Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 22, Problem 57P
To determine

The speed with which one has to drive to see a red light as green.

Expert Solution & Answer
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Answer to Problem 57P

The speed with which one has to drive to see a red light as green is 5×107 m/s .

Explanation of Solution

To occur a Doppler shift of magnitude in which a red light is seen as green, a relativistic relative velocity is required.

The observer is approaching the source and the source of light is stationary in this case. This implies the relative velocity is greater than zero.

vrel>0

Here, vrel is the relative velocity

Write the equation for the Doppler effect of EM waves.

fo=fs1+vrel/c1vrel/c

Here, fo is the frequency of the wave as seen by the observer, fs is the frequency of the source and c is the speed of EM waves in vacuum.

Divide both sides of the above equation by fs and square both sides of the equation.

(fofs)2=(fsfs1+vrel/c1vrel/c)2(fofs)2=1+vrel/c1vrel/c

Rearrange the above equation.

(fofs)2(1vrel/c)=1+vrel/c(fofs)2(fofs)2vrelc=1+vrelc(fofs)21=vrelc+(fofs)2vrelc(fofs)21=vrelc[1+(fofs)2]

Rewrite the above equation for vrel .

vrel=c(fofs)211+(fofs)2=c(fo/fs)21(fo/fs)2+1 (I)

Since frequency of the EM waves is inversely proportional to their wavelength, write .expression for fofs.

fofs=λsλo

Here, λs is the wavelength of the source and λo is the wavelength observed by the observer

Put the above equation in equation (I).

vrel=c(λs/λo)21(λs/λo)2+1 (II)

Conclusion:

Given that the wavelength of red is 630 nm and that of green is 530 nm . The value of c is 3.00×108 m/s .

Substitute 3.00×108 m/s for c , 630 nm for λs and 530 nm for λo in equation (II) to find vrel .

vrel=(3.00×108 m/s)(630 nm/530 nm)21(630 nm/530 nm)2+1=5×107 m/s

Therefore, the speed with which one has to drive to see a red light as green is 5×107 m/s .

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Chapter 22 Solutions

Physics

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