Physics - Geometrical Optics Question with Solution | TestHub

V = 8m
Magnification =$\frac{\text{v}}{\text{u}}=\frac{-1}{3}$
u = -24
also$\mu =\frac{\text{c}}{{\mathrm{V}}_{\mathrm{m}}}=\frac{{\mathrm{\lambda }}_{\mathrm{a}}}{{\mathrm{\lambda }}_{\mathrm{m}}}=\frac{3}{2}$
$\frac{1}{\text{f}}=\left(\mu -1\right)\left(\frac{1}{{\text{R}}_1}-\frac{1}{{\text{R}}_{\mathrm{2\; }}}\right)$
$\frac{1}{\text{f}}=\left(\mu -1\right)\left(\frac{1}{\infty }-\frac{1}{-\text{R}}\right)$
$\frac{1}{\text{f}}=\left(\frac{3}{2}-1\right)\frac{1}{\text{R}}$
f = 2R
from lens formula
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{8}+\frac{1}{24}=\frac{1}{\text{2R}}$
$\frac{4}{\text{24}}=\frac{1}{\text{2R}}$
$\Rightarrow \text{R}=3\text{m}$