/*! 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 15 A ray of light is incident on a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 33: Problem 15

A ray of light is incident on a plane surface separating two sheets of glass with refractive indexes 1.70 and 1.58 . The angle of incidence is \(62.0^{\circ},\) and the ray originates in the glass with \(n=1.70 .\) Compute the angle of refraction.

Short Answer

Expert verified
The angle of refraction is approximately \(71.82^{\circ}\).

Step by step solution

01

Understand Snell's Law

Snell's Law relates the angle of incidence and angle of refraction for a wave impinging on an interface between two media with different refractive indices, using the formula: \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \). Here, \( n_1 = 1.70 \) and \( n_2 = 1.58 \), and \( \theta_1 = 62.0^{\circ} \). We need to find \( \theta_2 \).
02

Apply Snell's Law

Insert the given values into Snell's Law: \( 1.70 \sin(62.0^{\circ}) = 1.58 \sin(\theta_2) \). This equation will allow us to solve for \( \theta_2 \).
03

Solve for \( \sin(\theta_2) \)

Calculate \( \sin(62.0^{\circ}) \) using a calculator to find \( 0.88295 \). Then substitute back into the equation: \( 1.70 \, \times \, 0.88295 = 1.58 \sin(\theta_2) \). Simplify to find \( \sin(\theta_2) = \frac{1.70 \, \times \, 0.88295}{1.58} \approx 0.949 \).
04

Calculate \( \theta_2 \)

Now, use the inverse sine function to find \( \theta_2 \): \( \theta_2 = \arcsin(0.949) \). Using a calculator, find that \( \theta_2 \approx 71.82^{\circ} \).
05

Double-Check Calculation

Finally, re-check the insertion of values and computation to ensure accuracy, reaffirming that the angle of refraction \( \theta_2 \) is approximately \( 71.82^{\circ} \).

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 91影视!

Key Concepts

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

Snell's Law
Snell's Law is a fundamental principle in the study of light and optics. It describes how light bends, or refracts, when it passes from one medium to another. This law is named after the Dutch mathematician, Willebrord Snellius. Snell's Law is represented by the equation:\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]Here:
  • \(n_1\) is the refractive index of the first medium.
  • \(\theta_1\) is the angle of incidence, the angle between the incoming ray and the normal to the interface.
  • \(n_2\) is the refractive index of the second medium.
  • \(\theta_2\) is the angle of refraction, the angle between the refracted ray and the normal.
This relationship tells us how the angle of the light ray changes as it enters a different medium. By understanding this law, we can predict how light will behave at the boundary between two materials. This principle is critical in designing lenses and understanding optical phenomena like rainbows.
Refractive Index
The refractive index of a material indicates how much the speed of light is reduced inside it compared to the speed in a vacuum. It is a dimensionless number typically greater than 1. Light travels slower in materials with higher refractive indices.
  • A vacuum has a refractive index of exactly 1.
  • In air, the refractive index is slightly more than 1, approximately \(1.0003\).
  • Glass has a higher refractive index, often ranging between 1.5 to 1.7, depending on its composition.
When light passes from a medium with a higher refractive index, like glass, into one with a lower refractive index, like air, it speeds up and bends away from the normal. Conversely, light slows and bends towards the normal when moving into a medium with a higher refractive index. This bending of light is a foundational concept in optics.
Angle of Refraction
The angle of refraction is the angle between the refracted light ray and the normal to the surface at the point of refraction. It is crucial in determining how much a light ray bends when passing through mediums with different refractive indices.
  • When light enters a denser medium (higher refractive index), it bends towards the normal, resulting in a smaller angle of refraction compared to the angle of incidence.
  • However, when light exits a denser medium into a less dense one, the angle of refraction will be larger than the angle of incidence, causing the light to bend away from the normal.
Using Snell's Law, one can calculate the specific angle of refraction when the angle of incidence and both refractive indices are known. This calculation is essential for understanding optical devices like lenses and fiber optics, as they rely on precise control of light refraction.

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

The refractive index of a certain glass is \(1.66 .\) For what incident angle is light reflected from the surface of this glass completely polarized if the glass is immersed in (a) air and (b) water?

We define the index of refraction of a material for sound waves to be the ratio of the speed of sound in air to the speed of sound in the material. Snell's law then applies to the refraction of sound waves. The speed of a sound wave is 344 \(\mathrm{m} / \mathrm{s}\) in air and 1320 \(\mathrm{m} / \mathrm{s}\) in water. (a) Which medium has the higher index of refraction for sound? (b) What is the critical angle for a sound wave incident on the surface between air and water? (c) For total internal reflection to occur, must the sound wave be traveling in the air or in the water? (d) Use your results to explain why it is possible to hear people on the opposite shore of a river or small lake extremely clearly.

A ray of light is traveling in a glass cube that is totally immersed in water. You find that if the ray is incident on the glass-water interface at an angle to the normal larger than \(48.7^{\circ}\) , no light is refracted into the water. What is the refractive index of the glass?

(a) A tank containing methanol has walls 2.50 \(\mathrm{cm}\) thick made of glass of refractive index \(1.550 .\) Light from the outside air strikes the glass at a \(41.3^{\circ}\) angle with the normal to the glass. Find the angle the light makes with the normal in the methanol. (b) The tank is emptied and refilled with an unknown liquid. If light incident at the same angle as in part (a) enters the liquid in the tank at an angle of \(20.2^{\circ}\) from the normal, what is the refractive index of the unknown liquid?

Light is incident normally on the short face of a \(30^{\circ}-\) \(60^{\circ}-90^{\circ}\) prism (Fig. \(\mathrm{P} 33.52 ) . \mathrm{A}\) drop of liquid is placed on the hypotenuse of the prism. If the index of refraction of the prism is \(1.62,\) find the maximum index that the liquid may have if the light is to be totally reflected.
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Electrostatics

Read Explanation

Engineering Physics

Read Explanation

Particle Model of Matter

Read Explanation

Magnetism

Read Explanation
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.