Chapter 35, Problem 35.46P

(a)

To determine

The critical angle for total internal reflection for light in the diamond incident on the interface between the diamond and outside air.

Answer to Problem 35.46P

The critical angle of refraction at air-diamond interface is $24.418°$.

Explanation of Solution

Given info: The condition for light ray travelling between air and a diamond is shown below.

Figure (I)

From Snell’s law of refraction to air-diamond interface to find the critical angle is,

$\begin{array}{c}{n}_1\sin {\theta }_1=n_2\sin {\theta }_2\\ \sin {\theta }_1=\frac{n_2}{n_1}\sin {\theta }_2\\ {\theta }_1={\sin }^{-1}\left(\frac{n_2\sin {\theta }_2}{n_1}\right)\end{array}$ (1)

Here,

${\theta }_1$ is the incident angle.

$n_1$ is refractive index of diamond.

$n_2$ is refractive index of air.

${\theta }_2$ is the refracted angle.

The value of ${\theta }_2$ at critical angle is $90°$, the refractive index of air is 1 , refractive index of diamond is $2.419$ and angle of incidence ${\theta }_1$ become ${\theta }_c$.

Substitute 1 for $n_2$, $2.419$ for $n_1$ and $90°$ for ${\theta }_2$ in equation (1) and find ${\theta }_{c.}$.

$\begin{array}{c}{\theta }_c={\sin }^{-1}\left(\frac{\left(1\right)\sin \left(90°\right)}{2.419}\right)\\ ={\sin }^{-1}\left(\frac{1}{2.419}\right)\\ ={\sin }^{-1}\left(0.413\right)\\ =24.418\end{array}$

Thus, the critical angle of refraction at air-diamond interface is $24.418°$.

Conclusion:

Therefore, the critical angle of refraction at air-diamond interface is $24.418°$.

(b)

To determine

To show: The light travelling towards point $P$ in the diamond is totally reflected.

Answer to Problem 35.46P

The angle of incidence is more than the critical angle all light is reflected from point P.

Explanation of Solution

Given info: The condition for light ray travelling between air and a diamond is shown in figure (I).

The critical angle for total internal reflection for light in the diamond incident on the interface between the diamond and outside air is $24.418°$ and angle of incidence as shown in Figure (1) is $35°$, which is more than the critical angle. As per the Snell’s law If the incident angle is more than the critical angle than all the light is totally reflected from the surface.

Conclusion:

Therefore, the angle of incidence is more than the critical angle all light is reflected from point P.

(c)

To determine

The critical angle for total internal reflection for light in the diamond when the diamond is immersed in the water.

Answer to Problem 35.46P

The critical angle of incidence at water-diamond interface is $33.439$.

Explanation of Solution

Given info: The condition for light ray travelling between air and a diamond is shown in figure (I).

From Snell’s law of refraction to water-diamond interface to find the critical angle is,

$\begin{array}{c}{n}_1\sin {\theta }_1=n_2\sin {\theta }_2\\ \sin {\theta }_1=\frac{n_2}{n_1}\sin {\theta }_2\\ {\theta }_1={\sin }^{-1}\left(\frac{n_2\sin {\theta }_2}{n_1}\right)\end{array}$ (2)

Here,

${\theta }_1$ is the incident angle.

$n_1$ is refractive index of diamond.

$n_2$ is refractive index of air.

${\theta }_2$ is the refracted angle.

The value of ${\theta }_2$ at critical angle is $90°$, the refractive index of water is $1.333$, refractive index of diamond is $2.419$ and angle of incidence ${\theta }_1$ become ${\theta }_c$.

Substitute 1 for $n_2$, $2.419$ for $n_1$ and $90°$ for ${\theta }_2$ in equation (2) and find ${\theta }_{c.}$.

$\begin{array}{c}{\theta }_c={\sin }^{-1}\left(\frac{\left(1.333\right)\sin \left(90°\right)}{2.419}\right)\\ ={\sin }^{-1}\left(\frac{1.333}{2.419}\right)\\ ={\sin }^{-1}\left(0.5510\right)\\ =33.439\end{array}$

Thus, the critical angle of incidence at water-diamond interface is $33.439$.

