/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 The drawing shows a ray of light... [FREE SOLUTION] | 91Ó°ÊÓ

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The drawing shows a ray of light traveling from point \(A\) to point \(B,\) a distance of \(4.60 \mathrm{~m}\) in a material that has an index of refraction \(n_{1}\). At point \(B\), the light encounters a different substance whose index of refraction is \(n_{2}=1.63 .\) The light strikes the interface at the critical angle of \(\theta_{c}=48.1^{\circ} .\) How much time does it take for the light to travel from \(A\) to \(B\) ?

Short Answer

Expert verified
Calculate the speed of light in the first medium and use it to determine the travel time over 4.60 m.

Step by step solution

01

Understanding the Index of Refraction

The index of refraction, represented as \( n \), determines how much the speed of light is reduced inside a material. It is defined as \( n = \frac{c}{v} \), where \( c \) is the speed of light in vacuum (~\( 3 \, \times \, 10^8 \, \text{m/s} \)) and \( v \) is the speed of light in that material.
02

Using the Critical Angle

The critical angle \( \theta_c \) is the angle of incidence beyond which light cannot pass through the interface and instead undergoes total internal reflection. It is calculated using \( \sin \theta_c = \frac{n_2}{n_1} \). Given \( \theta_c = 48.1^\circ \) and \( n_2 = 1.63 \), solve for \( n_1 \).
03

Calculating the Speed of Light in Material 1

With the obtained \( n_1 \), find the speed of light in the first material: \( v_1 = \frac{c}{n_1} \). This speed helps in determining how long it takes for light to travel a given distance in the material.
04

Finding the Time to Travel from A to B

Time is calculated using \( t = \frac{d}{v} \), where \( d \) is the distance traveled and \( v \) is the speed of light in the first material. Substitute \( d = 4.60 \, \text{m} \) and the speed \( v_1 \) calculated earlier to find \( t \).
05

Conclusion

By substituting all the values and solving, you can find the exact time required for the light to travel from point \( A \) to point \( B \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Index of Refraction
The index of refraction, often represented by the symbol \( n \), is a fundamental concept in the study of optics. It provides a measure of how much the speed of light is reduced when it travels through a particular material compared to its speed in a vacuum. The formula used to calculate the index of refraction is \( n = \frac{c}{v} \), where \( c \) is the speed of light in a vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \), and \( v \) is the speed of light in the material. This reduction in speed leads to the bending of light as it crosses from one medium to another, a phenomenon known as refraction. By knowing the index of refraction, one can predict how light will behave when entering or exiting different substances, making it crucial for lens design, optical instruments, and understanding natural phenomena like rainbows.
Critical Angle
The critical angle is a specific angle of incidence in which a ray of light traveling from a denser to a less dense medium no longer refracts out of the material but instead travels along the boundary of the two mediums. This angle is significant because any angle larger than this leads to a phenomenon called total internal reflection. To find the critical angle, we use the equation \( \sin \theta_c = \frac{n_2}{n_1} \), where \( n_1 \) is the index of refraction of the initial medium, and \( n_2 \) is the index of refraction of the second medium, which is typically less dense. In our exercise, the critical angle \( \theta_c \) is given as \( 48.1^{\circ} \), which can be used with the known index of refraction of the second medium to determine \( n_1 \). Understanding the critical angle is vital in optical fibers and some natural optical effects.
Total Internal Reflection
Total internal reflection (TIR) occurs when light attempts to pass from a denser medium to a less dense medium but instead reflects entirely back into the denser medium. This phenomenon happens when the angle of incidence exceeds the critical angle. Unlike ordinary reflection, TIR can lead to almost 100% of light being reflected, and no light passes into the second medium. This principle is critically important in the design of fiber optics, allowing for efficient data transmission over long distances without losing signal strength. In nature, this can also be seen in the shimmering effect observed just beneath the water surface when looking upwards. Total internal reflection is a key principle behind many modern communication technologies.
Speed of Light
The speed of light is a universal constant denoted by \( c \) and is approximately \( 3 \times 10^8 \, \text{m/s} \) in a vacuum. However, when light travels through different materials, its speed decreases, depending on the material's index of refraction. The speed of light in a material is given by the formula \( v = \frac{c}{n} \), where \( n \) is the material's index of refraction. The unique feature of light is that while its speed changes in different media, its frequency remains constant. This property accounts for the bending or refraction of light at media interfaces. Understanding how light's speed changes is critical for designing lenses, optical communication devices, and understanding everyday optical effects.
Ray of Light
A ray of light is a simple concept but forms the foundation of understanding complex optical phenomena. Represented as a straight line, a light ray is an idealization used to model the propagation of light through different media. When discussing rays, key attributes include their direction, angle of incidence, and how they interact with surfaces. Rays of light help us visualize how light bends or refracts when moving through layers of materials with different refractive indices. Diagrams using rays are utilized to predict lens behavior, the formation of images, and to solve problems related to the focus of lenses and mirrors. The behavior of light rays is governed by optical laws such as Snell's Law and the Law of Reflection, making them an essential tool in both educational and practical optics applications.

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Most popular questions from this chapter

A layer of liquid \(B\) floats on liquid \(A\). A ray of light begins in liquid \(A\) and undergoes total internal reflection at the interface between the liquids when the angle of incidence exceeds \(36.5^{\circ} .\) When liquid \(B\) is replaced with liquid \(C,\) total internal reflection occurs for angles of incidence greater than \(47.0^{\circ} .\) Find the ratio \(n_{B} / n_{C}\) of the refractive indices of liquids \(B\) and \(C\).

An object is placed to the left of a lens, and a real image is formed to the right of the lens. The image is inverted relative to the object and is onehalf the size of the object. (a) What kind of lens, converging or diverging, is used to produce this image? (b) How is the height \(h_{\mathrm{i}}\) of the image related to the height \(h_{0}\) of the object? Don't forget to take into account the fact that the image is inverted relative to the object. (c) What is the ratio \(d_{i} / d_{0}\) of the image distance to the object distance? For the situation described in the Concept Questions, the distance between the object and the image is \(90.0 \mathrm{~cm}\). (a) How far from the lens is the object? (b) What is the focal length of the lens?

Consult Interactive Solution \(\underline{26.51}\) at to review the concepts on which this problem depends. A camera is supplied with two interchangeable lenses, whose focal lengths are \(35.0\) and \(150.0 \mathrm{~mm}\). A woman whose height is \(1.60 \mathrm{~m}\) stands \(9.00 \mathrm{~m}\) in front of the camera. What is the height (including sign) of her image on the film, as produced by (a) the \(35.0\) -mm lens and \((b)\) the \(150.0-\mathrm{mm}\) lens?

Concept Simulation 26.4 at provides the option of exploring the ray diagram that applies to this problem. The distance between an object and its image formed by a diverging lens is \(49.0 \mathrm{~cm} .\) The focal length of the lens is \(-233.0 \mathrm{~cm}\). Find (a) the image distance and (b) the object distance.

Two identical diverging lenses are separated by \(16 \mathrm{~cm} .\) The focal length of each lens is \(-8.0 \mathrm{~cm} .\) An object is located \(4.0 \mathrm{~cm}\) to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.

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