at the end of a long tunnel
Background: This is another one from you the public
From: BUCK NEKKID
Sent: Thursday, November 07, 2002 9:30 PM
Subject: My Stupid Question
Here is my stupid question, a question which has intrigued me
Iwas 5. Get a really long straight tube, about 11,160,000 miles
oflight approx 186,000 miles per second times 60 seconds) long. Seal
end. Coat the interior with some sort of miracle mirror material
100% reflectivity. Fire a high power visible laser into it for 30
The beam should now be approximately half the length of the tube.
tube. Please answer this: How long will the light keep bouncing
tube? If I open the tube will the light be visible for 30 seconds
or will the
light have filled out the tube and take a full 60 seconds to empty?
this idea be useful? I was a strange child, and probably shouldn't
read so many National Geographic magazines.
Well this one has us stumped...we asked Aladdin but he just said
*woof* so it looks like we will have to call a college campus on
this one and the next =) Well we got some user input on this one and
we can't argue because no one wanted to talk to us :( But until then
here is his input:
Info: Let me make sure I understand the question first:
At one end (spot A) we shine a laser for 30 seconds into a tube that
is so long it takes light a minute to go through. Is the person
asking what you see at the other end (spot B)? If that's the case,
you just have to remember that light is traveling at a finite speed,
so once the laser is turned off, the person at the other end will
not see *anything*, since the first photons (think of them as light
particles) are only halfway down the tube. So 30 seconds after the
laser shut off, the person at B will see the first part of the
laser; see it lit for 30 seconds, then nothing again.
You simply can only see light if it:
a) Is traveling towards you
b) Has had time enough to reach you
Oh, and I'm assuming a straight tube with the laser being shined in
the middle -- no need for the mirrored coating. If the light is
bouncing off mirrors, that means it was pointed at an angle with
respect to the tube. So replace all "30"s above with "30*cos (a)"
where "a" is the angle the laser was at, because the light is now
traveling at longer distances since some distance is "wasted" going
back and forth in the tube.
What's funny is that this question doesn't touch on the fascinating
thing about light -- that its speed is constant, even in different
frames of reference (think different speeds). Check this out:
The above problem could have been done with anything... sound waves
for example. You blow a whistle, and someone will hear it later on
at the other end.
Now let's say we've got a noisemaker in the middle of the tube, and
a person at spot A and a person at spot B. Will they both hear the
sound at the same time? Of course... the sound is going in opposite
directions, but it's still going the same distance.
Now, let's say the tube, and the people, are all on a train going
really fast (where the person in the back is at spot A, so think of
the train going left to right, with A on the left). Again, the sound
goes off in the middle of the tube... will they hear it at the same
The person a spot A "catches up" to where the noisemaker made its
noise, and so the sound didn't travel as far. While the sound had to
travel it's usual distance, plus some extra, for person B since by
the time the sound got to where person B *was*, the train moved and
person B was now up ahead a little ways.
Okay, that was easy stuff. Now for light... instead of a noisemaker
we have two laser pointers pointed in opposite directions.
Scenario 1 is the same. No one moving, the light travels the same
distance from the middle of the tube to each observer. They see it
at the same time.
Scenario 2 is different. The light still travels less distance to
the person at spot A, and more distance to the person at spot B, but
if there was some guy sitting still off the train track watching the
beam of light, he would report that they see them at the *same
time*. It's freaky. But light always travels at the same speed to
anyone who's measuring it. I could go into, but I'm at work, and
this email is long enough...
Can't argue will
not argue and probably don't understand it, is what we are saying
here at S.Q.A. although we are interested in these types of
questions, so don't be afraid to ask...Still trying to get a hold of
someone that will take us seriously.