Problem 64 - College Physics

Question

The critical angle for a certain plastic and air interface is \(39^{\circ} .\) If the angle of incidence is adjusted so the reflected light has maximum polarization, what would be the angle of refraction?

Step-by-Step Solution

Answer

The angle of refraction is the Brewster angle for this plastic-air interface.

Step 1: Understanding the Concepts

To solve this problem, we need to understand the concepts of critical angle and Brewster's angle. The critical angle is the angle of incidence above which total internal reflection occurs. Brewster's angle is the angle of incidence at which light with a particular polarization is perfectly transmitted through a surface, with no reflection.

Step 2: Determine Brewster's Angle

The Brewster's angle can be found using the equation \( \theta_B = \tan^{-1}(n_2 / n_1) \), where \( n_1 \) is the refractive index of the incident medium (plastic) and \( n_2 \) is the refractive index of the refracting medium (air). We need to determine these refractive indices.

Step 3: Use Critical Angle to Find Refractive Indices

The critical angle \( \theta_c \) is given by \( 39^{\circ} \). Using the equation \( \theta_c = \sin^{-1}(n_2 / n_1) \), and knowing \( n_2 = 1 \) (refractive index of air), we find \( n_1 = 1 / \sin(39^{\circ}) \).

Step 4: Calculate Brewster's Angle

Substitute the values into the Brewster's angle equation: \( \theta_B = \tan^{-1}(1 / n_1) \). Since we found \( n_1 = 1 / \sin(39^{\circ}) \), we substitute to find \( \theta_B \).

Step 5: Calculate Angle of Refraction

When light is polarized maximally at Brewster's angle, the angle of refraction \( \theta_r \) can be found using Snell's Law: \( n_1 \sin(\theta_B) = n_2 \sin(\theta_r) \). Since \( n_2 = 1 \) for air, solve for \( \theta_r \).

Step 6: Solve for Angle of Refraction

Use the equation derived from Snell's Law: \( \sin(\theta_r) = n_1 \sin(\theta_B) \). Calculate \( \theta_r \) from \( \sin^{-1}(n_1 \sin(\theta_B)) \).

Key Concepts

Brewster's Angle

Brewster's angle is a fascinating concept in optics. When light strikes the boundary between two media, it can reflect or refract. Brewster's angle is the specific angle of incidence at which light with a certain polarization is perfectly transmitted through the boundary without reflecting.
This phenomenon occurs because the reflected light becomes completely polarized at this angle. To find Brewster's angle (\( \theta_B \)), we use the formula:

• \[ \theta_B = \tan^{-1}\left(\frac{n_2}{n_1}\right) \]

Here, \( n_1 \) is the refractive index of the initial medium and \( n_2 \) is that of the second medium.
In our problem, the light travels from plastic to air, with \( n_2 = 1 \) for air. Using Brewster’s formula, we determine the angle at which the reflected light is completely polarized. This tells us how light can travel efficiently through different materials without unwanted reflections.

Critical Angle

The critical angle is crucial in understanding phenomena like total internal reflection. It is the minimum angle of incidence above which light does not pass through to the second medium, but rather reflects entirely within the first medium.
This occurs when light travels from a medium with a higher refractive index to one with a lower refractive index, like moving from plastic to air. The critical angle (\( \theta_c \)) is calculated by:

• \[ \theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right) \]

For our exercise, the critical angle of \( 39^{\circ} \) indicates the transition point beyond which total internal reflection happens.
With \( n_2 = 1 \) for air, substituting in gives us the refractive index of plastic, \( n_1 \). Understanding the critical angle assists in determining angles of incidence needed to achieve maximum internal reflection, a useful concept in fiber optics and various scientific applications.

Angle of Refraction

The angle of refraction, denoted as \( \theta_r \), is prominent in Snell's Law, which provides the relationship between the angles and indices of refraction in two media. Snell’s Law is expressed as:

• \[ n_1 \sin(\theta_i) = n_2 \sin(\theta_r) \]

Here, \( \theta_i \) is the angle of incidence on the boundary.
For our problem, when light hits the plastic-air interface at Brewster's angle, Snell’s Law helps us calculate the resulting angle of refraction in the air.
Knowing \( n_1 \) and \( \theta_B \) allows us to rearrange Snell's Law to find:

• \[ \sin(\theta_r) = \frac{n_1 \sin(\theta_B)}{n_2} \]

Given \( n_2 = 1 \) for the air, solving this gives \( \theta_r \), the angle where light transitions optimally from the plastic to the air, ensuring effective conveyance across media boundaries.