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A beam of light strikes a sheet of glass at an angle of \(57.0^{\circ}\) with the normal in air. You observe that red light makes an angle of \(38.1^{\circ}\) with the normal in the glass, while violet light makes a \(36.7^{\circ}\) angle. (a) What are the indexes of refraction of this glass for these colors of light? (b) What are the speeds of red and violet light in the glass?

Short Answer

Expert verified
Red light index: 1.52, speed: \(1.97 \times 10^8\) m/s. Violet light index: 1.54, speed: \(1.95 \times 10^8\) m/s.

Step by step solution

01

Identify Known Values

First, identify all the information provided in the problem. We have the angle of incidence (in air) of the light beam, which is \(57.0^{\circ}\). The angles of refraction in glass are \(38.1^{\circ}\) for red light and \(36.7^{\circ}\) for violet light.
02

Apply Snell's Law for Red Light

Use Snell's Law: \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), where \( n_1 \) is the index of refraction of air (approximately 1), \( \theta_1 = 57.0^{\circ} \), and \( \theta_2 = 38.1^{\circ} \) for red light. Solve for \( n_2 \) (index of refraction for red light):\[ n_2 = \frac{\sin 57.0^{\circ}}{\sin 38.1^{\circ}} \approx 1.52 \]
03

Apply Snell's Law for Violet Light

Similarly for violet light, using Snell's Law with \( \theta_2 = 36.7^{\circ} \):\[ n_2 = \frac{\sin 57.0^{\circ}}{\sin 36.7^{\circ}} \approx 1.54 \]
04

Calculate Speed of Red Light in Glass

The speed of light in a medium is given by \( v = \frac{c}{n} \), where \( c \) is the speed of light in vacuum (approximately \(3.00 \times 10^8 \) m/s) and \( n \) is the index of refraction. For red light, \( n \approx 1.52 \):\[ v_{\text{red}} = \frac{3.00 \times 10^8}{1.52} \approx 1.97 \times 10^8 \text{ m/s} \]
05

Calculate Speed of Violet Light in Glass

Using the same formula for violet light with \( n \approx 1.54 \):\[ v_{\text{violet}} = \frac{3.00 \times 10^8}{1.54} \approx 1.95 \times 10^8 \text{ m/s} \]

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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 denoted as "n," is a crucial concept in optics. It tells us how much the speed of light decreases when it passes through a material. When light travels from one medium to another, its speed changes, and this change affects how light bends, which is known as refraction.
The index of refraction can be calculated using Snell's Law, where:
  • \( n_1 \) is the index of refraction of the first medium (such as air).
  • \( \theta_1 \) is the angle of incidence, the angle at which the light hits the surface.
  • \( n_2 \) is the index of refraction of the second medium (such as glass).
  • \( \theta_2 \) is the angle of refraction within the second medium.
Using the formula \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), we can solve for \( n_2 \). For example, if light travels from air (\( n \approx 1.0 \)) into glass, the index of refraction tells us how much slower light moves in the glass. For red light in glass, this was found to be approximately 1.52, while for violet light, it is about 1.54.
Speed of Light in Glass
The speed of light in different materials is a key factor in optics, dictating how light propagates through the medium. In a vacuum, light travels at its maximum speed of approximately \(3.00 \times 10^8 \) meters per second. However, when light enters different materials like glass, its speed is reduced. This reduction in speed is directly related to the index of refraction of the material.
To find the speed of light in a material, you can use the formula \( v = \frac{c}{n} \), where:
  • \( v \) is the speed of light in the material,
  • \( c \) is the speed of light in a vacuum,\(3.00 \times 10^8 \),
  • \( n \) is the index of refraction of the material.
For red light in a glass with \( n = 1.52 \), the calculated speed of light is approximately \(1.97 \times 10^8 \text{ m/s}\). Similarly, for violet light, using an index of 1.54, the speed becomes roughly \(1.95 \times 10^8 \text{ m/s}\). This change in speed affects how the light refracts within the material.
Angle of Incidence
The angle of incidence is an essential concept in understanding how light behaves when it strikes a surface. It is defined as the angle between the incoming light ray and the normal (an imaginary line perpendicular to the surface) at the point of contact.
In our example, the angle of incidence for both colors (red and violet light) was given as \(57.0^{\circ}\). This angle serves as a starting point in applying Snell's Law, helping to determine how much the light will bend as it enters a different medium, like glass.
  • The larger the angle of incidence, the more significant the refraction effect might be, depending on the mediums involved.
  • The relationship between the angle of incidence and angle of refraction is what Snell's Law clearly defines.
Thus, understanding the angle of incidence is essential for calculating the resulting path of the light as it passes from air to glass, helping to determine the index of refraction and how it changes with different wavelengths of light (such as red and violet).

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Most popular questions from this chapter

A glass plate 2.50 \(\mathrm{mm}\) thick, with an index of refraction of 1.40 , is placed between a point source of light with wavelength 540 \(\mathrm{nm}\) (in vacuum) and a screen. The distance from source to screen is 1.80 \(\mathrm{cm} .\) How many wavelengths are there between the source and the screen?

Three polarizing filters are stacked, with the polarizing axis of the second and third filters at \(23.0^{\circ}\) and \(62.0^{\circ}\) , respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 75.0 \(\mathrm{W} / \mathrm{cm}^{2}\) after it passes through the stack. If the incident intensity is kept constant, what is the intensity of the light after it has passed through the stack if the second polarizer is removed?

You sight along the rim of a glass with vertical sides so that the top rim is lined up with the opposite edge of the bottom (Fig. P33.49a). The glass is a (Fhin-walled, hollow cylinder 16.0 \(\mathrm{cm}\) high. The diameter of the top and bottom of the glass is 8.0 \(\mathrm{cm} .\) While you keep your eye in the same position, a friend fills the glass with a transparent liquid, and you then see a dime that is lying at the center of the bottom of the glass \((\) Fig. P33.49b). What is the index of refraction of the liquid?

A beam of light has a wavelength of 650 \(\mathrm{nm}\) in vacuum. (a) What is the speed of this light in a liquid whose index of refraction at this wavelength is 1.47\(?\) (b) What is the wavelength of these waves in the liquid?

A certain birefringent material has indexes of refraction \(n_{1}\) and \(n_{2}\) for the two per- pendicular components of linearly 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 \(\lambda_{0}=589 \mathrm{nm} .\)

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