Physics - Geometrical Optics Question with Solution | TestHub

$\because \mu =\frac{\mathrm{R}\mathrm{e}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{d}\mathrm{e}\mathrm{p}\mathrm{t}\mathrm{h}}{\mathrm{a}\mathrm{p}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{ }\mathrm{d}\mathrm{e}\mathrm{p}\mathrm{t}\mathrm{h}}$

Where,$\mu$refractive index of medium

$\Rightarrow \mathrm{A}\mathrm{p}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{ }\mathrm{d}\mathrm{e}\mathrm{p}\mathrm{t}\mathrm{h}=\frac{\mathrm{R}\mathrm{e}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{d}\mathrm{e}\mathrm{p}\mathrm{t}\mathrm{h}}{\mu }$

But,$\mu \propto \frac{1}{\lambda } \because {\lambda }_v<{\lambda }_g<{\lambda }_y<{\lambda }_r$

Thus,${\mu }_v$is greater to all of them

$\therefore$Apparent depth is minimum for violet.

It means that the violet letter appear raised more.