Physics - Geometrical Optics Question with Solution | TestHub

${\text{R}}^2=2^2+{\left(\text{R}-\text{0.1}\right)}^2 \Rightarrow {\text{R}}^2=4+{\text{R}}^2+\text{0.01}-\text{0.2}R$

$0\text{.}2\text{R}=4\text{.}01 \Rightarrow \text{R}=\frac{4\text{.}01}{\text{0}\text{.}20}$

$=\frac{401}{20}\Rightarrow \text{R}=20\text{cm}$

from lens formula

$\frac{1}{\text{f}}=\left({}_{\text{a}}{\mathrm{\mu }}_{\text{g}}-1\right)(\frac{1}{{\text{R}}_1}-\frac{1}{{\text{R}}_2})\Rightarrow \frac{1}{\text{f}}=\left(1\text{.}5-1\right)(\frac{1}{20}-\frac{1}{\infty })$

$\frac{1}{\text{f}}=\frac{1}{40}\Rightarrow \text{f}=40\text{cm}$