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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 refraction 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 \(2.26 \times 10^8 \; m/s\).

Step by step solution

01

Identify Known Values

We know the angle of incidence, \( \theta_1 \), is \( 63.0^{\circ} \) and the angle of refraction, \( \theta_2 \), is \( 47.0^{\circ} \). We also know the speed of light in air, \( c \), is approximately \( 3.00 \times 10^8 \; m/s \).
02

Apply Snell's Law

Use Snell's Law: \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \). Here, \( n_1 = 1 \) for air. Therefore: \( \sin(63.0^{\circ}) = n_2 \sin(47.0^{\circ}) \).
03

Calculate the Index of Refraction

Rearrange Snell's Law to find \( n_2 \): \[ n_2 = \frac{\sin(63.0^{\circ})}{\sin(47.0^{\circ})} \]. Calculate this to get \( n_2 \approx 1.33 \).
04

Determine the Speed of Light in Material

The speed of light in the material, \( v \), is given by \( v = \frac{c}{n_2} \). Substitute the values: \[ v = \frac{3.00 \times 10^8 \; m/s}{1.33} \].
05

Calculate the Result

Calculate the result: \[ v \approx 2.26 \times 10^8 \; m/s \]. This is the speed of light in the material.

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Key Concepts

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

Angle of Incidence
When light travels from one medium into another, it strikes the boundary at a certain angle. This angle, known as the angle of incidence, is measured between the incoming light ray and an imaginary line that is perpendicular to the surface at the point of contact, known as the normal. For example, in the exercise where light travels through air and hits a material, the angle of incidence is given as \(63.0^{\circ}\). Understanding the angle of incidence is crucial as it determines how much the light will bend when entering a new medium.

Here are some additional points about the angle of incidence:
  • It is always measured against the normal, not the surface itself. This helps provide consistency when applying Snell's Law.
  • The angle of incidence is vital in predicting how light behaves as it transitions from one medium to another.
  • Different angles of incidence can lead to different refractive outcomes, affecting phenomena like reflection and refraction.
Angle of Refraction
As light enters a new medium, it bends. This bending of light at the interface between two different media is primarily because light travels at different speeds in different materials. The angle at which light refracts, or bends, is called the angle of refraction.

In the given exercise, the angle of refraction is \(47.0^{\circ}\). This angle is measured between the refracted ray and the normal in the new medium. Understanding the angle of refraction is essential for predicting the path of the light ray inside the material.

Key aspects of the angle of refraction include:
  • It is analogous to the angle of incidence but is involved with the outgoing light beam.
  • The angle of refraction depends on the refractive indices of the two media, as per Snell's Law.
  • In the example, using Snell鈥檚 Law helps calculate the refractive index of the new medium based on the given angles.
Speed of Light in Material
The speed of light is not constant across different materials. Instead, it changes based on the material's optical density, which is quantified using the refractive index. In the problem, the speed of light in air is approximately \(3.00 \times 10^8 \; m/s\), but it slows down when passing through a denser material.

The formula derived from Snell's Law helps calculate the speed of light in the material, using the equation:\[ v = \frac{c}{n_2} \]where \(c\) is the speed of light in vacuum or air, and \(n_2\) is the refractive index obtained from the angles of incidence and refraction.

Noteworthy points about the speed of light in materials include:
  • A higher refractive index usually means the light will travel more slowly through that material.
  • In the exercise, the index of refraction found was approximately 1.33, indicating the speed of light slows down in this new medium compared to air.
  • Knowing the speed of light in materials is vital for applications ranging from fiber optics to lens design.

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Most popular questions from this chapter

The contacts worn by a farsighted person allow her to see objects clearly that are as close as \(25.0 \mathrm{cm},\) even though her uncorrected near point is \(79.0 \mathrm{cm}\) from her eyes. When she is looking at a poster, the contacts form an image of the poster at a distance of \(217 \mathrm{cm}\) from her eyes. (a) How far away is the poster actually located? (b) If the poster is \(0.350 \mathrm{m}\) tall, how tall is the image formed by the contacts?

A layer of oil \((n=1.45)\) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid. The angle of incidence is \(64.0^{\circ},\) and the angle of refraction is \(53.0^{\circ} .\) What is the index of refraction of the unknown liquid?

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

The near point of a naked eye is \(32 \mathrm{cm} .\) When an object is placed at the near point and viewed by the naked eye, it has an angular size of 0.060 rad. A magnifying glass has a focal length of \(16 \mathrm{cm},\) and is held next to the eye. The enlarged image that is seen is located \(64 \mathrm{cm}\) from the magnifying glass. Determine the angular size of the image.
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