Question

I was just thinking about the case,

The whole cenario is in space there was an space craft it is attached to a mirror say 1.5 *10⁸ m below the space craft by many taut rods. Space craft is moving horizontally with high constant velocity. A person in space craft throws a ray of light to the attached mirror, velocity of light is 3*10⁸ m/sec. So time taken by ray of light to come back at initial position is 1sec.

There was another observer w.r.t him ray of light follows a v-shaped path where the edge of v-shaped path is 1.5*10⁸ m below so each part of path must be greater then 1.5*10⁸ m this mean overall distance is increased ! So time taken also increased!

Both person start as well as stop their stop watch simultaneously.

But it read different time period how?

What is happening here?

Answer 1

Since the distamce is increased,and the velocity being constant (3*10^8 m/s) the time for the light beam to reach the initail position will surely increase.Then what do you mean by starting and stoping the stopwatch simultaneosly.?The stopwatch of the first person reads a time of exact 1 second whereas the stopwatch of the second person reads a time greater than 1 second.

Hope you understood what i said...!

But the same is not practical.because, a beam of light will surely scatter when travelling such a long distance of 3*10^8 m .