Problem 52 - University Physics with Modern Physics

Question

Light is incident normally on the short face of a \(30^{\circ}-\) \(60^{\circ}-90^{\circ}\) prism (Fig. \(\mathrm{P} 33.52 ) . \mathrm{A}\) drop of liquid is placed on the hypotenuse of the prism. If the index of refraction of the prism is \(1.62,\) find the maximum index that the liquid may have if the light is to be totally reflected.

Step-by-Step Solution

Answer

The maximum refractive index of the liquid is approximately 1.403.

Step 1: Understand the Problem and Essential Parameters

We are given a right-angled triangle prism with angles of 30°, 60°, and 90°. Light is incident normally on the short face of the prism, and we need to find the maximum index of refraction for a liquid on the hypotenuse such that total internal reflection occurs. The refractive index of the prism is 1.62.

Step 2: Determine Critical Angle for Total Internal Reflection

Total internal reflection occurs when light tries to move from a medium with higher refractive index to a medium with lower refractive index. To find the critical angle \( \theta_c \), use the formula \( \sin \theta_c = \frac{n_2}{n_1} \), where \( n_2 \) is the refractive index of the liquid and \( n_1 \) is that of the prism. Since \( \theta_c \) occurs at the hypotenuse, use \( \theta_c = 90^\circ - 30^\circ = 60^\circ \).

Step 3: Apply Snell's Law at the Prism-Liquid Interface

Snell's law states \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \). For the critical angle condition, \( \theta_2 = 90^\circ \). Therefore, \( 1.62 \sin 60^\circ = n_2 \sin 90^\circ \). This simplifies to \( n_2 = 1.62 \sin 60^\circ \).

Step 4: Solve for the Maximum Refractive Index

Calculating the critical angle gives us \( \sin 60^\circ = \frac{\sqrt{3}}{2} \). Substitute this into the equation from Step 3: \( n_2 = 1.62 \times \frac{\sqrt{3}}{2} = 1.62 \times 0.866 = 1.403 \). Therefore, the maximum permissible refractive index of the liquid is approximately 1.403.

Key Concepts

Prism Optics

Prism optics is a fascinating area of study that focuses on how light interacts as it enters and exits a prism. A prism is a transparent optical element with flat, polished surfaces that refract light. Specifically, the topic of prism optics delves into:

• Refraction: This is the bending of light as it passes through different media. As light enters the prism's surface, it slows down or speeds up depending on the material's refractive index.
• Dispersion: Prism separates light into its component colors or spectrum, as seen famously with rainbows.
• Internal reflection: Light can bounce inside the prism, central to the concept of total internal reflection.

In our particular problem, a right-angled prism is used with angles of 30°, 60°, and 90°. The light enters normally, meaning it proceeds directly through without initial refraction. However, when it reaches the hypotenuse, the situation becomes more interesting as it involves reflection.

Critical Angle

The critical angle is a crucial concept in optics, especially regarding total internal reflection (TIR). This angle is defined as the angle of incidence beyond which light remains trapped inside the medium because reflection becomes total and no refraction occurs. For light moving from a medium with a high refractive index to a lower one (such as from prism to air or liquid), TIR can take place. The essential equations governing this process include:

• The mathematical expression for the critical angle \( \theta_c \):\[\sin \theta_c = \frac{n_2}{n_1}\]where \( n_1 \) is the refractive index of the initial medium and \( n_2 \) is the refractive index of the second medium.
• In the specific case of a 30°-60°-90° prism, the trial of total internal reflection occurs at an angle of 60° because the critical angle isn't exceeded until this interaction.

Understanding this concept is pivotal for determining how much a liquid on the hypotenuse can alter the results—a factor the exercise question explores. If the liquid's refractive index ends up higher than the derived \( n_2 \), TIR ceases.

Refractive Index

The refractive index of a material is a measure of how much it can bend the light. It is denoted by \( n \), and tells us about the speed of light in the material compared to its speed in a vacuum. Mathematically, it's expressed as:

• \[ n = \frac{c}{v}\]where \( c \) is the speed of light in vacuum and \( v \) is the speed of light in the medium.
• A higher refractive index means that light travels slower in the material—this results in more bending at the boundary.

In the exercise, we have a prism with a refractive index of 1.62. The solution focuses on the maximum refractive index a liquid might have on the hypotenuse for TIR to occur. By calculating the maximum \( n_2 \), which notably must be lower than the prism's refractive index to sustain TIR, we derived it to be about 1.403. This ensures any introduction of liquid over this limit deters the phenomenon and results in light leaking out rather than reflecting totally.