a) A distant spaceship carrying a red headlight,

emitting light with wavelength 600 nm, is heading
towards the earth. Observers on the earth detect the
light from the headlight as ultraviolet light with
wavelength 200 nm. What is the velocity of the
spaceship?
b) After the spaceship passed through the earth it
continues its flight. What frequency of light do
earth observers see?

a) To determine the velocity of the spaceship, we can use the principle of Doppler effect. The equation for the Doppler effect is:

\(\frac{Δλ}{λ_0} = \frac{v}{c}\)

Where:
Δλ = Change in wavelength
λ_0 = Initial wavelength
v = Velocity of the source (spaceship)
c = Speed of light in vacuum

In this case, the initial wavelength (λ_0) is 600 nm, and the observed wavelength (Δλ) is 200 nm. We need to find the velocity (v) of the spaceship.

Rearranging the equation, we get:
\(v = \frac{Δλ}{λ_0} \times c\)

Substituting the values, we have:
\(v = \frac{200\,nm}{600\,nm} \times 3.00 \times 10^8\,m/s\)

Calculating this value, the velocity of the spaceship is approximately 1.00 x 10^8 m/s.

b) After the spaceship has passed through the Earth, the frequency of light observed by Earth observers can be calculated using the equation:

\(f' = \frac{f}{(1 + \frac{v}{c})}\)

Where:
f' = Observed frequency
f = Initial frequency
v = Velocity of the source (spaceship)
c = Speed of light in vacuum

The initial frequency (f) corresponds to the initial wavelength (λ_0) from the previous question. The velocity of the spaceship (v) is the same as calculated earlier, and we need to find the observed frequency (f').

The relationship between frequency and wavelength is given by:
\(f = \frac{c}{λ}\)

Rearranging this equation, we can express the observed frequency as:
\(f' = \frac{c}{λ_0 + Δλ}\)

Substituting the values, we have:
\(f' = \frac{3.00 \times 10^8\,m/s}{600\,nm + 200\,nm}\)

Calculating this value, the observed frequency of light seen by Earth observers is approximately 4.29 x 10^14 Hz.