Using the mirror formula, we get

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}$

$$\begin{gathered} \frac{1}{0.75}=\frac{1}{v}+\frac{1}{-10} \\ -\frac{100}{75}=\frac{1}{v}-\frac{1}{10} \\ \mathrm{or}\frac{1}{v}=\frac{1}{10}-\frac{100}{75} \\ \mathrm{or}\frac{1}{v}=\frac{1}{10}-\frac{4}{3} \\ \mathrm{or}\frac{1}{v}=\frac{3}{30}-\frac{40}{30} \\ \mathrm{or}\frac{1}{v}=\frac{-37}{30} \\ \mathrm{or}v=\frac{-30}{37} \\ \mathrm{or}v=0.81\mathrm{m} \end{gathered}$$
Thus, the person's image will be formed at a distance of 0.81 m from the mirror.