A prism has an apex angle of 60° and an index of refraction of \( \sqrt{2\;} \). A light ray strikes one face at an angle of incidence of 45°. Determine:
a) The angle of emergence of the light ray;
b) The total deviation suffered by the light ray.

Problem data:

• Apex angle of the prism: Â = 60°;
• Angle of incidence of the light ray: î₁ = 45°;
• Index of refraction of the prism: \( n_2=\sqrt{2\;} \).

Problem diagram:

Assuming that the prism is in the air, the index of refraction of air is 1 (n₁ = 1).

Figure 1

The quantities with index 1 refer to the outside of the prism, and the quantities with index 2 refer to the interior of the prism.

Solution:

a) The light ray is refracted into the prism. To find the angle \( \hat r_1 \) that it makes with the normal at the face of the prism, we apply Snell's law (Figure 2).

\[ \begin{gather} n_1\sin{\hat i}_1=n_2\sin{\hat r}_2 \\[5pt] 1\times \sin 45°=\sqrt{2}\sin{\hat r}_2 \\[5pt] 1\times \frac{\sqrt{2}}{2}=\sqrt{2}\sin{\hat r}_2 \\[5pt] \sin{\hat r}_2=\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{2} \\[5pt] \sin{\hat r}_2=\frac{1}{2} \\[5pt] {\hat r}_2=\arcsin\left(\frac{1}{2}\right) \\[5pt] {\hat r}_2=30° \end{gather} \]

Figure 2

Using the equation relating the apex angle Â, the angle of refraction at the first face \( {\hat r}_2 \) and the angle of incidence at the second face \( {\hat i}_2 \) (Figure 3)

\[ \bbox[background:#99CCFF; padding:10px; border-radius:10px;] {\hat A={\hat r}_2+{\hat i}_2} \]

\[ \begin{gather} 60°=30°+{\hat i}_2 \\[5pt] {\hat i}_2=60°-30° \\[5pt] {\hat i}_2=30° \end{gather} \]

Figure 3

The light ray is refracted out of the prism at the second face. To find the angle \( {\hat r}_1 \) that it makes with the normal to the face of the prism, we apply Snell's law again (Figure 4)

\[ \begin{gather} n_2\sin{\hat i}_2=n_1\sin{\hat r}_1 \\[5pt] \sqrt{2}\times \sin 30°=1\times \sin{\hat r}_1 \\[5pt] \sqrt{2}\times \frac{1}{2}=\sin{\hat r}_1 \\[5pt] \sin{\hat r}_1=\frac{\sqrt{2}}{2} \\[5pt] {\hat r}_1=\arcsin\left(\frac{\sqrt{2}}{2}\right) \end{gather} \]

\[ \bbox[background:#FFCCCC; padding:10px; border-radius:10px;] {{\hat r}_1=45°} \]

Figure 4

b) The total angle of deviation is given by (Figure 5)

\[ \bbox[background:#99CCFF; padding:10px; border-radius:10px;] {\Delta={\hat i}_1+{\hat r}_1-\hat A} \]

\[ \Delta =45°+45°-60° \]

\[ \bbox[background:#FFCCCC; padding:10px; border-radius:10px;] {\Delta =30°} \]

Figure 5