Physics - Geometrical Optics Question with Solution | TestHub

First refraction,$\mathrm{u}=-40\mathrm{ }\mathrm{c}\mathrm{m};\mathrm{f}=+20\mathrm{c}\mathrm{m}$
From lens formula,$\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$
$\frac{1}{\mathrm{v}}-\frac{1}{-40}=\frac{1}{20}\Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{20}-\frac{1}{40}$
$\mathrm{v}=+40\mathrm{c}\mathrm{m}$
${\mathrm{m}}_1=-1$
First reflection,$\mathrm{u}=-20\mathrm{c}\mathrm{m};\mathrm{f}=-10\mathrm{c}\mathrm{m}$
From mirror formula,$\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$
$\frac{1}{\mathrm{v}}-\frac{1}{20}=\frac{1}{-10}$
$\frac{1}{\mathrm{v}}=\frac{1}{-10}+\frac{1}{20}\Rightarrow \mathrm{v}=-20\mathrm{c}\mathrm{m}$
${\mathrm{m}}_2=-1$
Second refraction,$\mathrm{u}=-40\mathrm{c}\mathrm{m},\mathrm{f}=+20\mathrm{c}\mathrm{m}$
From lens formula,$\frac{1}{\mathrm{v}}-\frac{1}{-40}=\frac{1}{20}\Rightarrow \mathrm{v}=+40\mathrm{ }\mathrm{c}\mathrm{m}$
${\mathrm{m}}_3=-1$
${\mathrm{m}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}={\mathrm{m}}_1\times {\mathrm{m}}_2\times {\mathrm{m}}_3=-1$
Finally, image will be at distance$40\mathrm{ }\mathrm{c}\mathrm{m}\mathrm{ }$from lens and is of same size as object.