Physics - Geometrical Optics Question with Solution | TestHub

$\mathrm{R}=60 \mathrm{cm}$, ${\mathrm{R}}_2=20 \mathrm{cm}$

Let ${\mathrm{P}}_1$ be power of lens of water ,${\mathrm{P}}_2$ be power of glass lens, and ${\mathrm{P}}_3$ be power of concave mirror

${\mathrm{P}}_{\mathrm{eq}}= 2{\mathrm{P}}_1 + 2{\mathrm{P}}_2 + {\mathrm{P}}_3$

$\frac{1}{-{\text{f}}_{\text{eq}}}=\frac{2}{{\text{f}}_1}+\frac{2}{{\text{f}}_2}-\frac{1}{{\text{f}}_3}\mathrm{ \; \; \; \; \; \; \; \; \; \; \; ....(1)}$

for the lens of water

$\frac{1}{{\text{f}}_1}=\left(\frac{4}{3}-1\right)\left(\frac{1}{\infty }-\frac{1}{-60}\right)$

$\frac{1}{{\text{f}}_1}=\frac{1}{180}$

for glass lens

$\frac{1}{{\text{f}}_2}=(\frac{3}{2}-1)(\frac{1}{-60}-\frac{1}{-20})$

$\frac{1}{{\text{f}}_2}=\frac{1}{60}$

$\text{For concave mirror}{\text{f}}_3=\frac{\text{R}}{2}$

${\mathrm{f}}_3=-10 \mathrm{cm}$

$\frac{1}{-{\text{f}}_{\text{eq}}}=\frac{2}{\text{180}}+\frac{2}{\text{60}}-\frac{1}{-\text{10}}$

$\frac{1}{-{\text{f}}_{\text{eq}}}=\frac{2\text{6}}{\text{180}}\Rightarrow {\text{f}}_{\text{eq}}=\frac{-\text{90}}{\text{13}}\text{cm}$