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Chapter 26: Problem 3

The refractive indices of materials \(A\) and \(B\) have a ratio of \(n_{\mu} / n_{B}=1.33\). The speed of light in material \(A\) is \(1.25 \times 10^{8} \mathrm{m} / \mathrm{s}\). What is the speed of light in material \(B ?\)

Short Answer

Expert verified
The speed of light in material B is \(1.6625 \times 10^8 \ m/s\).

Step by step solution

01

Understanding the Relationship

The refractive index of a material is defined by the equation \( n = \frac{c}{v} \), where \( c \) is the speed of light in vacuum (approximately \( 3 \times 10^8 \ m/s \)), and \( v \) is the speed of light in the material. We are given the ratio \( \frac{n_A}{n_B} = 1.33 \), where \( n_A \) and \( n_B \) are the refractive indices of materials A and B respectively.
02

Express Ratio in Terms of Speeds

From the refractive index definition \( n = \frac{c}{v} \), we can express it as \( n_A = \frac{c}{v_A} \) and \( n_B = \frac{c}{v_B} \). The given ratio becomes \( \frac{n_A}{n_B} = \frac{v_B}{v_A} = 1.33 \), which implies \( v_B = 1.33 \times v_A \).
03

Calculate Speed in Material B

We have \( v_A = 1.25 \times 10^8 \ m/s \). Using the equation \( v_B = 1.33 \times v_A \), substitute \( v_A \) into the equation: \( v_B = 1.33 \times 1.25 \times 10^8 \). Multiplying these values gives \( v_B = 1.6625 \times 10^8 \ m/s \).

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Key Concepts

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

Speed of Light
The speed of light is a fundamental constant often represented by the symbol \( c \), which stands for its speed in a vacuum. This speed is approximately \( 3 \times 10^8 \) meters per second (m/s). This is the fastest speed at which information or matter can travel.
A vacuum is a theoretical space devoid of matter, meaning light can travel through it without any obstruction or interference from particles.
  • In materials or media other than a vacuum, light's speed is reduced due to interactions with particles. This reduction in speed varies depending on the material's properties, such as density and composition.
  • This interactive behavior of light with materials is quantified by the refractive index.
  • Understanding these properties helps us design lenses, prisms, and other devices that manipulate light for various applications, from eyeglasses to cameras.
Altering the speed of light, whether by a medium's properties or technological means, is vital for fields like optics and photonics.
Ratio of Refractive Indices
The ratio of refractive indices describes the relationship between two materials' abilities to bend light. The refractive index \( n \) is calculated using the equation \( n = \frac{c}{v} \), where \( v \) is the speed of light in that particular material. When comparing two materials, such as \( A \) and \( B \), the ratio \( \frac{n_A}{n_B} \) helps determine how much slower light travels in one material compared to the other.
  • A ratio greater than 1 indicates that light travels more slowly in material \( A \) than in \( B \).
  • The given ratio \( \frac{n_A}{n_B} = 1.33 \) shows that light travels faster in material \( B \) than material \( A \), since the refractive index is inversely proportional to the speed of light in the material.
  • Using this ratio, we can deduce other properties, such as the speed of light in unknown conditions, for instance in material \( B \).
This understanding is crucial in optics, allowing for precise control over light paths and intensities in various engineering and scientific applications.
Light Propagation in Materials
Light propagation refers to how light moves through different materials, influenced by the material's refractive index. When light enters a new medium, its speed and direction can change due to the material's optical density. This change in speed is governed by the refractive index \( n \), with the fundamental relationship \( n = \frac{c}{v} \).
The refractive index indicates how much a material can slow down the light relative to its speed in a vacuum.
  • When light moves from a medium with a lower refractive index to a higher one, it bends towards the normal line, slowing down as it passes through the denser medium.
  • Conversely, moving from a higher to a lower refractive index, light speeds up and bends away from the normal.
  • This behavior is critical across optical technologies, including lenses, fiber optics, and vision correction devices.
Designers must account for the speed changes and direction alterations in light when engineering systems or materials that rely on its propagation characteristics.

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Most popular questions from this chapter

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 one- half the size of the object. 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?

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 image sensor, as produced by (a) the \(35.0-\mathrm{mm}\) lens and (b) the \(150.0-\mathrm{mm}\) lens?

The drawing shows a ray of light traveling through three materials whose surfaces are parallel to each other. The refracted rays (but not the reflected rays) are shown as the light passes through each material. A ray of light strikes the \(a-b\) interface at a \(50.0^{\circ}\) angle of incidence. The index of refraction of material \(a\) is \(n_{a}=1.20 .\) The angles of refraction in materials \(b\) and \(c\) are, respectively, \(45.0^{\circ}\) and \(56.7^{\circ} .\) Find the indices of refraction in these two media.

Mars subtends an angle of \(8.0 \times 10^{-5} \mathrm{rad}\) at the unaided eye. An astronomical telescope has an cyepiece with a focal length of \(0.032 \mathrm{m} .\) When Mars is viewed using this telescope, it subtends an angle of \(2.8 \times 10^{-3}\) rad. Find the focal length of the telescope's objective lens.

The near point of a naked eye is \(25 \mathrm{cm} .\) When placed at the near point and viewed by the naked eye, a tiny object would have an angular size of \(5.2 \times 10^{-5}\) rad. When viewed through a compound microscope, however, it has an angular size of \(-8.8 \times 10^{-3}\) rad. (The minus sign indicates that the image produced by the microscope is inverted.) The objective of the microscope has a focal length of \(2.6 \mathrm{cm},\) and the distance between the objective and the eyepiece is \(16 \mathrm{cm}\). Find the focal length of the eyepiece.
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