Physics - Geometrical Optics Question with Solution | TestHub

Here,${\mathrm{\mu }}_{\mathrm{g}}=1.5=\frac{3}{2}, {\mathrm{\mu }}_w ᷿=\frac{4}{3}$

$R_1=R,\Rightarrow R_2=-R$

From, lens maker's formula

$\frac{1}{f}=\left(\mathrm{\mu }-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e} \mathrm{\mu }=\frac{{\mathrm{\mu }}_L}{{\mathrm{\mu }}_M}$

$\therefore \frac{1}{f_aₐ}=\left(\frac{{\mathrm{\mu }}_{\mathrm{g}}}{{\mathrm{\mu }}_aₐ}-1\right)\left(\frac{1}{R}-\frac{1}{\left(-R\right)}\right)$



$\Rightarrow \frac{1}{f_a}=\frac{1}{R}$

Also

$\frac{1}{f_w ᷿}=\left(\frac{{\mathrm{\mu }}_{\mathrm{g}}}{{\mathrm{\mu }}_wₐ}-1\right)\left(\frac{1}{R}-\frac{1}{\left(-R\right)}\right)$



$\mathrm{O}\mathrm{r}\mathrm{ }\mathrm{ }\frac{1}{f_w}=\frac{1}{4R}$

Or$f_w ᷿=4R$