Chapter 23, Problem 85GP

To determine

The index of refraction of the prism.

Answer to Problem 85GP

Solution:

Since the ray is totally internally reflected at the opposite side, we will use the concept of critical angle. Critical angle for glass air is ${42}^0$.

Explanation of Solution

If the ray strike at the angle greater than critical angle, then only it will be totally internally reflected.

Given:

Apex angle of prism $A={65}^0$

Incident angle of the ray at the first side is $i_1={45}^0$

Formula used:

(1)Apex angle $A=\left(r_1+r_2\right),$ where r₁ and r₂ are refraction angle on both the sides.

(2)Critical angle of glass –air interface is ${42}^0$.

(3)Snell’s law $n_1\sin i=n_2\sin r$, where $n_1$ and $n_2$ are refractive index of two mediums.

Calculation:

Assume angle of refraction at air prism side is $r_1$ and at prism –air side is $r_2$

For internal reflection $r_2$ should be more than ${42}^0$, minimum should be ${42}^0$.

So, from apex angle formula

$\begin{array}{l}A=\left(r_1+r_2\right)=r_1+{42}^0\hfill \\ r_1=A-{42}^0={65}^0-{42}^0={23}^0\hfill \end{array}$

from Snell’s law at air-prism interface on first side

$\begin{array}{l}{n}_1\sin i_1=n_2\sin r_1\hfill \\ 1\times \sin {45}^0=n_2\sin {23}^0\hfill \\ 1\times \left(1/\surd 2\right)=n_2\times 0\cdot 391\hfill \\ n_2=0\cdot 7071/0\cdot 391=1\cdot 808=1\cdot 81\hfill \end{array}$