## Length contraction & light

Hey, sorry if this is a stupid question but I'm getting really confused...
If I am standing on the Earth and watch a spaceship moving past at say 0.8*c from left to right, I will observe the spaceship to contract in its direction of motion. If a person on the spaceship shines a torch also from left to right, what happens to the wavelength of the light as seen from my reference frame? Does it also contract?

Recognitions:
In the ship's frame, the wave peaks move at c, so the distance between peaks is just c/f, where f is the frequency that the peaks are being emitted by the torch as seen in the ship's frame. In your frame, the peaks still move at c, but the frequency becomes $$f \sqrt{1 - v^2/c^2}$$ due to time dilation, and meanwhile the torch is moving at velocity v so it will have moved a distance of $$v / (f \sqrt{1 - v^2/c^2})$$ between emitting successive peaks, so the the distance between peaks should either be $$c / (f \sqrt{1 - v^2/c^2}) - v / (f \sqrt{1 - v^2/c^2})$$ or $$c / (f \sqrt{1 - v^2/c^2}) + v / (f \sqrt{1 - v^2/c^2})$$ depending on whether the torch is shining in the direction of the ship's motion or in the opposite direction. This simplifies to $$(c \pm v)/(f \sqrt{1 - v^2/c^2})$$. You could also get this from the relativistic doppler shift equation, $$f_{observed} = f_{emitted} \sqrt{1 - v^2/c^2} / (1 - v/c)$$, keeping in mind that the wavelength you observe is $$c / f_{observed}$$.
To compare the wavelength you see with the wavelength seen by the ship-observer, just divide $$(c \pm v)/(f \sqrt{1 - v^2/c^2})$$ (the wavelength seen by you) by c/f (the wavelength seen by the ship-observer), which gives $$(1 \pm v/c) / \sqrt{1 - v^2/c^2}$$ for the factor that the wavelength changes in your frame. This is not the same as the Lorentz contraction factor $$\sqrt{1 - v^2/c^2}$$.