Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What must be the relative speed between source and receiver if the wavelength of an EM wave as measured by the receiver is twice the wavelength as measured by the source? Are source and observer moving closer together or farther apart?

Short Answer

Expert verified
Answer: The relative speed between the source and the receiver is (3/5)c, and they are moving closer together.

Step by step solution

01

Understand the Doppler Effect formula

The Doppler effect formula for electromagnetic waves, when the source and the receiver are moving along the line of sight, is given by: λ' = λ√((c + v_r) / (c - v_r)) Here, λ' is the wavelength measured by the receiver, λ is the wavelength measured by the source, c is the speed of light, and v_r is the relative velocity. We know that λ' = 2λ, so we can rewrite the equation as follows: 2λ = λ√((c + v_r) / (c - v_r))
02

Solve for v_r

We will now solve the equation for the relative velocities between the source and the receiver: 2λ = λ√((c + v_r) / (c - v_r)) Divide both sides by λ: 2 = √((c + v_r) / (c - v_r)) Square both sides: 4 = (c + v_r) / (c - v_r) Now, cross-multiply and simplify: 4(c - v_r) = (c + v_r) 4c - 4v_r = c + v_r Rearrange the terms to isolate v_r: 5v_r = 3c v_r = (3/5)c
03

Determine the direction of motion

Now that we have determined the relative velocity, we can determine the direction of motion: v_r = (3/5)c Since the relative velocity, v_r, is positive, the source and the receiver are moving closer together.
04

Final answer

The relative speed between the source and the receiver is (3/5)c, and they are moving closer together.

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 Vaia!

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

Light of wavelength \(659.6 \mathrm{nm}\) is emitted by a star. The wavelength of this light as measured on Earth is \(661.1 \mathrm{nm} .\) How fast is the star moving with respect to Earth? Is it moving toward Earth or away from it?
How long does it take light to travel from this text to your eyes? Assume a distance of \(50.0 \mathrm{cm}\).
A police car's radar gun emits microwaves with a frequency of $f_{1}=36.0 \mathrm{GHz}$. The beam reflects from a speeding car, which is moving away at \(43.0 \mathrm{m} / \mathrm{s}\) with respect to the police car. The frequency of the reflected microwave as observed by the police is \(f_{2} .\) (a) Which is larger, \(f_{1}\) or \(f_{2} ?\) (b) Find the frequency difference \(f_{2}-f_{1}\) [Hint: There are two Doppler shifts. First think of the police as source and the speeder as observer. The speeding car "retransmits" a reflected wave at the same frequency at which it observes the incident wave. \(]\)
The magnetic field in a microwave traveling through vacuum has amplitude $4.00 \times 10^{-11} \mathrm{T}\( and frequency \)120 \mathrm{GHz}$. Find the amplitude and frequency of the electric field.
In astronomy it is common to expose a photographic plate to a particular portion of the night sky for quite some time in order to gather plenty of light. Before leaving a plate exposed to the night sky, Matt decides to test his technique by exposing two photographic plates in his lab to light coming through several pinholes. The source of light is \(1.8 \mathrm{m}\) from one photographic plate and the exposure time is \(1.0 \mathrm{h}\). For how long should Matt expose a second plate located \(4.7 \mathrm{m}\) from the source if the second plate is to have equal exposure (that is, the same energy collected)?
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.

Sign-up for free