/*! 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 32 The source in Young's experiment... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 32

The source in Young's experiment emits at two wavelengths. On the viewing screen, the fourth maximum for one wavelength is located at the same spot as the fifth maximum for the other wavelength. What is the ratio of the two wavelengths?

Short Answer

Expert verified
The ratio of the two wavelengths in Young's experiment is \(\frac{\lambda_1}{\lambda_2} = \frac{5}{4}\).

Step by step solution

01

Write the conditions for maxima for both wavelengths

According to the given situation, we can write the conditions for maxima for both wavelengths as: \(d\ sin(\theta_1)=4\lambda_1\) and \(d\ sin(\theta_2)=5\lambda_2\) Since both maxima coincide, the angles 胃1 and 胃2 must be equal, so we have: \(d\ sin(\theta_1)=d\ sin(\theta_2)\)
02

Set up the equation relating the two wavelengths

Using the conditions for maxima, we can set up the equation relating the two wavelengths 位1 and 位2: \(4\lambda_1=5\lambda_2\)
03

Solve for the ratio of the two wavelengths

Now, we will solve for the ratio: \(\frac{\lambda_1}{\lambda_2}=\frac{5\lambda_2}{4\lambda_1}\) Divide both sides by 位1: \(\frac{\lambda_1}{\lambda_2} \times \frac{1}{\lambda_1} = \frac{5\lambda_2}{4\lambda_1} \times \frac{1}{\lambda_1}\) Which simplifies to: \(\frac{\lambda_1}{\lambda_2} = \frac{5}{4}\) So, the ratio of the two wavelengths is 5:4.

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.

Understanding Interference in Young's Double-Slit Experiment
When we talk about interference in the context of Young's Double-Slit Experiment, we're looking at a fascinating optical phenomenon. Interference happens when two waves overlap and combine, either amplifying or canceling each other. In Young's experiment, light waves from two slits meet and form a pattern of alternating bright and dark bands on a screen. The bright bands, known as maxima, occur where waves constructively interfere, meaning their crests and troughs align perfectly, reinforcing each other.
  • The positions of these maxima are determined by the condition that the path difference between the two waves is an integer multiple of the wavelength. This path difference is connected to the slit spacing (d), the angle of observation (\(\theta\)), and the wavelength (\(\lambda\)).
  • Constructive interference at the screen is given by the formula \[d\cdot \sin(\theta) = m\cdot \lambda\]
  • The parameter \(m\) is an integer, representing the order of the maximum (e.g., first, second, third maximum, etc.).
In the original exercise, two different wavelengths produce masixma that coincide, requiring an understanding of their interference patterns.
Delving into Wavelength Ratios
One of the key tasks in problems like Young's Double-Slit Experiment is understanding how different wavelengths relate to each other in terms of their interference patterns. A wavelength ratio indicates how one wavelength compares to another.In our scenario, the fourth maximum of one wavelength corresponds to the fifth maximum of another. This is mathematically described by setting up equations based on their maxima conditions and equating them since they occur at the same angle.
  • The relation for maxima of these two light waves are \(d\cdot\sin(\theta_1) = 4\lambda_1\) and \(d\cdot\sin(\theta_2) = 5\lambda_2\).
  • Since they coincide, \(\theta_1 = \theta_2\), leading us to equate \(4\lambda_1 = 5\lambda_2\).
  • Solving this gives the ratio \(\frac{\lambda_1}{\lambda_2} = \frac{5}{4}\).
Thus, the wavelengths are in the ratio of 5:4, meaning one wavelength is 1.25 times longer than the other.
Exploring Optical Physics and Its Principles
Optical physics is a branch of physics focused on the study of light and its interactions with matter. It includes phenomena like reflection, refraction, diffraction, and interference 鈥� all crucial for understanding Young's Double-Slit Experiment. Young's experiment is a cornerstone in optical physics because it provides clear evidence of the wave nature of light. Using just coherent light beams and slits, interference patterns demonstrate how light waves superpose.
  • Reflection and refraction describe the bending of light rays, essential for designing lenses and understanding how our eyes focus.
  • Diffraction is important for explaining the spread of light waves when encountering obstacles, not unlike the slit setup in Young's experiment.
  • Interference demonstrates that light can constructively and destructively combine, forming colorful patterns or intensity variations.
These principles have broad applications, ranging from the development of optical devices to the exploration of quantum mechanics. Understanding such concepts helps connect theory to technological advancements.

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

In a double-slit experiment, the fifth maximum is 2.8 \(\mathrm{cm}\) from the central maximum on a screen that is \(1.5 \mathrm{m}\) away from the slits. If the slits are \(0.15 \mathrm{mm}\) apart, what is the wavelength of the light being used?

To save money on making military aircraft invisible to radar, an inventor decides to coat them with a nonreflective material having an index of refraction of \(1.20,\) which is between that of air and the surface of the plane. This, he reasons, should be much cheaper than designing Stealth bombers. (a) What thickness should the coating be to inhibit the reflection of \(4.00-\mathrm{cm}\) wavelength radar? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

An inventor notices that a soap bubble is dark at its thinnest and realizes that destructive interference is taking place for all wavelengths. How could she use this knowledge to make a nonreflective coating for lenses that is effective at all wavelengths? That is, what limits would there be on the index of refraction and thickness of the coating? How might this be impractical?

A hydrogen gas discharge lamp emits visible light at four wavelengths, \(\lambda=410,434,486,\) and \(656 \mathrm{nm}\) (a) If light from this lamp falls on a \(N\) slits separated by \(0.025 \mathrm{mm},\) how far from the central maximum are the third maxima when viewed on a screen \(2.0 \mathrm{m}\) from the slits? (b) By what distance are the second and third maxima separated for \(l=486 \mathrm{nm}\) ?

Red light \((\lambda=710 . \mathrm{nm})\) illuminates double slits separated by a distance \(d=0.150 \mathrm{mm} .\) The screen and the slits are \(3.00 \mathrm{m}\) apart. (a) Find the distance on the screen between the central maximum and the third maximum. (b) What is the distance between the second and the fourth maxima?
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Physics of Motion

Read Explanation

Electricity

Read Explanation

Modern Physics

Read Explanation

Oscillations

Read Explanation

Energy Physics

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.