Question

In the figure, a ray is incident on one face of a triangular glass prism in air. The angle of incidence $\mathit{\theta }$is chosen so that the emerging ray also makes the same angle $\mathit{\theta }$with the normal to the other face. Show that the index ofrefraction of the glass prism is given by

$\mathit{\psi }\mathit{n}\mathbf{=}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\mathbf{\psi }\mathbf{+}\mathbf{\varphi }\mathbf{)}}{\mathbf{s}\mathbf{i}\mathbf{n}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\varphi }}$

where$\mathit{\varphi }$ is the vertex angle of the prism and $\mathit{\psi }$is thedeviation angle, the total angle through which the beam is turned in passing through the prism. (Under these conditions the deviation angle has the smallest possible value, which is called theangle of minimum deviation.)

Figure:

Short answer

The index of refraction of the glass prism in given by$n=\frac{\sin \frac{1}{2}(\psi +\varphi )}{\sin \frac{1}{2}\varphi }$ is shown in theexplanation part.

Step-by-step solution

Given

The figure is given in which a ray is incident on one face of a triangular glass prism in the air.

Understanding the concept

Write for the angle of refraction in terms of thevertex angle from the geometry of the prism and the ray diagram. Thenusing the formula for theexterior angle of triangle,solve for the angle of incidence in terms of thevertex angle. Then, using Snell’s law of refraction, show that the index of refraction of the glass prism in given by $\mathit{n}\mathbf{=}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\mathbf{\psi }\mathbf{+}\mathbf{f}\mathbf{)}}{\mathbf{s}\mathbf{i}\mathbf{n}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{f}}$.

Formula:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Show that the index of refraction  n of the glass prism is given by n=sin12(ψ+f)sin12f 

Consider the diagram for the condition as:

Let${\theta }_2$be the angle of refraction.

From the figure, we can write

$\begin{array}{l}{\theta }_2+\alpha ={90}^0\\ {\theta }_2={90}^0-\alpha \end{array}$

Consider the equation for the angle as:

$\varphi +2\alpha ={180}^0$

Hence,

$\alpha =\frac{({180}^0-\varphi )}{2}$

Solve for the second angle as:

$\begin{array}{l}{\theta }_2=90°-\frac{(180°-\varphi )}{2}\\ {\theta }_2=\frac{(180°-180°+\varphi )}{2}\end{array}$

${\theta }_2=\frac{\varphi }{2}$

Using the formula for the exterior angle of triangle and solve as:

$\begin{array}{c}\psi =2(\theta -{\theta }_2)\\ \psi =2(\theta -\frac{\varphi }{2})\\ 2\theta =\psi +\varphi \\ \theta =\frac{1}{2}(\varphi +\psi )\end{array}$

Snell’s law given as:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Here, $n_1=1,{\theta }_1=\theta ,n_2=n,{\theta }_2={\theta }_2$. Therefore,

$\begin{array}{l}\sin \theta =n\sin {\theta }_2\\ n=\frac{\sin \theta }{\sin {\theta }_2}\\ n=\frac{\sin \frac{1}{2}(\varphi +\psi )}{\sin \frac{\varphi }{2}}\end{array}$