Chapter 22, Problem 48AP

(a)

To determine

The maximum value of angle for which the refracted ray will undergo total internal reflection at the left vertical face of the block.

Answer to Problem 48AP

For any angle less than equal to $90°$ will occur the total internal reflection.

Explanation of Solution

The following ray diagram show path of the light.

Formula to calculate the angle ${\theta }_3$ is,

${\theta }_3={\sin }^{-1}\left(\frac{n_a}{n_p}\right)$

• ${\theta }_3$ is the angle
• $n_a\text{and}{n}_p$ are the refractive index of air and polystyrene

Substitute $1$ for $n_a$. $1.49$ for $n_p$ to find ${\theta }_3$.

$\begin{array}{c}{\theta }_3={\sin }^{-1}\left(\frac{1}{1.49}\right)\\ =42.2°\end{array}$

Formula to calculate the angle ${\theta }_2$ is,

${\theta }_2=\left(90°-{\theta }_3\right)$

• ${\theta }_2$ is the angle

Substitute $42.2°$ for ${\theta }_3$ to find ${\theta }_2$.

$\begin{array}{c}{\theta }_2=\left(90°-42.2°\right)\\ =47.8°\end{array}$

Formula to calculate the angle ${\theta }_1$ is,

$\sin {\theta }_1=\left(\frac{n_p\sin {\theta }_2}{n_a}\right)$

• ${\theta }_1\text{and}{\theta }_2$ are the angle
• $n_a\text{and}{n}_p$ are the refractive index of air and polystyrene

Substitute $1$ for $n_a$. $1.49$ for $n_p$, $47.8°$ for ${\theta }_2$ to find ${\theta }_1$.

$\begin{array}{c}\sin {\theta }_1=\left(\frac{\left(1.49\right)\sin \left(47.8\right)}{\left(1\right)}\right)\\ =1.1\end{array}$

Thus, for any angle less than equal to $90°$ will occur the total internal reflection.

Conclusion:

Therefore, for any angle less than equal to $90°$ will occur the total internal reflection.

(b)

To determine

The maximum value of angle for which the refracted ray will undergo total internal reflection at the left vertical face of the block when surrounding is water.

Answer to Problem 48AP

For any angle less than equal to $90°$ will occur the total internal reflection.

Explanation of Solution

The ray diagram is given below:

Formula to calculate the angle ${\theta }_3$ is,

${\theta }_3={\sin }^{-1}\left(\frac{n_w}{n_p}\right)$

• ${\theta }_3$ is the angle
• $n_w\text{and}{n}_p$ are the refractive index of water and polystyrene

Substitute $1.33$ for $n_a$. $1.49$ for $n_p$ to find ${\theta }_3$.

$\begin{array}{c}{\theta }_3={\sin }^{-1}\left(\frac{1.33}{1.49}\right)\\ =63.5°\end{array}$

Formula to calculate the angle ${\theta }_2$ is,

${\theta }_2=\left(90°-{\theta }_3\right)$

• ${\theta }_2$ is the angle

Substitute $63.5°$ for ${\theta }_3$ to find ${\theta }_2$.

$\begin{array}{c}{\theta }_2=\left(90°-63.5°\right)\\ =26.5°\end{array}$

Formula to calculate the angle ${\theta }_1$ is,

${\theta }_1={\sin }^{-1}\left(\frac{n_p\sin {\theta }_2}{n_w}\right)$

• ${\theta }_1\text{and}{\theta }_2$ are the angle
• $n_w\text{and}{n}_p$ are the refractive index of water and polystyrene

Substitute $1.33$ for $n_w$. $1.49$ for $n_p$, $26.5°$ for ${\theta }_2$ to find ${\theta }_1$.

$\begin{array}{c}{\theta }_1={\sin }^{-1}\left(\frac{\left(1.49\right)\sin \left(26.5\right)}{\left(1.33\right)}\right)\\ =29.9°\end{array}$

Thus, for any angle less than equal to $90°$ will occur the total internal reflection.

Conclusion:

Therefore, for any angle less than equal to $90°$ will occur the total internal reflection.

(c)

To determine

what will happen when the polystyrene block will immerse in carbon disulfide.

Answer to Problem 48AP

When the polystyrene block will immerse in carbon disulfide the total internal reflection will not occur.

Explanation of Solution

The total internal reflection will not occur when the polystyrene block will immerse in carbon disulfide as the refractive index of the polystyrene is less than that of the carbon disulfide.

Conclusion:

Therefore, when the polystyrene block will immerse in carbon disulfide the total internal reflection will not occur.