/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Light in vacuum is incident on t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Light in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of \(32.0^{\circ}\) with the normal to the surface, while in the glass it makes an angle of \(16.0^{\circ}\) with the normal. What is the index of refraction of the glass?

Short Answer

Expert verified
The index of refraction of the glass is approximately 1.923.

Step by step solution

01

Understanding Snell's Law

To find the index of refraction of the glass, we will use Snell's Law, which relates the angles of incidence and refraction to the indices of refraction of two media. Snell's Law is expressed as: \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \), where \(n_1\) and \(\theta_1\) are the index of refraction and angle in the first medium (vacuum), and \(n_2\) and \(\theta_2\) are those in the second medium (glass).
02

Assign Given Values

In this problem, the light is incident from vacuum, so \( n_1 = 1.0 \). The angle of incidence, \( \theta_1 \), is given as \( 32.0^{\circ} \), and the angle of refraction, \( \theta_2 \), is \( 16.0^{\circ} \). We need to solve for \( n_2 \), the index of refraction of the glass.
03

Apply Snell's Law

Substitute the given values into Snell's Law: \( 1.0 \cdot \sin(32.0^{\circ}) = n_2 \cdot \sin(16.0^{\circ}) \). Simplify this equation to find \( n_2 \).
04

Calculate Sine Values

Compute the sine values for the angles: \( \sin(32.0^{\circ}) \approx 0.5299 \) and \( \sin(16.0^{\circ}) \approx 0.2756 \).
05

Solve for the Index of Refraction

Now with the sine values calculated, the equation becomes: \( 0.5299 = n_2 \cdot 0.2756 \). Solve for \( n_2 \) by dividing both sides by \( 0.2756 \), giving \( n_2 = \frac{0.5299}{0.2756} \approx 1.923 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Angle of Incidence
The concept of the angle of incidence is foundational in understanding how light interacts with different surfaces. This angle is defined as the angle between the incident light ray and the normal (a line perpendicular) to the surface at the point of contact. It is important in the study of optics and Snell's Law.
An incidents example is when a beam of light traveling through the air, or in this case, a vacuum, approaches another medium, such as glass. The angle of incidence helps determine how much the light will "bend" or change direction when it enters the second medium.
In the scenario with our exercise, the angle of incidence is given as 32 degrees. This angle tells us how the light initially approaches the glass slab. It's a critical component because together with Snell's Law, it lets us eventually calculate the refractive index of the glass.
Angle of Refraction
After light enters a new medium, such as glass, the angle at which it travels within this medium compared to the normal line is known as the angle of refraction. This angle is influenced by factors such as the properties of the new medium and the original angle of incidence.
In the exercise example, when light crosses from a vacuum into glass, it slows down due to the optical density of the glass. This results in the bending of the light path, measurable as the angle of refraction. For the given problem, this angle is 16 degrees, illustrating a significant change from its original path.
Understanding the angle of refraction is crucial, as it allows us to quantify this change and, using Snell's Law, also connect it to a medium’s index of refraction, which in turn affects how much the light bends.
Index of Refraction
The index of refraction, or refractive index, is a measure of how much light bends as it passes from one medium into another. It's a dimensionless number that indicates how much slower light travels in that medium compared to in a vacuum.
It arises in Snell's Law, represented by the symbol "n." Using it alongside angles of incidence and refraction, we can solve for unknown quantities, such as in our exercise where we determined the index of refraction of glass. The relationship is given by the formula: \[ n_1 \sin{\theta_1} = n_2 \sin{\theta_2} \]
In the original exercise, the index of refraction for the vacuum was given a value of 1.0, allowing us to solve for the glass's index using the angles provided. With all the given data, the refractive index of the glass was calculated to be approximately 1.923. This tells us that light travels almost twice as slowly in glass as it does in a vacuum.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A small spaceship with a mass of only \(1.5 \times 10^{3} \mathrm{~kg}\) (including an astronaut) is drifting in outer space with negligible gravitational forces acting on it. If the astronaut turns on a \(25 \mathrm{~kW}\) laser beam, what speed will the ship attain in \(45.0\) day because of the momentum carried away by the beam?

It has been proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of foil. How large must the surface area of the sail be if the radiation force is to be equal in magnitude to the Sun's gravitational attraction? Assume that the mass of the ship \(+\) sail is \(1800 \mathrm{~kg}\), that the sail is perfectly reflecting, and that the sail is oriented perpendicular to the Sun's rays. See Appendix C for needed data. (With a larger sail, the ship is continuously driven away from the Sun.)

Figure 33-45 depicts a simplistic optical fiber: a plastic core \(\left(n_{1}=1.58\right)\) is surrounded by a plastic sheath \(\left(n_{2}=1.46\right)\). A light ray is incident on one end of the fiber at angle \(\theta\). The ray is to undergo total internal reflection at point \(A\), where it encounters the core-sheath boundary. (Thus there is no loss of light through that bound- ary.) What is the maximum value of \(\theta\) that allows total internal reflection at \(A\) ?

Sunlight just outside Earth's atmosphere has an intensity of \(1.40 \mathrm{~kW} / \mathrm{m}^{2}\). Calculate (a) \(E_{m}\), (b) \(B_{m}\), (c) \(E_{\mathrm{rms}}\), and (d) \(B_{\mathrm{rms}}\) for sunlight there, assuming it to be a plane wave.

Some neodymium-glass lasers can provide \(100 \mathrm{TW}\) of power in \(2.2 \mathrm{~ns}\) pulses at a wavelength of \(0.26 \mu \mathrm{m}\). How much energy is contained in a single pulse?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.