Where laser beam will fall (see the attachment)

[jubair.pk]

Please refer the attached setup.

* The laser source is moving at very high speed (say, V=c/2).
* The slit is very close to laser source.
* The screen is far away from the laser source.

I would like to know,

On the 'screen', where the laser beam will fall. A or B?


-Jubair

 

[ghwellsjr]

B, according to the way you drew the picture.

 

[jambaugh]

Imagine as the bit of light which hits point B as it is leaving the laser. Recall the laser has a certain length and so it takes some time for the bit of light to traverse that length. In that time the laser has moved so the direction (in our frame) the light must travel to start from the back of the laser, then exit the front of the laser is the direction in line to B.

BTW, the results are the same whether you're talking Laser moving at c/2 or gun moving at half the muzzle velocity, or pitcher throwing a 90mph fast ball from a train moving 45mph. There's no "special" relativity issues here.

 

[Dale]

Is the slit intended to cause a diffraction pattern or just to effectively turn the laser on and off at specific times? If the latter then you may want to replace it with a switch.

Assuming you are not interested in diffraction then the two previous posters are correct about B. Even if you are interested in diffraction the peak will still be at B but some light could diffract to A.

 

[lightarrow]

If you consider the problem in the frame of reference of the laser source, everything becomes simpler: the source is stationary and the screen moves down. Since light speed is finite, in the time interval between the instant a light pulse exits the source and the instant it arrives on the screen, the screen has moved down, so the light pulse will hit it in a point above A.
You can easily see that this time interval is L/c where L = 100 miles. So the screen has moved down of c/2 * L/c = L/2 = 50 miles. So the point B hit by the light is 50 miles above the point A.

 

[grav-universe]

Individual photons will follow an angled path, like that of B. According to the frame of the screen, whatever height the laser is at when the photon strikes the screen is the same height the photon will strike. However, although the path of a single photon is angled, a steady stream of photons fired by the laser will form a straight horizontal line projecting from the laser at all times, because all of the photons are following the same angle at the same vertical speed of the laser.

 

[grav-universe]

In the frame of the laser, imagine a pipe that extends directly outward from the "nozzle" of the laser. Since the photon travels in a straight line from the laser, it will not hit the edges of the pipe. Now look at how the same thing will occur in the frame of the screen. The pipe still extends outward from the laser and of course both frames must agree that the photons do not strike the edge of the pipe. Since the pipe is moving with the laser according to the frame of the screen, a photon that is fired from the laser must travel at an angle with the same vertical speed of the pipe and laser in order to not strike the side of the pipe. A stream of photons, however, will form a straight horizontal line with the pipe at all times, although the individual photons are traveling at an angle.

 

[grav-universe]

If you consider the problem in the frame of reference of the laser source, everything becomes simpler: the source is stationary and the screen moves down. Since light speed is finite, in the time interval between the instant a light pulse exits the source and the instant it arrives on the screen, the screen has moved down, so the light pulse will hit it in a point above A.
You can easily see that this time interval is L/c where L = 100 miles. So the screen has moved down of c/2 * L/c = L/2 = 50 miles. So the point B hit by the light is 50 miles above the point A.

That is true in the frame of the laser. The length of the screen will be contracted by sqrt[1 - (v/c)^2] = .866 in the frame of the laser also, however, so in the frame of the screen, B will be (50 miles) / .866 = 57.7 miles above A. Another way to see this in the frame of the screen is that the horizontal distance to the screen and the distance the laser travels vertically will become the sides of a right triangle while the distance the photon travels becomes the hypotenuse. Since the photon travels twice the distance that the laser does in the same time, the distances will form a 30, 60, 90 degree triangle. The distance of B above A in the frame of the screen, then, is (tan 30 degrees) * 100 miles = 57.7 miles.

 

[lightarrow]

That is true in the frame of the laser. The length of the screen will be contracted by sqrt[1 - (v/c)^2] = .866 in the frame of the laser also, however, so in the frame of the screen, B will be (50 miles) / .866 = 57.7 miles above A. Another way to see this in the frame of the screen is that the horizontal distance to the screen and the distance the laser travels vertically will become the sides of a right triangle while the distance the photon travels becomes the hypotenuse. Since the photon travels twice the distance that the laser does in the same time, the distances will form a 30, 60, 90 degree triangle. The distance of B above A in the frame of the screen, then, is (tan 30 degrees) * 100 miles = 57.7 miles.

Certainly, thanks for this correction.