Question

The radius of curvature of a convex mirror used as a rear view mirror in a moving car is 12.0 m. A truck is coming from behind it at a distance of 3.54 m. Calculate (a) position, and (b) size of the image relative to the size of the truck. What will be the nature of the image?

Answer 1

Given,
The mirror is convex.

Distance of the object from the mirror 'u' = -3.54 cm
Radius of curvature of the mirror 'R' = 12 cm

Focal length of the convex mirror 'f' = 6 cm

We have to find the position of the image 'v' and its magnification 'm'.

Using the mirror formula, we get

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \Rightarrow \frac{1}{1}=\frac{1}{v}+\frac{1}{-3.5} \\ \Rightarrow \frac{1}{1}+\frac{10}{35}=\frac{1}{v} \\ \Rightarrow \frac{35}{35}+\frac{10}{35}=\frac{1}{v} \\ \Rightarrow \frac{45}{35}=\frac{1}{v} \\ \Rightarrow v=\frac{35}{45}=\frac{7}{9}=0.77\mathrm{m} \end{gathered}$$

Thus, the distance of the image 'v' is 0.77 m behind the mirror.

Now, using the magnification formula, we get

$$\begin{gathered} m=\frac{-v}{u} \\ m=\frac{-0.77}{-3.5}=\frac{1}{4.5} \\ \mathrm{P}\mathrm{ositive}\mathrm{magnification}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{image}\mathrm{is}\mathrm{virtual}\mathrm{and}\mathrm{erect}. \end{gathered}$$

Thus, the image is virtual, erect and smaller in size.