Problem 38 - College Physics

Question

A light beam traveling upward in a plastic material with an index of refraction of 1.60 is incident on an upper horizontal air interface. (a) At certain angles of incidence, the light is not transmitted into air. The cause of this is (1) reflection, (2) refraction, (3) total internal reflection. Explain. (b) If the angle of incidence is \(45^{\circ},\) is some of the beam transmitted into air? (c) Suppose the upper surface of the plastic material is covered with a layer of liquid with an index of refraction of \(1.20 .\) What happens in this case?

Step-by-Step Solution

Answer

(a) 3 (total internal reflection); (b) No, it is not transmitted; (c) Some light is transmitted into the liquid.

Step 1: Understanding Total Internal Reflection

Total internal reflection occurs when the light beam travels from a medium with a higher index of refraction to a medium with a lower index of refraction, and the angle of incidence exceeds the critical angle. Since air has an index of refraction lower than plastics, (1) total internal reflection is the reason light isn't transmitted into air.

Step 2: Calculate Critical Angle

Use Snell's Law to calculate the critical angle for light passing from plastic to air. Snell's Law is \[n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\]where \(n_1 = 1.60\) for plastic, \(n_2 = 1.00\) for air, and \(\theta_2 = 90^{\circ}\) at the critical angle, \(\theta_c\). Solving, \[1.60 \sin(\theta_c) = 1.00 \sin(90^\circ)\] \[\sin(\theta_c) = \frac{1}{1.60}\] \[\theta_c \approx 38.68^\circ\].

Step 3: Evaluate Incident Angle of 45°

Since the critical angle from plastic to air is approximately \(38.68^{\circ}\) and the angle of incidence is \(45^{\circ}\), which is greater than the critical angle, total internal reflection occurs, and the beam is not transmitted.

Step 4: Consider Adding a Liquid Layer

When a layer of liquid with an index of refraction of \(1.20\) is added, calculate the critical angle between plastic and liquid: \[1.60 \sin(\theta_c) = 1.20 \sin(90^\circ)\] \[\sin(\theta_c) = \frac{1.20}{1.60}\] \[\theta_c \approx 48.59^\circ\]. Since \(45^{\circ}\) is less than this angle, some transmission into the liquid occurs, and further critical angle analysis would be needed for air.

Key Concepts

Snell's Law

Snell's Law is a fundamental principle in the world of optics, especially when studying the behavior of light as it crosses boundaries between different media. It provides a mathematical formula that allows us to understand how light bends or changes direction due to varying speeds in different substances. Snell's Law is given by the equation:

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

Here,

• \(n_1\) and \(n_2\) represent the indices of refraction of the two media.
• \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction, respectively.

The law helps predict how much light will refract when moving from medium A to medium B. It's essential for calculating critical angles, which determine conditions under which total internal reflection occurs.

Index of Refraction

The index of refraction, often denoted as \(n\), quantifies how much light slows down as it enters a medium compared to its speed in a vacuum. It's defined by the formula:

• \[n = \frac{c}{v}\]

where \(c\) is the speed of light in a vacuum and \(v\) is the speed of light in the medium. Different materials have different indices of refraction:

• Air has an index close to 1.00.
• Common glass might have an index around 1.5.
• Plastic, like in this exercise, may have an index of 1.60.

A higher index indicates that light travels more slowly in that material. This concept is crucial when predicting how light will behave when transitioning between different substances.

Critical Angle

The critical angle is the angle of incidence above which total internal reflection occurs when light moves from a denser to a rarer medium. It is found using Snell’s Law by setting the angle of refraction to \(90^{\circ}\), where the light just grazes along the surface:

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

It occurs only when \(n_1 > n_2\). For example:- In the exercise, light moves from plastic \((n = 1.60)\) to air \((n = 1.00)\).- Calculating gives a critical angle of approximately \(38.68^{\circ}\).If the angle of incidence exceeds the critical angle, light reflects entirely back into the denser medium.

Light Transmission

Light transmission refers to the passage of light through a material. Whether light transmits, reflects, or refracts at the boundary between different media depends largely on the angle of incidence and the relative indices of refraction.For angles less than the critical angle:

• Light is partially reflected and partially refracted.

When the angle of incidence surpasses the critical angle:

• Total internal reflection occurs, and no light transmits through the second medium.

In the exercise example, a beam hitting the air at an angle greater than \(38.68^{\circ}\) results in total internal reflection back into the plastic.

Physics

Physics serves as the backbone of understanding light behavior. Concepts like Snell's Law, indices of refraction, and critical angles are essential in optical physics. They explain everyday phenomena such as the making of mirages, fiber optics, and camera lens design. Applying physics:

• Predict the path of light in complex systems.
• Design efficient optical devices and systems.

Understanding these core concepts facilitates deeper exploration of light's nature. Through exercises like the one described, students learn how principles of physics manifest in practical scenarios involving light and vision.