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Chapter 22: Problem 60

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.

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Most popular questions from this chapter

Light polarized in the \(x\) -direction shines through two polarizing sheets. The first sheet's transmission axis makes an angle \(\theta\) with the \(x\) -axis and the transmission axis of the second is parallel to the \(y\) -axis. (a) If the incident light has intensity \(I_{0},\) what is the intensity of the light transmitted through the second sheet? (b) At what angle \(\theta\) is the transmitted intensity maximum?
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