Chapter 23, Problem 74P

(a)

To determine

The ray diagram with all rays emerging from the prism.

Answer to Problem 74P

Explanation of Solution

The angle of prism is $45°$ and refractive index of the glass is $1.50$.

Write the law of reflection

  ${\theta }_{\text{i}}={\theta }_{\text{r}}$                                                                                                                        (I)

Here, ${\theta }_{\text{i}}$ is the angle of incidence and ${\theta }_{\text{r}}$ is the angle of reflection.

Write the expression for ${\theta }_{\text{i}}$ using the small isosceles triangle below the incident ray

  ${\theta }_{\text{i}}+45°=90°$                                                                                                           (II)

Find ${\theta }_{\text{i}}$ using the above equation

  ${\theta }_{\text{i}}=45°$

Write the expression for critical angle

  ${\theta }_{\text{c}}={\sin }^{-1}\frac{n_{\text{t}}}{n_{\text{i}}}$                                                                                                          (III)

Here, ${\theta }_{\text{c}}$ is the critical angle, $n_{\text{t}}$ is the refractive index of the refractive medium and $n_{\text{i}}$ is the refractive index of the incident medium.

Substitute $1.00$ for $n_{\text{t}}$ and $1.50$ for $n_{\text{t}}$ in (III) to find ${\theta }_{\text{c}}$

  $\begin{array}{c}{\theta }_{\text{c}}={\sin }^{-1}\frac{1.00}{1.50}\\ =41.8°\end{array}$

Since ${\theta }_{\text{i}}>{\theta }_{\text{c}}$, there is no transmitted ray because of total internal reflection and the ray travels downward.

(b)

To determine

The ray diagram when the prism is immersed in water.

Answer to Problem 74P

The ray diagram when the prism is immersed in water is given in figure 2.

Explanation of Solution

The refractive index of the water is $1.33$.

Substitute $1.33$ for $n_{\text{t}}$ and $1.50$ for $n_{\text{i}}$ in (III) to find ${\theta }_{\text{c}}$

  $\begin{array}{c}{\theta }_{\text{c}}={\sin }^{-1}\frac{1.00}{1.33}\\ =62.5°\end{array}$

Since ${\theta }_{\text{i}}<{\theta }_{\text{c}}$, there is a transmitted ray at an angle given by Snell’s law.

Write the expression for Snell’s law

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$                                                                                             (IV)

Here, ${\theta }_1$ is the incident angle, ${\theta }_2$ is the angle of refraction, $n_1$ is the refractive index of the refractive medium and $n_2$ is the refractive index of the incident medium.

Rearrange for ${\theta }_2$

  ${\theta }_2={\sin }^{-1}\left(\frac{n_1}{n_2}\sin {\theta }_1\right)$                                                                                            (V)

Substitute $1.50$ for $n_1$, $1.33$ for $n_2$ and $45°$ for ${\theta }_1$ in (V) to find ${\theta }_2$

  $\begin{array}{c}{\theta }_2={\sin }^{-1}\left(\frac{1.50}{1.33}\sin 45°\right)\\ =53°\end{array}$

Thus, the angle of refraction is $53°$.

(c)

To determine

The ray diagram when the prism is immersed in sugar solution.

Answer to Problem 74P

The ray diagram when the prism is immersed in sugar solution is given in figure 3.

Explanation of Solution

The refractive index of the sugar solution is $1.50$ which is same as the refractive index of the prism. Hence no refraction occurs and the ray emerges without any change in direction.

The ray diagram for prism immersed in sugar solution is given below