Physics - Geometrical Optics Question with Solution | TestHub

R. I. of lens,

$\mu =\frac{c}{v}=\frac{3\times {10}^8}{2\times {10}^8}=\frac{3}{2}$

From $\mathrm{\Delta ACO},$ radius of curvature of lens is,

${\mathrm{R}}^2={\mathrm{R}}_1^2+(\mathrm{R}-0.25{)}^2$

$\therefore {\mathrm{R}}^2=(2.5{)}^2+(\mathrm{R}-0.25{)}^2$

$\therefore {\mathrm{R}}^2=6.25+{\mathrm{R}}^2+0.0625-0.5\mathrm{R}$

$\therefore 0.5\mathrm{R}=6.3125$

$\therefore \mathrm{R}=12.625\mathrm{cm}$

$\mathrm{F}=\frac{\mathrm{R}}{\mu -1}=\frac{12.625}{\frac{3}{2}-1}=25.25 \mathrm{cm}$