Chapter 23, Problem 84GP

To determine

The angle at which light ray emerges from the opposite face.

Answer to Problem 84GP

Solution:

Angle at which light emerge from other side of the prism is 56∙25⁰.

Explanation of Solution

In this question Snell’s law would be used twice to find the incidence angle at other side of the prism.

Given:

prism is equilateral, $A={60}^0$

Incidence angle at first side of prism is $i_1={45}^0$

Refractive index of prism $n_2=1\cdot 54$

Formula used:

(1)Snell’s law $n_1\sin i=n_2\sin r$, where i is incidence angle in one medium and r is refracted angle in other medium. n₁ and n₂ are refractive index of both medium.

(2)Angle of prism $A=\left(r_1+r_2\right)$, where A is prism angle and r₁ and r₂ are refracted angle on both the sides of prism.

Calculation:

For air-prism interface

$\begin{array}{l}{n}_1\sin i_1=n_2\sin r_1\hfill \\ 1\times \sin {45}^0=1\cdot 54\times \sin r_1\hfill \\ \sin r_1=\left(1/\surd 2\right)/1\cdot 54=0\cdot 7071/1\cdot 54=0\cdot 459\hfill \\ r_1={\sin }^{-1}\left(0\cdot 459\right)=27\cdot {32}^0\hfill \\ \therefore A=\left(r_1+r_2\right)\hfill \\ r_2=A-r_1{}_{}={60}^0-27\cdot {32}^0=32\cdot {68}^0\hfill \end{array}$

So now for prism- air interface on other side

$\begin{array}{l}{n}_2\sin r_2=n_1\sin i_2\hfill \\ 1\cdot 54\times \sin \left(32\cdot {68}^0\right)=1\times \sin i_2\hfill \end{array}$

$\begin{array}{l}\sin i_2=1\cdot 54\times 0\cdot 5399=0\cdot 8315\\ i_2={\sin }^{-1}\left(0\cdot 8315\right)=56\cdot {25}^0\end{array}$