Question

A dentist's mirror has a radius of curvature of 3 cm. How far must it be placed from a small dental cavity to give a virtual image of the cavity that is magnified five times?

Answer 1

Given,

The dentist's mirror is a concave mirror.

Radius of curvature 'R' = -3 cm

Therefore, the$\mathrm{focal}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{mirror}\mathrm{f}=\frac{\mathrm{R}}{2}=\frac{-3}{2}=-1.5\mathrm{cm}$

Given that magnification 'm' = 5

Therefore, using the magnification formula, we get

$$\begin{gathered} m=-\frac{v}{u} \\ \mathrm{v}=-\mathrm{mu} \end{gathered}$$

v=-5u
Now, using the mirror formula, we get
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$$\begin{gathered} \frac{1}{-1.5}=\frac{1}{-5u}+\frac{1}{u} \\ \Rightarrow \frac{1}{-1.5}=\frac{1}{-5u}+\frac{1}{u} \\ \Rightarrow \frac{-1}{1.5}=\frac{1}{-5u}+\frac{5}{5u}=\frac{4}{5u} \\ \Rightarrow u=\frac{\left(-1.5\right)\times 4}{5}=\frac{-6}{5} \\ \Rightarrow u=-1.2\mathrm{cm} \end{gathered}$$
The dentist should place the object at a distanc of 1.2 cm from the mirror.