Physics - Geometrical Optics Question with Solution | TestHub

By Snell's law
$1\sin 60=\mathrm{ }\mathrm{n}\sin {\mathrm{r}}_1\Rightarrow \sin {\mathrm{r}}_1=\frac{1}{2}{\mathrm{r}}_1={30}^{\mathrm{o}}$...(i)
By differentiating 'w.r.t' n
$\mathrm{O}=\sin {\mathrm{r}}_1+\mathrm{n}\cos {\mathrm{r}}_1\left(\frac{\mathrm{d}{\mathrm{r}}_1}{\mathrm{d}\mathrm{n}}\right)$
$=\frac{1}{2}+\sqrt{3}\mathrm{ }\left(\sqrt{\frac{3}{2}}\right)\frac{\mathrm{d}{\mathrm{r}}_1}{\mathrm{d}\mathrm{n}}$
$=\frac{\mathrm{d}{\mathrm{r}}_1}{\mathrm{d}\mathrm{n}}=\frac{1}{3}$...(ii)
By applying Snell's law
$\mathrm{n}\sin {\mathrm{r}}_2=1\sin \mathrm{\theta }$
$\mathrm{n}\sin \left(60-{\mathrm{r}}_1\right)=1\sin \mathrm{\theta }$
$\left[\therefore \mathrm{A}+{\mathrm{r}}_1+r_2\right]$
By differentiating 'w.r.t' n
$\sin (60-{\mathrm{r}}_1)-\mathrm{n}\cos \left(60-{\mathrm{r}}_1\right)\frac{\mathrm{d}{\mathrm{r}}_1}{\mathrm{d}\mathrm{n}}=\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{\theta }\frac{\mathrm{d}\mathrm{\theta }}{\mathrm{d}\mathrm{n}}\mathrm{ }$
By substituting value of '${\mathrm{r}}_1$' and$\frac{\mathrm{d}{\mathrm{r}}_1}{\mathrm{d}\mathrm{n}}$from (i) and (ii)
$\frac{\mathrm{d}\mathrm{\theta }}{\mathrm{d}\mathrm{n}}=2$