Chapter 35, Problem 35.68AP

(a)

To determine

The maximum angle of incidence for which the phenomena of total internal reflection occurs at the left vertical surface.

Answer to Problem 35.68AP

The total internal reflection will occur for all the values of angle of incidence and the maximum value for angle of incidence is $90°$.

Explanation of Solution

Given Info: The light of wavelength $589\text{nm}$ is incident on the polystyrene block of refractive index $n$ at an angle $\theta$. The incident ray gets refracted from the horizontal surface by angle ${\theta }_1$ and strikes the left vertical surface at angle ${\theta }_2$ as shown in figure below.

The refractive index of the polystyrene is $1.49$ and the refractive index of air is $1.00$.

Figure (I)

For total internal reflection from the left vertical surface the angle of incidence for the vertical surface must be greater than the critical angle.

$\left({\theta }_2>{\theta }_c\right)$ (1)

The formula to calculate the critical angle is,

${\theta }_c={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$ . (2)

Here,

${\theta }_c$ is the critical angle.

$n_1$ is the refractive index of the polystyrene.

$n_2$ is the refractive index of the air.

Substitute $1.49$ for $n_1$ and $1.00$ for $n_2$ in the equation (2).

$\begin{array}{c}{\theta }_c={\sin }^{-1}\left(\frac{1.00}{1.49}\right)\\ =42.2°\end{array}$

From equation (1) the value of the angle ${\theta }_2$ must be greater than $42.2°$.

From Figure (1) in triangle $\text{ABC}$ the value of the angle ${\theta }_1$ is,

${\theta }_1=90°-{\theta }_2$

Substitute $42.2°$ for ${\theta }_2$ in the above equation.

$\begin{array}{c}{\theta }_1=90°-42.2°\\ =47.8°\end{array}$

Thus the value of ${\theta }_1$ is $47.8°$.

Formula to calculate the value of $\theta$ from Snell’s law is,

$n_{\text{air}}\sin \theta =n\sin {\theta }_1$

Here,

$n_{\text{air}}$ is the refractive index of air.

$n$ is the refractive index of polystyrene.

Substitute $1.00$ for $n_{\text{air}}$ $1.49$ for $n$ and $47.8°$ for ${\theta }_1$ in the above equation.

$\begin{array}{c}{n}_{\text{air}}\sin \theta =n\sin {\theta }_1\\ \left(1.00\right)\sin \theta =\left(1.49\right)\sin \left(47.8°\right)\\ \sin \theta =1.10\end{array}$

As the angle ${\theta }_1$ must be less than the $47.8°$, $\left({\theta }_1\le 47.8°\right)$, thus in the above equation the,

$\sin \theta \le 1.10$ (3)

Thus any value of $\theta$ satisfy the equation (3) total internal reflection will occur at the left vertical surface, and the maximum angle will be less than $90°$.

Conclusion:

Therefore, total internal reflection will occur at the left vertical surface for any angle less than $90°$.

(b)

To determine

The angle of incidence if the polystyrene slab is immersed in water.

Answer to Problem 35.68AP

The total internal reflection will occur for angle of incidence $30.3°$ when the polystyrene slab is immersed in water.

Explanation of Solution

Explanation

Given info:  The refractive index of water is $1.33$.

For the polystyrene slab surrounded in water the value of critical angle is,

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

Substitute $1.33$ for $n_2$ and $1.49$ for $n_1$ in the above equation.

$\begin{array}{c}{\theta }_c={\sin }^{-1}\left(\frac{1.33}{1.49}\right)\\ =63.2°\end{array}$

The value of ${\theta }_2$ is $63.2°$.

From figure (1) in triangle $\text{ABC}$ the value of ${\theta }_1$ is,

${\theta }_1=90°-{\theta }_2$

Substitute $63.2°$ for ${\theta }_2$ in the above equation.

$\begin{array}{c}{\theta }_1=90°-63.2°\\ =26.8°\end{array}$

From Snell’s Law, calculate the value of the angle of incidence.

  $n_{\text{water}}\sin \theta =n\sin {\theta }_1$

Here,

$n_{\text{water}}$ is the refractive index of water.

$n$ is the refractive index of polystyrene.

Substitute $1.33$ for $n_{\text{water}}$ and $1.49$ for $n$ in the above equation.

$\begin{array}{c}{n}_{\text{water}}\sin \theta =n\sin {\theta }_1\\ \left(1.33\right)\sin \theta =1.49\sin \left(26.8°\right)\\ \theta =30.3°\end{array}$

Thus when the polystyrene slab is immersed in water the angle of incidence is $30.3°$.

Conclusion:

Therefore, the total internal reflection will occur for angle of incidence $30.3°$ when the polystyrene slab is immersed in water.

(c)

To determine

The angle of incidence for the phenomena of total internal reflection if the polystyrene slab is immersed in carbon disulphide.

Answer to Problem 35.68AP

The refractive index of the carbon disulphide is more than the refractive index of the polystyrene, so the total internal refraction is not possible.

Explanation of Solution

Explanation

Given info:  The refractive index of carbon disulphide is $1.628$.

The phenomena of total internal reflection only takes place when the light ray is travelling form a higher refractive index material to a low refractive index material.

For the case where the polystyrene slab is immersed in the carbon disulphide the light ray at the interface of the left vertical wall will not undergo the total internal reflection as at the interface the light will be propagating form a lower refractive index material to a higher refractive material and for the phenomena of total internal reflection to occur the light must be travelling form higher refractive index material to a low refractive index material.

Hence, the phenomena of total internal reflection will not take place when the slab is immersed in carbon disulphide.

Conclusion:

Therefore, the phenomena of total internal reflection will not take place when the slab is immersed in carbon disulphide.