Chapter 22, Problem 42P

(a)

To determine

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

Answer to Problem 42P

The critical angle for total internal reflection for light in the diamond incident on the interface between the diamond and the outside air is $24.42°$.

Explanation of Solution

Given Info: The refractive index of diamond is $2.419$.

Explanation:

Formula to calculate the critical angle is,

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

• ${\theta }_c$ is the critical angle
• $n_a\text{and }\text{}{n}_d$ are the refractive index of air and diamond

Substitute $1$ for $n_a$, $2.419$ for $n_d$ to find ${\theta }_c$.

$\begin{array}{c}{\theta }_c={\sin }^{-1}\left(\frac{\left(1\right)}{\left(2.419\right)}\right)\\ =24.42°\end{array}$

Thus, the critical angle for total internal reflection for light in the diamond incident on the interface between the diamond and the outside air is $24.42°$.

Conclusion:

Therefore, the critical angle for total internal reflection for light in the diamond incident on the interface between the diamond and the outside air is $24.42°$.

(b)

To determine

That the light is travelling towards point P in the diamond is totally reflected.

Answer to Problem 42P

The light is travelling towards point P in the diamond is totally reflected.

Explanation of Solution

When the light beam is travelling towards the point P, the incident angle will be $35°$.

Since the incident angle is greater than the critical angle, total internal reflection will occur.

Thus, the light is travelling towards point P in the diamond is totally reflected.

Conclusion:

Therefore, the light is travelling towards point P in the diamond is totally reflected.

(c)

To determine

The critical angle at the diamond water interface.

Answer to Problem 42P

The critical angle at the diamond water interface is $33.44°$.

Explanation of Solution

Formula to calculate the critical angle at the diamond water interface is,

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

• ${\theta }_c$ is the critical angle
• $n_w\text{and}{n}_d$ are the refractive index of water and diamond

Substitute $1.333$ for $n_w$, $2.419$ for $n_d$, to find the value ${\theta }_c$,

$\begin{array}{c}{\theta }_c={\sin }^{-1}\left(\frac{1.333}{2.419}\right)\\ =33.44°\end{array}$

Thus, the critical angle at the diamond water interface is $33.44°$.

Conclusion:

Therefore, the critical angle at the diamond water interface is $33.44°$.

(d)

To determine

The explanation when the diamond is immersed in water, does the light ray entering the top surface undergo total internal reflection.

Answer to Problem 42P

The total internal reflection will occur.

Explanation of Solution

When the light beam is travelling towards the point P, the incident angle will be, $35°$.

Since the incident angle is greater than the critical angle, total internal reflection will occur.

Thus, the total internal reflection will occur at point P.

Conclusion:

Therefore, the total internal reflection will occur at point P.

(e)

To determine

The direction of rotation of the diamond about an axis perpendicular to the page.

Answer to Problem 42P

The direction of rotation of the diamond about an axis perpendicular to the page is $\text{clockwise}$.

Explanation of Solution

To have light leaved the diamond at point P, we have to diminish the edge of rate now. Along these lines, we ought to turn the diamond clockwise, along these lines conveying the typical line nearer to the occurrence beam at point P.

Thus, the direction of rotation of the diamond about an axis perpendicular to the page is $\text{clockwise}$.

Conclusion:

Therefore, the direction of rotation of the diamond about an axis perpendicular to the page is $\text{clockwise}$.

(f)

To determine

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

Answer to Problem 42P

The find the angle of rotation at which light will first exit the diamond at point P is $2.9°$.

Explanation of Solution

Calculating the angle of refraction is,

$\begin{array}{c}{\theta }_r=\left(35-33.44\right)\\ =1.6°\end{array}$

• ${\theta }_r$ is the angle of refraction

Formula to calculate the angle of rotation is,

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

• $\theta$ is the angle of rotation
• $n_d\text{and}{n}_w$ are the refractive index of diamond and water

Substitute $2.419$ for $n_d$, $1.333$ for $n_w$, $1.6°$ for ${\theta }_r$, to find the value of $\theta$.

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(2.419\right)\sin \left(1.6\right)}{\left(1.333\right)}\right)\\ =2.9°\end{array}$

Thus, the find the angle of rotation at which light will first exit the diamond at point P is $2.9°$.

Conclusion:

Therefore, the find the angle of rotation at which light will first exit the diamond at point P is $2.9°$.