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\(\bullet\) 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
The indexes of refraction are 1.357 for red and 1.406 for violet. Speeds are approximately 2.21 x 10鈦 m/s for red light and 2.13 x 10鈦 m/s for violet light in glass.

Step by step solution

01

Determine the Index of Refraction for Red Light

We will use Snell's law to find the index of refraction for red light. Snell's law states \(n_1 \sin \theta_1 = n_2 \sin \theta_2\). Here, the air's index of refraction \(n_1 = 1\), and the angle in air \(\theta_1 = 57.0^{\circ}\). The angle in the glass for red light \(\theta_2 = 38.1^{\circ}\). So, \(1 \cdot \sin(57.0^{\circ}) = n_2 \cdot \sin(38.1^{\circ})\). We solve for \(n_2\) (red):\[n_{red} = \frac{\sin(57.0^{\circ})}{\sin(38.1^{\circ})} = \frac{0.8387}{0.6181} \approx 1.357\]
02

Determine the Index of Refraction for Violet Light

Similarly, using Snell鈥檚 law for violet light, with \(\theta_2 = 36.7^{\circ}\), we calculate:\[n_2 \cdot \sin(36.7^{\circ}) = 1 \cdot \sin(57.0^{\circ})\]Thus, the index of refraction for violet light is:\[n_{violet} = \frac{\sin(57.0^{\circ})}{\sin(36.7^{\circ})} = \frac{0.8387}{0.5964} \approx 1.406\]
03

Calculate the Speed of Red Light in the Glass

The speed of light in a medium is given by \(v = \frac{c}{n}\), where \(c\) is the speed of light in a vacuum \((3 \times 10^8\, \text{m/s})\), and \(n\) is the index of refraction.For red light:\[v_{red} = \frac{3 \times 10^8}{1.357} \approx 2.21 \times 10^8\, \text{m/s}\]
04

Calculate the Speed of Violet Light in the Glass

Using the same formula for violet light:\[v_{violet} = \frac{3 \times 10^8}{1.406} \approx 2.13 \times 10^8\, \text{m/s}\]

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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 is a fundamental concept in understanding how light behaves when it travels from one medium to another. Essentially, this index measures how much light bends, or refracts, as it enters a different medium. This bending happens due to a change in the speed of light.
The index of refraction, denoted as \( n \), is defined by the equation:
  • \( n = \frac{c}{v} \)
where \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the medium.
A higher index means light slows down more and bends more sharply. This is why different colors, such as red and violet, have different indexes in the same medium like glass. The difference in refraction is also why prisms can create rainbows by splitting white light into its component colors.
Speed of Light
The speed of light is a constant in a vacuum, approximately \( 3 \, \times \, 10^8 \, \text{m/s} \). However, when light travels through a medium, like glass or water, its speed decreases. This reduction in speed compared to a vacuum directly relates to the medium's index of refraction.
To find the speed of light in any medium, use the formula:
  • \( v = \frac{c}{n} \)
where \( c \) is the speed of light in a vacuum, and \( n \) is the index of refraction of the medium. Thus,mdifferent mediums will have different light speeds, impacting how light behaves as it enters and exits these materials.
For instance, in glass, red light travels faster than violet light because the red light has a lower index of refraction. Hence, the speed at which light travels in a medium and its refraction index are tightly intertwined.
Optics
Optics is the study of light, its behavior, and properties, and how it interacts with different materials. It involves understanding phenomena like reflection, refraction, and diffraction. Optics is paramount in designing various devices such as eyeglasses, cameras, and telescopes, which rely on controlling light.
In optics, Snell's Law is critical for explaining how light refracts at the boundary between two media. It is mathematically represented by:
  • \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \)
where \( n_1 \) and \( n_2 \) are the indices of refraction for the initial and second medium, while \( \theta_1 \) and \( \theta_2 \) are the respective angles of incidence and refraction.
By mastering the principles of optics and Snell鈥檚 Law, we gain insights into numerous real-world applications, from fiber optic communications to corrective eyewear, enhancing our ability to manipulate light for practical uses.
Angle of Incidence
The angle of incidence is vital in determining how light behaves when it hits a surface or another medium. It is the angle formed between the incoming light ray and an imaginary line perpendicular to the surface called the normal.
The significance of the angle of incidence is that it affects the angle at which light will refract, or bend, upon entering a new medium. Snell's Law utilizes this angle to calculate the refraction, helping us understand how lenses and various optical phenomena work.
  • For example, if light hits glass at an angle, like in our exercise where the angle of incidence is \( 57.0^{\circ} \), different colors of light are refracted at different angles due to their unique indices of refraction.
The understanding of incidence angles allows scientists and engineers to design lenses that focus light precisely, ensuring clarity in everything from eyeglasses to sophisticated microscopes and telescopes.

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Most popular questions from this chapter

\(\bullet$$\bullet \mathrm{A}\) sinusoidal electromagnetic wave of frequency \(6.10 \times 10^{14} \mathrm{Hz}\) travels in vacuum in the \(+x\) -direction. The magnetic field is parallel to the \(y\) -axis and has amplitude \(5.80 \times 10^{-4} \mathrm{T} .\) (a) Find the magnitude and direction of the electric field. (b) Write the wave functions for the electric and magnetic fields in the form of Equations \((23.3).\)

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\(\bullet\) At the floor of a room, the intensity of light from bright overhead lights is 8.00 \(\mathrm{W} / \mathrm{m}^{2} .\) Find the radiation pressure on a totally absorbing section of the floor.

\(\bullet$$\bullet\) A beam of polarized light passes through a polarizing filter. When the angle between the polarizing axis of the filter and the direction of polarization of the light is \(\theta\) , the intensity of the emerging beam is \(I\) . If you instead want the intensity to be \(I / 2,\) what should be the angle (in terms of \(\theta )\) between the polarizing angle of the filter and the original direction of polarization of the light?

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