Chapter 34, Problem 18P

A triangular glass prism with apex angle 60.0° has an index of refraction of 1.50. (a) Show that if its angle of incidence on the first surface is θ₁ = 48.6°, light will pass symmetrically through the prism as shown in Figure 34.16. (b) Find the angle of deviation δ_min for θ₁ = 48.6°. (c) What If? Find the angle of deviation if the angle of incidence on the first surface is 45.6°. (d) Find the angle of deviation if θ₁ = 51.6°.

To determine

To show: Light will pass symmetrically through the prism if the angle of incidence on the first surface ${\theta }_1=48.6°$.

Answer to Problem 18P

The light will pass symmetrically through the prism, if the angle of incidence on the first surface ${\theta }_1=48.6°$.

Explanation of Solution

Given information: The apex angle is $60\text{º}$, apex refraction is $1.50$ and the angle of refraction at first interface is $48.6\text{º}$.

The diagram for the given condition is shown below.

Figure (1)

Apply Snell’s law of refraction at the first interface.

The Snell’s law of refraction is,

$n_1\sin {\text{θ}}_1=n_2\sin {\text{θ}}_2$ (1)

Here,

$n_1$ is the index of refraction of air.

$n_2$ is the index of refraction of the medium.

${\text{θ}}_1$ is angle of refraction at the first interface.

${\text{θ}}_2$ is angle of refraction in medium.

Substitute $1$ for $n_1$, $1.50$ for $n_2$ and $48.6\text{º}$ for ${\text{θ}}_1$ in equation (1).

$\begin{array}{c}1\times \sin 48.6\text{º}=1.50\times \sin {\text{θ}}_2\\ \sin {\text{θ}}_2=0.50\\ {\text{θ}}_2=30\text{º}\end{array}$

Apply Snell’s law of refraction at the second interface.

The Snell’s law of refraction is,

$n_2\sin {\text{θ}}_2=n_1\sin {\text{θ}}_4$ (2)

Here,

$n_1$ is the index of refraction of air.

$n_2$ is the index of refraction of the medium.

${\text{θ}}_4$ is angle of refraction at the second interface.

${\text{θ}}_2$ is angle of refraction in medium.

Substitute $1$ for $n_1$, $1.50$ for $n_2$ and $\text{30º}$ for ${\text{θ}}_2$ in equation (2).

$\begin{array}{c}1.50\times \sin 30\text{º}=1\times \sin {\text{θ}}_4\\ {\text{θ}}_4\approx 48.6\text{º}\end{array}$

Since, ${\text{θ}}_1={\text{θ}}_4$, so light will pass symmetrically through the prism.

Conclusion:

Therefore, the light will pass symmetrically through the prism.

To determine

The angle of minimum deviation ${\delta }_{\text{min}}$ for ${\text{θ}}_1=48.6\text{º}$.

Answer to Problem 18P

The angle of minimum deviation ${\delta }_{\text{min}}$ for ${\text{θ}}_1=48.6\text{º}$ is $37.2\text{º}$.

Explanation of Solution

Given information: The apex angle is $60\text{º}$, apex refraction is $1.50$ and the angle of refraction at first interface is $48.6\text{º}$.

The angle of minimum deviation ${\delta }_{\text{min}}$ is,

${\delta }_{\text{min}}={\text{θ}}_1+{\text{θ}}_4-\Phi$ (3)

Here,

${\delta }_{\text{min}}$ is the angle of minimum deviation.

$\Phi$ is the apex angle.

${\theta }_1$ is the orientation angle at first interface.

${\text{θ}}_4$ is the orientation angle at second interface

Substitute $60\text{º}$ for $\Phi$ $48.6\text{º}$ for ${\text{θ}}_4$ and $48.6\text{º}$ for ${\text{θ}}_1$ in equation (3).

$\begin{array}{c}{\delta }_{\text{min}}=48.6\text{º}+48.6\text{º}-60\text{º}\\ =37.2\text{º}\end{array}$

Conclusion:

Therefore, the orientation angle in the proper frame is $37.2\text{º}$.

To determine

The angle of minimum deviation ${\delta }_{\text{min}}$ for ${\text{θ}}_1=45.6\text{º}$.

Answer to Problem 18P

The angle of minimum deviation ${\delta }_{\text{min}}$ for ${\text{θ}}_1=45.6\text{º}$ is $31.2\text{º}$.

Explanation of Solution

Given information: The apex angle is $60\text{º}$, apex refraction is $1.50$ and the angle of refraction at first interface is $45.6\text{º}$.

Apply Snell’s law of refraction at the first interface.

The Snell’s law of refraction is,

$n_1\sin {\text{θ}}_1=n_2\sin {\text{θ}}_2$

Here,

$n_1$ is the index of refraction of air.

$n_2$ is the index of refraction of the medium.

${\text{θ}}_1$ is angle of refraction at the first interface.

