Relative to the speed of light Einstein's relativity tells us that light always travels at the speed of light relative to me, no matter how fast I'm going. Right? This really confuses me though. If light travels from A to B in one second while I'm standing at A, then where does light travel to when I'm going from A to B for one second as well, at half the speed of light? Surely the light is not further than B, is it? So what does it mean that light always travels with the speed of light relative to me?
 A: When you are moving from A to B, the distance between A and B shrinks relative to you. This is known as length contraction.
The equation is as follows:
$$L' = L \sqrt{1 - v^2/c^2}$$
where L' is the length you will see at move. L is the length at the resting reference frame. v is your speed, and c the speed of light. In your case:
$$L' = L \sqrt{1 - (\frac{1}{2}c)^2/c^2} = L \frac{\sqrt{3}}{2}$$
So your distance between A and B would only be a fraction of the distance. And you will see the light travelling that distance in the speed of light.
But still, would the light beam not travel this in just a single second? It won't! Because in your reference frame the time is not the same as the time at the rest system where the distance between A and B is L. To calculate how long it will take the light to travel from A and B in your moving time frame, we need to calculate the time difference between those two reference frames.
To calculate the time dilation you have the equation taken from here.
$$ \Delta t' = \frac{\Delta t}{\sqrt{1 - v^2 / c^2}}$$
In your case:
$$ \Delta t' = \frac{1}{\sqrt{1 - (\frac{c}{2})^2 / c^2}} = \frac{1}{\sqrt{3/4}}= \frac {2}{\sqrt{3}} sec $$
So as you can see, its going to take the light less time to cross that short distance in your moving reference frame.
