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Chapter 33: Problem 13

In a material having an index of refraction \(n\) , a light ray has frequency \(f,\) wavelength \(\lambda,\) and speed \(v\) . What are the frequency, wavelength, and speed of this light (a) in vacuum and (b) in a material having refractive index \(n^{\prime} ?\) In each case, express your answers in terms of only \(f, \lambda, v, n,\) and \(n^{\prime} .\)

Short Answer

Expert verified
In vacuum: frequency \(f\), wavelength \(\frac{c}{f}\), speed \(c\). In new medium: frequency \(f\), wavelength \(\lambda \cdot \frac{n}{n'}\), speed \(v \cdot \frac{n}{n'}\).

Step by step solution

01

Understand the relationship in a medium

In a material with refractive index \(n\), the speed of light is \(v = \frac{c}{n}\), where \(c\) is the speed of light in vacuum. The frequency \(f\) of light remains constant irrespective of the medium.
02

Solving for light properties in vacuum

In vacuum, the speed of light is \(c\). Since frequency remains unchanged, \(f_{vacuum} = f\). The wavelength in vacuum can be written using the formula \(\lambda_{vacuum} = \frac{c}{f}\). Express the wavelength in terms of given variables: \(\lambda = \frac{v}{f}\), so \(v = c\Rightarrow\lambda_{vacuum} = \frac{c}{f}\).
03

Establish relationships for a new medium with index \(n'\)

In a medium with a different refractive index \(n'\), the light speed changes to \(v_{n'} = \frac{c}{n'}\) but the frequency \(f\) remains the same as in vacuum or the original medium. The wavelength becomes \(\lambda_{n'} = \frac{v_{n'}}{f}\) or \(\lambda_{n'} = \frac{c}{n'f}\).
04

Final expressions in terms of \(f, \lambda, v, n, n'\)

For vacuum: \(f_{vacuum} = f\), \(\lambda_{vacuum} = \frac{c}{f}\), \(v_{vacuum} = c\).For the material with refractive index \(n'\): \(f_{n'} = f\), \(\lambda_{n'} = \frac{v}{f} \cdot \frac{n}{n'}\), \(v_{n'} = \frac{v \cdot n}{n'}\).

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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 as \( n \), is a measure of how much a medium can bend (or refract) light. It is a dimensionless number calculated as the ratio of the speed of light in vacuum \( c \) to the speed of light in the medium \( v \). The formula is \( n = \frac{c}{v} \). When light enters a medium, the change in speed results in the bending of the light path. This bending is dependent on the difference in the refractive indices of the two media.
  • Higher \( n \): The light bends more, travels slower.
  • Lower \( n \): The light bends less, travels faster.
  • \( n \) of vacuum is 1 since light travels without deviation at maximum speed.
Understanding this concept is crucial as it affects how light behaves as it transitions from one medium to another, affecting applications like lenses, glasses, and even optical fibers.
Wavelength and Frequency
Wavelength \( \lambda \) and frequency \( f \) are fundamental properties of light waves. In a given medium, the speed of light \( v \) is connected to these two properties through the equation \( v = f \cdot \lambda \). Here, \( \lambda \) represents the distance between consecutive peaks of a wave, and \( f \) is the number of wave cycles that pass a fixed point per second. Importantly, frequency remains constant when light changes from one medium to another, meaning that all the changes in speed due to different media only affect the wavelength.
  • In vacuum: the speed \( c \), wavelength \( \lambda_{vacuum} = \frac{c}{f} \).
  • In a medium: wavelength adjusts but frequency \( f \) does not change.
  • Knowing this helps predict how light will behave when it moves between different materials, crucial for optics and various technologies.
Speed of Light in Mediums
The speed of light is not a constant value in all mediums, even though its upper limit, the speed in a vacuum, is marked by the letter \( c \) (approximately \( 3 \times 10^8 \text{ m/s} \)). In any medium, the speed \( v \) is determined by the medium's refractive index \( n \) using the formula \( v = \frac{c}{n} \). This simply means the denser the material, or the higher its refractive index, the slower light travels.
  • For vacuum: \( v = c \).
  • For any medium: use \( v = \frac{c}{n} \).
  • Different mediums like water, glass, or air will affect \( v \) since their \( n \) values differ.
By understanding how light speed changes in different mediums, we gain insight into optical phenomena such as refraction and total internal reflection, important for lenses and devices like periscopes.

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Most popular questions from this chapter

Light with a frequency of \(5.80 \times 10^{14} \mathrm{Hz}\) travels in a block of glass that has an index of refraction of \(1.52 .\) What is the wave-length of the light (a) in vacuum and (b) in the glass?

Old photographic plates were made of glass with a light-sensitive emulsion on the front surface. This emulsion was some what transparent. When a bright point source is focused on the front of the plate, the developed photograph will show a halo around the image of the spot. If the glass plate is 3.10 \(\mathrm{mm}\) thick and the halos have an inner radius of \(5.34 \mathrm{mm},\) what is the index of refraction of the glass? (Hint: Light from the spot on the front surface is scattered in all directions by the emulsion. Some of it is then totally reflected at the back surface of the plate and returns to the front surface.)

A ray of light is incident on a plane surface separating two sheets of glass with refractive indexes 1.70 and 1.58 . The angle of incidence is \(62.0^{\circ}\) , and the ray originates in the glass with \(n=1.70 .\) Compute the angle of refraction.

A certain birefringent material has indexes of refraction \(n_{1}\) and \(n_{2}\) for the two perpendicular components of lincarly polarized light passing through it. The corresponding wavelengths are \(\lambda_{1}=\lambda_{0} / n_{1}\) and \(\lambda_{0} / n_{2}\) , where \(\lambda_{0}\) is the wavelength in vacuum. (a) If the crystal is to function as a quarter-wave plate, the number of wavelengths of each component within the material must differ by \(\frac{1}{4}\) . Show that the minimum thickness for a quarter-wave plate is $$ d=\frac{\lambda_{0}}{4\left(n_{1}-n_{2}\right)} $$ (b) Find the minimum thickness of a quarter-wave plate made of siderite \(\left(\mathrm{FeO} \cdot \mathrm{CO}_{2}\right)\) if the indexes of refraction are \(n_{1}=1.875\) and \(n_{2}=1.635\) and the wavelength in vacuum is \(\lambda_{0}=589 \mathrm{nm}\) .

A parallel beam of light in air makes an angle of \(47.5^{\circ}\) with the surface of a glass plate having a refractive index of 1.66 . (a) What is the angle between the reflected part of the beam and the surface of the glass? (b) What is the angle between the refracted beam and the surface of the glass?
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