Physics - Geometrical Optics Question with Solution | TestHub

$AB=6 \mathrm{cm}$

$r=3 \mathrm{cm}$
$O{A}^2=O{P}^2+P{A}^2$

$R^2={\left(R-t\right)}^2+r^2$

$R^2=R^2+t^2-2Rt+r^2$

$2Rt=t^2+r^2$

$R=\frac{t^2+r^2}{2t}=\frac{{\left(0·3\right)}^2+{\left(3\right)}^2}{2\times 0·3}$

$R\approx \frac{r^2}{2t}=\frac{{\left(3\right)}^2}{2\times 0·3}=\frac{9}{0·6}=\frac{90}{6}=15 \mathrm{cm}$

Use$\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{1}{f}=\left(1\text{.}5-1\right)\left(\frac{1}{15}\right)=\frac{1}{30}$

$f=30 \mathrm{cm}$