Question

The refracting angle of a prism ${60}^o$.The refractive index of the material of the prism is $\sqrt{\frac{7}{3}}$. The limiting angle of incidence of a ray that will be transmitted through the prism in this case will be :

• A. ${30}^0$
• B. ${45}^o$
• C. ${40}^o$
• D. ${50}^o$

Answer 1

Answer: (A)

${30}^0$

$\begin{array}{c}\text{Given}A={60}^{\circ },\mu =\sqrt{\frac{7}{3}}\\ \text{we know that}{\gamma }_1+r_2=A\\ \Rightarrow r_1+r_2={60}^{\circ }\end{array}$

$\begin{array}{c}\text{Snell's law at first refraction}\\ \begin{array}{c}\sin i=\mu \sin {\gamma }_1\\ \mathrm{Sin}{\gamma }_1=\sqrt{\frac{3}{7}}\sin i\end{array}\\ \text{Snell's law at second refraction}\end{array}$

$\begin{array}{cc}\mu \sin r_2& =1\\ & \Rightarrow \sin r_2=\sqrt{\frac{3}{7}}\\ r_2={\sin }^{-1}\left(\sqrt{\frac{3}{7}}\right)& \\ r_2=& {40.89}^{\circ }\Rightarrow r_1={19.1066}^{\circ }\end{array}$

$\begin{array}{cc}\Rightarrow \mathrm{Sin}{r}_1& =0.327\\ \sin i& =\sqrt{\frac{7}{3}}\sin r_1\\ \sin i& =1/2\\ i& ={30}^{\circ }\end{array}$