Question: Suppose the prism of Fig. 33-53 has apex angle $\mathbf{\varphi }\mathbf{=}{\mathbf{60}}^{\mathbf{o}}$and index of refraction$\mathbf{n}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{60}$ . (a) What is the smallest angle of incidence$\mathit{\theta }$ for which a ray can enter the left face of the prism and exit the right face? (b) What angle of incidence$\mathit{\theta }$ is required for the ray to exit the prism with an identical angle$\mathit{\theta }$ for its refraction, as it does in Fig. 33-53?

Figure:

The apex angle of the prism is, $\varphi =60.0^o$

The Refractive index is, $n=1.60$

Using the concept of total internal reflection and critical angle, we can determine the results.

Formula:

Critical angle,${\mathrm{\theta }}_{\mathrm{c}}={\sin }^{-1}\left(\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}\right)$

If the original ray is incident at point A, the prism vertex is at point B, and the point where the interior ray strikes the right surface of the prism is the point C.

The angle between the line AB and the interior ray is $\beta$the angle between the line AC and the interior ray is $\alpha$.

When the incident ray is at the minimum angle for which light can exit the prism, the light exits along the second face. That is, the angle of refraction at the second face is ${90}^{\circ }$and the angle of incidence there for the interior ray is the critical angle for the total internal reflection.

Let ${\theta }_1$ be the angle of incidence for the original incident ray; let be the angle of refraction at the first face, and let be the angle of incidence at the second face.

The law of refraction applied to point C, yields $n\sin {\theta }_3=1$

$\begin{array}{rcl}{\sin }_3& =& \frac{1}{n}\\ & =& \frac{1}{1.60}\\ & =& 0.625\end{array}$

The interior angles of the triangle ABC must sum to ${180}^{\circ }$.

So, $\alpha +\beta ={180}^{\circ }$

Now,

$\begin{array}{c}\alpha ={90}^{\circ }-{\theta }_3\\ =51.{32}^{\circ }\end{array}$

So,

$\begin{array}{c}\beta ={120}^{\circ }-51.{32}^{\circ }\\ =69.{68}^o\end{array}$

Thus,

$\begin{array}{rcl}{\theta }_2& =& {90}^{\circ }-\beta \\ & =& 21.{32}^{\circ }\end{array}$

The law of refraction, applied to point A , yields

.$\begin{array}{rcl}\sin {\theta }_1& =& n\sin {\theta }_2\\ \sin {\theta }_1& =& 1.60\times \sin 21.{32}^{\circ }\\ \sin {\theta }_1& =& 0.5817\\ {\theta }_1& =& {\sin }^{-1}0.5817\\ {\theta }_1\text{'}& =& 35.6^{\circ }\end{array}$

Thus, ${\theta }_1=35.6^{\circ }$

We apply the law of refraction to point C.

Since the angle of refraction, there is the same as the angle of incidence at A,

${\mathrm{nsin\theta }}_3=\mathrm{sin\theta }$

Now, $\alpha +\beta ={120}^o$

$\alpha ={90}^o-{\theta }_3$,

And $\beta ={90}^{\circ }-{\theta }_2$, as before.

This means ${\theta }_2+{\theta }_3={60}^{\circ }$

Thus, the law of refraction leads to

localid="1662998883723" $$\begin{gathered} {\mathrm{sin\theta }}_1=\mathrm{nsin}\left({60}^{\circ }-{\mathrm{\theta }}_2\right) \\ {\mathrm{sin\theta }}_1=\mathrm{nsin}{60}^{\circ }{\mathrm{cos\theta }}_2-\mathrm{nsin}{60}^{\circ }{\mathrm{cos\theta }}_2 \\ \mathrm{From}\mathrm{the}\mathrm{trigonometry}\mathrm{identity} \\ \sin (\mathrm{A}-\mathrm{B})=\mathrm{sinAcosB}-\mathrm{cosAsinB} \end{gathered}$$

Next, we apply the law of refraction to point A.

$$\begin{gathered} \sin {\theta }_1=n\sin {\theta }_2 \\ \sin {\theta }_2=\frac{\sin {\theta }_1}{n} \end{gathered}$$

Hence,

${\mathrm{cos\theta }}_2=\sqrt{1-{\sin }^2{\mathrm{\theta }}_2}=\sqrt{1-\left(\frac{1}{{\mathrm{n}}^2}\right){\sin }^2{\mathrm{\theta }}_1}$

Now,

$sin{\theta }_1=n_{1}sin{60}^0\sqrt{1-\left(\frac{1}{n^2}\right){sin}^2{\theta }_1}-cos60\sin {\theta }_1$

$\left(1+\cos {60}^{\circ }\right)\sin {\theta }_1=\sin {60}^{\circ }\sqrt{n^2-{\sin }^2{\theta }_1}$

Squaring on both sides and solving it for ${\theta }_1$ we get

$$\begin{gathered} \sin {\theta }_1=\frac{n\sin {60}^o}{\sqrt{(1+\cos \left({60}^{\circ }\right)+{\sin }^2{60}^0}} \\ \sin {\theta }_1=\frac{1.60\times \sin {60}^o}{\sqrt{(1+\cos \left({60}^{\circ }\right)+{\sin }^2{60}^0}} \end{gathered}$$

That gives

$\sin {\theta }_1=0.80$

And hence,

${\theta }_1=53.1^{\circ }$