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A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red light of wavelength 635 nm appear green to him, with a wavelength of 550 nm. The police officer writes out a traffic citation for speeding. How fast was the physicist traveling, according to his own testimony

Sagot :

Answer:

Explanation:

[tex]\lambda[/tex] = Observed wavelength = 550 nm

[tex]\lambda'[/tex] = Actual wavelength = 635 nm

c = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]

v = Velocity of the physicist

Doppler shift is given by

[tex]f=\sqrt{\dfrac{c+v}{c-v}}f'\\\Rightarrow \dfrac{c}{\lambda}=\sqrt{\dfrac{c+v}{c-v}}\dfrac{c}{\lambda'}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}=\dfrac{c+v}{c-v}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}=\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}(1-\dfrac{v}{c})=1+\dfrac{v}{c}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}(1-\dfrac{v}{c})=1+\dfrac{v}{c}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}-1=\dfrac{v}{c}(1+\dfrac{\lambda'^2}{\lambda^2})\\\Rightarrow v=\dfrac{c(\dfrac{\lambda'^2}{\lambda^2}-1)}{1+\dfrac{\lambda'^2}{\lambda^2}}[/tex]

[tex]\Rightarrow v=\dfrac{3\times 10^8\times (\dfrac{635^2}{550^2}-1)}{1+\dfrac{635^2}{550^2}}\\\Rightarrow v=42817669.77\ \text{m/s}[/tex]

The physicist was traveling at a velocity of [tex]42817669.77\ \text{m/s}[/tex].

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