For the given incident ray as shown in the figure, the condition of the total internal reflection of this ray and the minimum refractive index of prism will be:

• A

  $\frac{\sqrt{3}+1}{2}$
• B

  $\frac{\sqrt{2}+1}{2}$
• Correct answer
  C

  $\sqrt{\frac{3}{2}}$
• D

  $\sqrt{\frac{7}{6}}$

Solution

For given condition and use snell's law 1. $\sin 45^{\circ}=\mu \sin \left(90-q_C\right)$ $\begin{array}{ll} \frac{1}{\sqrt{2}}=\mu \cos \theta_C=\sqrt{\mu^2-1} \\ \Rightarrow \quad \mu^2=1+\frac{1}{2}=\frac{3}{2} \\ \Rightarrow \quad \mu=\sqrt{\frac{3}{2}} \end{array}$ $\begin{array}{ll} \therefore & \sin \theta_c=\frac{1}{\mu} \\ \Rightarrow \quad \cos \theta_c=\sqrt{\frac{\mu^2-1}{\mu}} \end{array}$

Hence, the correct answer is option: 'C'.