Problem 20 - University Physics with Modern Physics

Question

At the very end of Wagner's series of operas Ring of the Nibelung, Brunnhilde takes the golden ring from the finger of the dead Siegfried and throws it into the Rhine, where it sinks to the bottom of the river. Assuming that the ring is small enough compared to the depth of the river to be treated as a point and that the Rhine is 10.0 \(\mathrm{m}\) deep where the ring goes in, what is the area of the largest circle at the surface of the water over which light from the ring could escape from the water?

Step-by-Step Solution

Answer

The area is approximately 408.12 square meters.

Step 1: Understanding Snell's Law

First, we need to consider Snell's Law, which relates the angle of incidence and the angle of refraction between two different media with different refractive indices: \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \). Here, \( n_1 \) is the refractive index of water, \( n_2 \) is the refractive index of air, \( \theta_1 \) is the angle of incidence (inside the water), and \( \theta_2 \) is the angle of refraction (in the air). For water, \( n_1 \approx 1.33 \) and for air, \( n_2 \approx 1.00 \).

Step 2: Calculating the Critical Angle

The critical angle is the angle of incidence above which total internal reflection occurs, and no refraction happens. This angle is given by \( \sin \theta_c = \frac{n_2}{n_1} \). Substituting the values, we get \( \sin \theta_c = \frac{1.00}{1.33} \approx 0.752 \). Therefore, \( \theta_c = \arcsin(0.752) \approx 48.75^\circ \).

Step 3: Finding the Circular Area on the Surface

When light hits the surface at an angle equal to or less than the critical angle, it can escape the water. The light can escape within a cone whose axis is perpendicular to the water's surface, and the base of this cone is the circle on the water's surface. Using trigonometry, at the depth \( h = 10.0 \ \mathrm{m} \), the radius \( r \) of the circle is \( r = h \tan \theta_c \). Substitute \( \theta_c = 48.75^\circ \) and \( h = 10.0 \ \mathrm{m} \) to find \( r = 10.0 \times \tan(48.75^\circ) \approx 11.4 \ \mathrm{m} \).

Step 4: Calculating the Area of the Circle

The area \( A \) of the circle is given by the formula \( A = \pi r^2 \). Substituting \( r = 11.4 \ \mathrm{m} \), we find \( A = \pi \times (11.4)^2 \approx 408.12 \ \mathrm{m}^2 \).

Key Concepts

Critical Angle

The critical angle is a specific angle of incidence in which light moving from a denser medium (like water) to a less dense medium (like air) is refracted at 90 degrees along the boundary. When the angle of incidence exceeds this point, total internal reflection occurs. This phenomenon is vital in understanding how light behaves in different environments, like when light tries to move from the underwater setting of the river Rhine to the air above.
Understanding and determining the critical angle, like in the textbook problem, helps predict how light will travel through various mediums.
The formula used to calculate the critical angle is \( \sin \theta_c = \frac{n_2}{n_1} \), where \( n_1 \) and \( n_2 \) are the refractive indices of the two media involved. In simpler terms, it's the point where refraction shifts to reflection entirely.

• Water has a refractive index \( n_1 \) of about 1.33.
• Air generally has a refractive index \( n_2 \) of 1.00.

Substituting these values yields a critical angle of approximately 48.75 degrees. This angle signifies the threshold for light to escape the water and enter the air.

Total Internal Reflection

Total internal reflection occurs when light traveling in a medium with a higher refractive index (like water) hits the boundary of a medium with a lower refractive index (like air) at an angle greater than the critical angle.
Beyond this critical point, light doesn't pass through the boundary but instead reflects entirely back into the denser medium. This reflection is not just partial, like with typical reflections we see in mirrors, but total.
This principle is significant in numerous technologies, such as fiber optics, where light signals travel long distances through cables without escaping. It's the reason fiber optic cables are incredibly efficient in data transmission.
In the context of this exercise, if the ring emits light at angles exceeding the critical angle, that light will entirely reflect back into the water rather than reaching the air. Understanding this reflection helps derive the extent to which light from the ring might be visible from above.

Refractive Index

The refractive index is a measure of how much the speed of light is reduced within a medium. It's crucial in determining how rays of light bend, or refract, when passing from one medium into another.
Light travels fastest in a vacuum, and when it moves through any other medium, the speed reduces depending on the medium's refractive index. This translates into a bending effect known as refraction.

• Air has a refractive index of approximately 1.00, implying minimal reduction in light speed.
• Water has a refractive index of around 1.33, indicating a more substantial reduction in light speed.

The refractive index directly influences the angles calculated in Snell's Law, which predicts how light will travel upon meeting a boundary between two different media. It also plays a role in determining the critical angle and understanding total internal reflection.
Grasping the concept of the refractive index can also aid in comprehending other optical phenomena, making it a fundamental element in the study of optics.