Chapter 23, Problem 81GP

To determine

We must find the minimum incidence angle of ray if it is to emerge from the opposite side and not to be internally reflected.

Answer to Problem 81GP

Solution:

${67.35}^o$

Explanation of Solution

When there is maximum deviation, the ray would get totally internally reflected. So, for making the incidence ray minimum such as that the ray emerged from other side, we should keep deviation minimum.

Given:

Angle of prism $A=75^0$

Refractive index of prism $=n=1\cdot 58$

Formula used:

Snell’s law $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

In a triangle, internal angle= sum of opposite external angles

Calculation:

For minimum incidence angle, the emergent angle should be maximum i.e. 90^o

Using Snell’s law:

$\begin{array}{l}1.58\sin i_2=1.00\sin {90}^o\\ \Rightarrow i_2={\sin }^{-1}\left(\frac{1}{1.58}\right)={39.26}^o\\ r_1={75}^o-{39.26}^o={35.74}^o\\ n_1\sin i_1=n_2\sin r_1\\ 1.00\sin i_1=1.58\sin {35.74}^o\\ i_1={\sin }^{-1}0.922={67.35}^o\end{array}$