${\text{θ}}_2$ is angle of refraction in medium.

Substitute $1$ for $n_1$, $1.50$ for $n_2$ and $45.6\text{º}$ for ${\text{θ}}_1$ in above equation.

$\begin{array}{c}1\times \sin 45.6\text{º}=1.50\times \sin {\text{θ}}_2\\ \sin {\text{θ}}_2=0.476\\ {\text{θ}}_2=28.44\text{º}\end{array}$

Apply Snell’s law of refraction at the second interface.

The Snell’s law of refraction is,

$n_2\sin {\text{θ}}_2=n_1\sin {\text{θ}}_4$

Here,

$n_1$ is the index of refraction of air.

$n_2$ is the index of refraction of the medium.

${\text{θ}}_4$ is angle of refraction at the second interface.

${\text{θ}}_2$ is angle of refraction in medium.

Substitute $1$ for $n_1$, $1.50$ for $n_2$ and $\text{28}\text{.44º}$ for ${\text{θ}}_2$ in above equation.

$\begin{array}{c}1.50\times \sin 28.44\text{º}=1\times \sin {\text{θ}}_4\\ {\text{θ}}_4=45.6\text{º}\end{array}$

The angle of minimum deviation ${\delta }_{\text{min}}$ is,

${\delta }_{\text{min}}={\text{θ}}_1+{\text{θ}}_4-\Phi$ (3)

Here,

${\delta }_{\text{min}}$ is the angle of minimum deviation.

$\Phi$ is the apex angle.

${\theta }_1$ is the orientation angle at first interface.

${\text{θ}}_4$ is the orientation angle at second interface

Substitute $60\text{º}$ for $\Phi$ $45.6\text{º}$ for ${\text{θ}}_4$ and $45.6\text{º}$ for ${\text{θ}}_1$ in above equation.

$\begin{array}{c}{\delta }_{\text{min}}=45.6\text{º}+45.6\text{º}-60\text{º}\\ =31.2\text{º}\end{array}$

Conclusion:

Therefore, the orientation angle in the proper frame is $31.2\text{º}$.

To determine

The angle of minimum deviation ${\delta }_{\text{min}}$ for ${\text{θ}}_1=51.6\text{º}$.

Answer to Problem 18P

The angle of minimum deviation ${\delta }_{\text{min}}$ for ${\text{θ}}_1=51.6\text{º}$ is $43.2\text{º}$.

Explanation of Solution

Given information: The apex angle is $60\text{º}$, apex refraction is $1.50$ and the angle of refraction at first interface is $51.6\text{º}$.

Apply Snell’s law of refraction at the first interface.

The Snell’s law of refraction is,

$n_1\sin {\text{θ}}_1=n_2\sin {\text{θ}}_2$

Here,

$n_1$ is the index of refraction of air.

$n_2$ is the index of refraction of the medium.

${\text{θ}}_1$ is angle of refraction at the first interface.

${\text{θ}}_2$ is angle of refraction in medium.

Substitute $1$ for $n_1$, $1.50$ for $n_2$ and $51.6\text{º}$ for ${\text{θ}}_1$ in above equation.

$\begin{array}{c}1\times \sin 51.6\text{º}=1.50\times \sin {\text{θ}}_2\\ {\text{θ}}_2=31.5\text{º}\end{array}$

Apply Snell’s law of refraction at the second interface.

The Snell’s law of refraction is,

$n_2\sin {\text{θ}}_2=n_1\sin {\text{θ}}_4$

Here,

$n_1$ is the index of refraction of air.

$n_2$ is the index of refraction of the medium.

${\text{θ}}_4$ is angle of refraction at the second interface.

${\text{θ}}_2$ is angle of refraction in medium.

Substitute $1$ for $n_1$, $1.50$ for $n_2$ and $31.5\text{º}$ for ${\text{θ}}_2$ in above equation.

$\begin{array}{c}1.50\times \sin 31.5\text{º}=1\times \sin {\text{θ}}_4\\ {\text{θ}}_4=51.6\text{º}\end{array}$

The angle of minimum deviation ${\delta }_{\text{min}}$ is,

${\delta }_{\text{min}}={\text{θ}}_1+{\text{θ}}_4-\Phi$

Here,

${\delta }_{\text{min}}$ is the angle of minimum deviation.

$\Phi$ is the apex angle.

${\theta }_1$ is the orientation angle at first interface.

${\text{θ}}_4$ is the orientation angle at second interface

Substitute $60\text{º}$ for $\Phi$ $51.6\text{º}$ for ${\text{θ}}_4$ and $51.6\text{º}$ for ${\text{θ}}_1$ in above equation.

$\begin{array}{c}{\delta }_{\text{min}}=51.6\text{º}+51.6\text{º}-60\text{º}\\ =43.2\text{º}\end{array}$

Conclusion:

Therefore, the orientation angle in the proper frame is $43.2\text{º}$.