Ligth clock with spaceships side-by-side In the reference frame of an observer, two spaceships travel in a straight direction (e.g. x axis) at a very high velocity and side-by-side; the distance between them is always d (km) = c (km/s) x 0.1 (s).
At its time zero, spaceship one begins the emission of one photon each 0.1 (s) to the spaceship two (parallel to y axis).
The wavelength of one photon is λ(i) = 0.001 x λ(i-1) and the first photon has a wavelength of λ(0).
Question 1: Find the wavelength of the first photon detected by spaceship two.
Question 2: Suppose that the observer can see each photon. What does he will see?
This is not homework.
 A: To answer question 2:
If it fires a photon orthogonal to its heading, it will travel orthogonal to the source's heading from the POV of the source. From the POV of an outside observer, it will travel not orthogonal to the motion of the source, but slightly along its direction of motion.
This can be easily intuited because in its own frame, the source is at rest. This means when one spaceship fires a photon to the adjacent one, the photon will hit the other ship. From the point of view of an outside observer this must still hold true, but since both ships are moving, the outside observer will not see the photon as fired orthogonal to the direction of motion; it will be angled such that it has the same velocity as the spaceships in their direction of motion, and thus can eventually hit the second ship.
The speed of the photon must, obviously, still be $c$. To find the velocity of the photon orthogonal to the direction of motion of the spaceships, all you need is a little trigonometry.
$$c^2-v^2=x^2$$
If you know $c$ and $v$ (the ship's velocity), then solve for $x$ for the component of the photon's velocity orthogonal to the direction of motion as seen by the observer. Note, this only works out so well when the moving ship emits the photon perpendicular to its heading