Conclusion:

Therefore, the critical angle of incidence at water-diamond interface is $33.439$.

(d)

To determine

The ray incident at point $P$ undergoes total internal reflection or not when diamond is immersed in the water.

Answer to Problem 35.46P

Yes, the light ray undergoes total internal reflection at point $P$ when diamond is immersed in the water.

Explanation of Solution

Given info: The condition for light ray travelling between air and a diamond is shown in figure (I).

The critical angle for total internal reflection for light in the diamond incident on the interface between the diamond and water is $33.439$ and angle of incidence as shown in Figure (1) is $35°$, which is more than the critical angle. As per the Snell’s law If the incident angle is more than the critical angle than the total internal reflection will occur.

Thus, the light undergoes total internal reflection at $P$ when diamond is immersed in the water.

Conclusion:

Therefore, the light undergoes total internal reflection at $P$ when diamond is immersed in the water.

(e)

To determine

The direction in which the diamond is rotated such that the light at a point $P$ will exit the diamond.

Answer to Problem 35.46P

The diamond is rotated in clockwise direction on the axis perpendicular to the page through $O$.

Explanation of Solution

Given info: The condition for light ray travelling between air and a diamond is shown in figure (I).

The critical angle for total internal reflection for light in the diamond incident on the interface between the diamond and water is $33.439$ and angle of incidence as shown in Figure (1) is $35°$, which is more than the critical angle. As per the Snell’s law If the incident angle is more than the critical angle than the total internal reflection will occur.

The light will exit from the diamond only when the incident angle is less than the critical angle. So, to reduce the angle of incidence the diamond should be rotated in clockwise direction on the axis perpendicular to the plane of paper.

Thus, the light will exit at point $P$ when the angle of incidence is changed by rotating the diamond in clockwise direction.

Conclusion:

Therefore, the light will exit at point $P$ when the angle of incidence is changed by rotating the diamond in clockwise direction.

(f)

To determine

The angle of rotation at which the light first exit the diamond at point $P$.

Answer to Problem 35.46P

The diamond is rotated by $2.83°$ clockwise for the ray to exit at point $P$.

Explanation of Solution

Given info: The condition for light ray travelling between air and a diamond is shown in figure (I).

Let the angle is rotated clockwise by $\theta °$, the angle of incidence ${\theta }_1$ is changed , causing the angle of refraction ${\theta }_{}$ to change.

Apply Snell’s law at the water-diamond interface.

$\begin{array}{l}{n}_1\sin {\theta }_1=n_2\sin {\theta }_2\\ 1.333\sin {\theta }_1=2.419\sin {\theta }_2\end{array}$ (3)

The condition of the situation is shown below.

Figure (II)

The angle at the vertex $B$ as shown in Figure (1) is $35°$ because the extended line $AB$ is parallel to the $EF$ extended from the base of the diamond. From the sum of the interior angles of $ABP$ find angle ${\theta }_3$.

$\begin{array}{c}\left(90°-{\theta }_2\right)+\left(90-{\theta }_3\right)+35°=180\\ {\theta }_3=35°-{\theta }_2\end{array}$

The requirement is that the angle of incidence ${\theta }_3$ results in an angle of refraction $90°$.

Apply Snell’s law and find angle ${\theta }_2$.

$\begin{array}{c}{n}_1\sin {\theta }_3=n_2\sin 90°\\ 2.419\sin \left(35°-{\theta }_2\right)=1.333\\ 35°-{\theta }_2={\sin }^{-1}\left(\frac{1.33}{2.419}\right)\\ {\theta }_2=1.561°\end{array}$

Substitute $1.561°$ for ${\theta }_2$ in equation (3) and find.

$\begin{array}{c}1.333\sin \theta =2.419\sin \left(1.561°\right)\\ \sin \theta =\frac{3.776}{1.333}\\ \theta =2.83°\end{array}$

Thus, the diamond is rotated by $2.83°$ clockwise for the ray to exit at point $P$.

Conclusion:

Therefore, the diamond is rotated by $2.83°$ clockwise for the ray to exit at point $P$.