Okay, whenever I have trouble visualizing a concept, I always try to look up some proofs with the aid of a diagram. Let us do the same.
Let's assume that we have a clock with a laser, that 'ticks' every time the beam hits the laser after reflecting off:
___________ (That is the top surface that it reflects off)
| (That is the beam of light that is going to hit the surface and reflect off
____()______ (That is the bottom surface, and the thing in between is the part where the beam was blasted)
Let's say that the distance between the shooting surface and the reflecting surface is L and the speed of light is C. So the time taken : T = 2L/C.
Now let's assume that the surfaces are moving at some velocity, let's call it V.
_______A_________ (Again our top surface, but in motion)
/ \
/ \ (Since, the ship is in motion, the beam blaster thingy will move, and due
/ \ to relativity, the beam will be like that too, on an angle)
/ \
/ \
___()______()___
B D C
So the distance between the midpoint at the bottom and the beginning is point where it reaches the top will be: BD = vt/2 and the length is still L. Using pythagorean theorem, we can say that BD^2 + L^2 = AB^2, so (BD^2 + L^2)^0.5 = AB. Since AB and CD are identical, we can say that : 2*(BD^2 + L^2)^0.5 = AB = Ct(1). C is still the speed of light, and t(1) will be the time it took for it to travel that distance. Since the speed of light is a constant, we can say the above.
So now we have the following:
Ct(1) = 2*(BD^2 + L^2)^0.5
With a some math bashing, which you should be able to do, like square both sides and get t(1) on one side, you should get this:
t(1) = 2L / (C (1 - V^2/c^2))
But remember we said already that 2L/C was T? So now:
t(1) = T / (1 - c^2/V^2)
where t(1) is the time indicated by the faster person's clock, T is the time indicated by the slower person's clock, and V is the velocity of the faster person with respect to the slower person. Still don't get it, well here's a story for you :D
There are two twin brothers. On their thirtieth birthday, one of the brothers goes on a space journey in a superfast rocket that travels at 99% of the speed of light. The space traveller stays on his journey for precisely one year, whereupon he returns to Earth on his 31st birthday. On Earth, however, seven years have elapsed, so his twin brother is 37 years old at the time of his arrival. This is due to the fact that time is stretched by factor 7 at approx. 99% of the speed of light, which means that in the space traveller’s reference frame, one year is equivalent to seven years on earth. Yet, time appears to have passed normally to both brothers, i.e. both still need five minutes to shave each morning in their respective reference frame.
Still not? Then go to youtube, and write Physics for Smarties: Time Dilation. He does pretty much what I did, but with better diagrams :S. Anyways, hope I helped. :D