Aphysics professor claims in court that the reason he went through the red light (λ = 660 nm) was that, due to his motion, the red color was doppler shifted to green (λ = 510 nm). how fast was he going?

the speed is 0.53 10⁸ m / s

Explanation:

As say in the exercise, the Doppler effect must be applied, in this case because it is an electromagnetic radiation with speed 3 10⁸ m/s and nothing can go faster we must use the relativistic Doppler effect

        fo = fe √[(c-v) /(c + v)]

Where fo is the observed frequency, fe the emitted frequency, c the speed of light and v the relative speed of the observer and emit, it is positive move away

The light fulfills the relationship

        c = λ f

        f = c / λ

In this case v is negative since the source and the observer approach

Substituting in the equation

         c /λo = c /λe √[ (c + v) / (c-v)]

         λo = 510 10⁻⁹ m

         λe = 610 10⁻⁹ m

We calculate the speed

        (λe/λo)² = (c + v) / (c-v)

       (λe /λo)² (c-v) = c + v

       v +(λe /λo)² v = (λe /λo)² c -c

       v [1 +(λe /λo)²] = c [(λe /λo)²-1]

       v = c [(λe /λo)² -1] / [1 +(λe /λo)²]

       v = 3 10⁸ [(610/510)² -1] / [1+ (610/510)²]

       v = 3 10⁸ [1.43-1] / [1 + 1.43]

       v = 0.53 10⁸ m / s

 A little more about the speed of light

i will answer your question if you answer mine

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.