Physics - Geometrical Optics Question with Solution | TestHub

Location of image of $\mathrm{A}:-$
$\begin{array}{rl}& \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\Rightarrow \frac{1}{\mathrm{v}}-\frac{1}{-30}=\frac{1}{20}\Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{60}\Rightarrow \mathrm{v}=60\mathrm{cm}\\ & \therefore \mathrm{m}=2\end{array}$
Since size of object is small wrt the location hence
$\begin{array}{rl}& dv=m^{2}du\Rightarrow dv=4\times 1=4\mathrm{cm}\\ & h_i=m{h}_0\Rightarrow h_i(dy)=2\times 2=4\mathrm{cm}\end{array}$
$\therefore$ Angle made with principle axis $=-45^{\circ }$