For the given incident ray as shown in the figure, the condition of the total internal reflection of this…
- $\frac{\sqrt{3}+1}{2}$
- $\frac{\sqrt{2}+1}{2}$
- $\sqrt{\frac{3}{2}}$
- $\sqrt{\frac{7}{6}}$
Solution
For given condition and use snell's law
1. $\sin 45^{\circ}=\mu \sin \left(90-q_C\right)$
$\begin{array}{ll}
\frac{1}{\sqrt{2}}=\mu \cos \theta_C=\sqrt{\mu^2-1} \\
\Rightarrow \quad \mu^2=1+\frac{1}{2}=\frac{3}{2} \\
\Rightarrow \quad \mu=\sqrt{\frac{3}{2}}
\end{array}$
$\begin{array}{ll}
\therefore & \sin \theta_c=\frac{1}{\mu} \\
\Rightarrow \quad \cos \theta_c=\sqrt{\frac{\mu^2-1}{\mu}}
\end{array}$Asked in: NEET 2002