Physics - Geometrical Optics Question with Solution | TestHub

In air$\delta =\left(1.5-1\right)A=\frac{A}{2}$
In water$\delta =\left(\frac{{\mu }_g}{{\mu }_w}-1\right)A=\left(\frac{1.5}{1.33}-1\right)A=\frac{A}{8}$
$\frac{{\delta }_{\mathrm{in} \mathrm{water}}\mathrm{ }}{{\delta }_{\mathrm{in} \mathrm{air}}}=\frac{1}{4}$
We know that,${\delta }_{\text{in air}}=\left(a{\mu }_g-1\right)A$and${\delta }_{\text{in water}}=\left(w{\mu }_g-1\right)A$.
$\frac{{\delta }_w}{{\delta }_a}=\frac{\left(\frac{{\mu }_g}{{\mu }_w}-1\right)}{\left({\mu }_9-1\right)}$
$\frac{{\delta }_w}{{\delta }_a}=\frac{\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right)}{\left(\frac{3}{2}-1\right)}$
$\frac{{\delta }_w}{{\delta }_a}=\frac{\left(\frac{9}{8}-1\right)}{\left(\frac{3}{2}-1\right)}$
$\frac{{\delta }_w}{{\delta }_a}=\frac{\frac{1}{8}}{\frac{1}{2}}=\frac{1}{4}$