Problem 30 - University Physics with Modern Physics

Question

Light traveling in water strikes a glass plate at an angle of incidence of \(53.0^{\circ} ;\) part of the beam is reflected and part is refracted. If the reflected and refracted portions make an angle of \(90.0^{\circ}\) with each other, what is the index of refraction of the glass?

Step-by-Step Solution

Answer

The index of refraction of the glass is approximately 1.76.

Step 1: Understanding the problem

We are given that a light beam strikes a glass plate at an angle of incidence of \(53.0^{\circ}\), and the reflected and refracted beams make a \(90.0^{\circ}\) angle with each other. We need to find the index of refraction of the glass.

Step 2: Applying Snell's Law

Snell's Law is given by \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\), where \(n_1\) is the index of refraction of the medium from which the light comes (water, \(n_1 \approx 1.33\)), \(\theta_1\) is the angle of incidence, \(n_2\) is the index of refraction of the second medium (glass), and \(\theta_2\) is the angle of refraction. Here, \(\theta_1 = 53.0^{\circ}\).

Step 3: Geometry of the problem

Since the reflected and refracted beams make an angle of \(90.0^{\circ}\), the angle of reflection \(\theta_r\) is equal to the angle of incidence \(\theta_1 = 53.0^{\circ}\). Therefore, the angle of refraction \(\theta_2\) is \(\theta_r + \theta_2 = 90.0^{\circ}\), or \(\theta_2 = 37.0^{\circ}\).

Step 4: Solve for the index of refraction of glass

Using Snell's Law, substitute the known values: \(1.33 \sin(53.0^{\circ}) = n_2 \sin(37.0^{\circ})\). Calculate \(\sin(53.0^{\circ})\) and \(\sin(37.0^{\circ})\), then solve for \(n_2\): \[n_2 = \frac{1.33 \sin(53.0^{\circ})}{\sin(37.0^{\circ})}\].

Step 5: Calculation

Compute the trigonometric values and calculate \(n_2\): \[\sin(53.0^{\circ}) \approx 0.7986\] and \[\sin(37.0^{\circ}) \approx 0.6018\]. Thus, \[n_2 = \frac{1.33 \times 0.7986}{0.6018} \approx 1.76\].

Key Concepts

Index of Refraction

The index of refraction, often denoted as "\( n \)," is key to understanding how light behaves when it transitions between different mediums. It describes the way light bends, or refracts, once it passes from one material into another. This number tells us how much the speed of light changes in any given medium compared to its speed in a vacuum. The equation for the index of refraction is given by:

• \( n = \frac{c}{v} \)

where \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the medium.
When light moves from a fast medium (like air) to a slower medium (like glass), the index of refraction will be greater than 1. If the medium slows down the light significantly, the index becomes larger. For instance, if a material has an index of 1.76, like glass, light moves slower there than in air. This causes the light to change direction, which is crucial for understanding lenses, prisms, and various optical devices.

Angle of Incidence

The angle of incidence is the angle between the incoming light ray and the normal (an imaginary line perpendicular to the surface) at the point of incidence. In optics, especially when using Snell's Law, understanding and accurately measuring this angle is vital since it affects the path light will take as it enters another medium.

• Imagine you are looking at a pool of water; the angle at which sunlight hits the water surface is the angle of incidence.
• In the context of Snell's Law, this is the angle we plug into the formula: \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \).

For example, in our exercise, the light initially from water strikes the glass surface under an angle of incidence of \( 53.0^{\circ} \). Recognizing the importance of the angle of incidence simplifies solving problems that involve light transitions, such as those seen in reflections and refractions, giving you insight into real-world applications like fiber optics.

Angle of Refraction

The angle of refraction is all about the light's new direction once it travels through a different medium. It's measured from the normal line inside the second medium, affected by the medium's index of refraction. According to Snell's Law, this angle is determined by:

• \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \)

where \( \theta_2 \) represents the angle of refraction, \( n_1 \) and \( \theta_1 \) are the index and angle of the initial medium, and \( n_2 \) is the index of the new medium.
For our particular example, it's stated that the reflected and refracted angles add up to \( 90^{\circ} \), implying a simple relationship between the angle of incidence and the angle of refraction. Since the angle of incidence is \( 53.0^{\circ} \), the angle of refraction must be \( 37.0^{\circ} \) (as they sum up to \( 90^{\circ} \)). This helps us determine the unknown index of refraction of the glass and is fundamental in correctly applying Snell's Law for any two mediums where light travels between.