At the interface between two materials having refractive indices $n_1$ and $n_2$, the critical angle for reflection of an em wave is $\theta_{1 C}$. The $n_2$ material is replaced by another material having refractive index $n_3$ such that the critical angle at the interface between…

Question

At the interface between two materials having refractive indices $n_1$ and $n_2$, the critical angle for reflection of an em wave is $\theta_{1 C}$. The $n_2$ material is replaced by another material having refractive index $n_3$ such that the critical angle at the interface between $n_1$ and $n_3$ materials is $\theta_{2 \mathrm{C}}$. If $\mathrm{n}_3 \gt \mathrm{n}_2 \gt \mathrm{n}_1 ; \frac{\mathrm{n}_2}{\mathrm{n}_3}=\frac{2}{5}$ and $\sin \theta_{2 \mathrm{C}}-\sin \theta_{1 \mathrm{C}}=\frac{1}{2}$, then $\theta_{1 \mathrm{C}}$ is

Aicharya · Accepted Answer

$\begin{aligned} & \sin 	heta_{1 C}=\frac{n_1}{n_2} \ & \sin 	heta_{2 C}=\frac{n_1}{n_3} \ & \sin 	heta_{2 C}-\sin 	heta_{1 C}=\frac{1}{2}\end{aligned}$ $\begin{aligned} & \mathrm{n}_1 \frac{\mathrm{n}_2}{\mathrm{n}_3}-\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{1}{2} \ & \mathrm{n}_1 \frac{\mathrm{n}_2}{\mathrm{n}_3}-\mathrm{n}_1=\frac{\mathrm{n}_2}{2} \ & \mathrm{n}_1\left(\frac{2}{5}-1ight)=\frac{\mathrm{n}_2}{2} \ & \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{-5}{6} \ & =\sin ^{-1}\left(-\frac{5}{6}ight)\end{aligned}$

At the interface between two materials having refractive indices \(n_1\) and \(n_2\), the critical angle for…

Solution