Physics - Geometrical Optics Question with Solution | TestHub

Given $R=30 \mathrm{cm}$

So, $f=\frac{R}{2}=+15 \mathrm{cm}$

For convex mirror, virtual image is formed for real object. Thus, magnification $\left(m\right)=\frac{1}{2}$

The magnification can be written as

$m=-\frac{v}{u} ...\left(1\right)$

Equation (1) implies that

$$\begin{gathered} \frac{1}{2}=-\frac{v}{u} \\ \Rightarrow v=-\frac{u}{2} \end{gathered}$$

The relation between the object distance, image distance and the focal length of the mirror is given by

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u} ...\left(2\right)$

Equation (2) implies that

$$\begin{gathered} \frac{1}{15}=\frac{1}{{\displaystyle -\frac{u}{2}}}+\frac{1}{u} \\ =-\frac{1}{u} \\ \Rightarrow u=-15 \mathrm{cm} \end{gathered}$$