Physics - Geometrical Optics Question with Solution | TestHub

For reflection from a concave mirror,
$\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\Rightarrow \frac{1}{\mathrm{v}}-\frac{1}{15}=\mathrm{ }\frac{-1}{10}$
$\frac{1}{\mathrm{v}}=\frac{1}{15}-\frac{1}{10}=\frac{-1}{30}$
$\therefore \mathrm{v}=-30$
Magnification$\left({\mathrm{m}}_1\right)=-\frac{\mathrm{v}}{\mathrm{u}}=-2$
Now for refraction from lens,
$\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$
$\therefore$Magnification$\left({\mathrm{m}}_2\right)=\frac{\mathrm{v}}{\mathrm{u}}=-1$
$\therefore {\mathrm{M}}_1={\mathrm{m}}_1{\mathrm{m}}_2=2\mathrm{ }$
Now when the set-up is immersed in liquid, no effect for the image formed by mirror.
We have$\left({\mathrm{\mu }}_{\mathrm{L}}-1\right)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)=\frac{1}{10}$
$\Rightarrow \left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)=\frac{1}{5}$
When lens is immersed in liquid,
$\frac{1}{{\mathrm{f}}_{\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{s}}}=\left(\frac{{\mathrm{\mu }}_{\mathrm{L}}}{{\mathrm{\mu }}_{\mathrm{S}}}-1\right)\mathrm{ }\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)=\frac{2}{7}\times \frac{1}{5}=\frac{2}{35}$
$\therefore \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{{\mathrm{f}}_{\mathrm{L}\mathrm{i}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{d}}}$
$\Rightarrow \frac{1}{\mathrm{v}}=\frac{2}{35}-\frac{1}{20}=\frac{8-7}{140}=\frac{1}{140}$
$\therefore$Magnification$=-\frac{140}{20}=-7$
$\therefore {\mathrm{M}}_2=2\times 7=14$
$\therefore \left|\frac{{\mathrm{M}}_2}{{\mathrm{M}}_1}\right|=7$