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Chapter 26: Problem 15

A beam of light is traveling in air and strikes a material. The angles of incidence and reffaction are \(63.0^{\circ}\) and \(47.0^{\circ},\) respectively. Obtain the speed of light in the material.

Short Answer

Expert verified
The speed of light in the material is approximately \(2.33 \times 10^8\) m/s.

Step by step solution

01

Understand Snell's Law

Snell's Law relates the angles of incidence and refraction to the speeds in the two mediums: \( n_1\sin\theta_1 = n_2\sin\theta_2 \). Here, \( n_1 \) and \( n_2 \) are the refractive indices of air and the material, respectively. For air, \( n_1 \approx 1 \). We have \( \theta_1 = 63.0^\circ \) and \( \theta_2 = 47.0^\circ \)."
02

Calculate the Refractive Index of the Material

We rearrange Snell's Law to find \( n_2 \): \( n_2 = \frac{\sin\theta_1}{\sin\theta_2} \). Substitute the given angles: \( n_2 = \frac{\sin 63.0^\circ}{\sin 47.0^\circ} \). Use a calculator to solve: \( n_2 \approx 1.29 \)."
03

Find the Speed of Light in the Material

The refractive index is also given by \( n = \frac{c}{v} \), where \( c \approx 3 \times 10^8 \text{ m/s} \) is the speed of light in vacuum and \( v \) is its speed in the material. Rearrange to find \( v \): \( v = \frac{c}{n_2} \). Substitute \( n_2 = 1.29 \) and \( c = 3 \times 10^8 \text{ m/s} \): \( v = \frac{3 \times 10^8}{1.29} \approx 2.33 \times 10^8 \text{ m/s} \)."

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Key Concepts

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

Refractive Index
The refractive index is a fundamental concept in optics. It tells us how much light bends when it enters a new medium. When light travels from one medium to another, its speed changes, which causes it to bend. This bending is quantified by the refractive index.

Mathematically, the refractive index () is defined as the ratio of the speed of light in a vacuum ({c}) to the speed of light in the material ({v}).
  • Refractive index formula: = \( \frac{c}{v} \)
  • In a vacuum, the refractive index is always 1.
  • For other materials, it's greater than 1.
Understanding the refractive index is crucial when studying how light interacts with different materials, as it determines the amount of bending of light.
Speed of Light
The speed of light is one of the most important constants in physics. It is the speed at which light travels in a vacuum, and it's about \( 3 \times 10^8 \text{ m/s} \).

When light moves through other materials, like water or glass, it travels more slowly. This speed reduction is described by the refractive index of the material:
  • In a material, speed of light \( v = \frac{c}{n} \)
  • Here, \( c \) is the speed of light in vacuum, \( n \) is the refractive index.
The calculation of light speed in different materials is crucial in fields like optics and engineering, helping us design instruments and understand natural phenomena.
Angle of Incidence
The angle of incidence refers to the angle at which a light beam hits a surface. This angle is measured from the normal line, which is a line perpendicular to the surface.

  • The angle at which light approaches a new medium affects how much it bends.
  • In Snell's Law, represented as \( \theta_1 \).
  • Determines part of the refractive index calculation \( n_2 = \frac{\sin\theta_1}{\sin\theta_2} \).
Understanding the angle of incidence is important for predicting how light will behave when it hits a new medium, such as moving from air into water.
Angle of Refraction
The angle of refraction is the angle between the refracted beam of light and the normal line. This angle gives insight into the bending of light as it moves into a different medium.

  • In Snell's Law, denoted as \( \theta_2 \).
  • It is crucial for determining the new direction of the light beam.
  • Directly affects how much the light will bend upon entering the material.
Calculating the angle of refraction helps us understand how light spreads through materials, laying the groundwork for cutting lenses and analyzing optical devices.

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Most popular questions from this chapter

An object has an angular size of 0.0150 rad when placed at the near point (21.0 cm) of an eye. When the eye views this object using a magnifying glass, the largest possible angular size of the image is 0.0380 rad. What is the focal length of the magnifying glass?

An object is in front of a converging lens \((f=0.30 \mathrm{m}) .\) The magnification of the lens is \(m=4.0 .\) (a) Relative to the lens, in what direction should the object be moved so that the magnification changes to \(m=-4.0 ? \quad\) (b) Through what distance should the object be moved?

A farsighted person can read printing as close as 25.0 cm when she wears contacts that have a focal length of 45.4 cm. One day, she forgets her contacts and uses a magnifying glass, as in Figure 26.39b. Its maximum angular magnification is 7.50 for a young person with a normal near point of 25.0 cm. What is the maximum angular magnification that the magnifying glass can provide for her?

The drawing shows a rectangular block of glass \((n=1.52)\) surrounded by liquid carbon disulfide \((n=1.63) .\) A ray of light is incident on the glass at point A with a \(30.0^{\circ}\) angle of inci- dence. At what angle of refraction does the ray leave the glass at point \(\mathrm{B} ?\)

To focus a camera on objects at different distances, the converging lens is moved toward or away from the image sensor, so a sharp image always falls on the sensor. A camera with a telephoto lens \((f=200.0 \mathrm{mm})\) is to be focused on an object located first at a distance of 3.5 \(\mathrm{m}\) and then at 50.0 \(\mathrm{m}\) . Over what distance must the lens be movable?
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