Trigonometry (?) help

I've been think for an hour about this and came to some basic conclusions, but I am unable to go any further;

There are two mirrors, each 1m in dimension, standing 1m apart. Between them is released a beam of light. What is the maximum angle one of the mirrors can deviate from being parallel to the other one, such that the beam will not escape the space between the mirrors for 0.001s.

So I have calculated how the angles at which the beam will be reflected are related to the angle of deviation, and some initial lengths, but I don't know what to do with this one later. Help please?

The light will bounce back and forth a number of times. What you need to do is use the speed of light and the distance between the mirrors to figure out how many bounces you'll get. Then apply trig to get the maximum angle.

>>2
Ummm, what?

The question is a little ambiguous, so I'm going to assume that this picture is basically what you meant:

http://img33.imageshack.us/my.php?image=15694489.jpg

(Hooray MS Paint!) The blue lines are the mirrors, and d is the deviation angle.  The yellow line is the light, which starts at the midpoint of the line joining the centers of the mirrors.  r is the distance from the midpoints of the mirrors to the intersection of the lines through the mirrors.  You can get an expression for r in terms of d alone by using trig.

Now at each step, instead of bending the line, leave the line straight and reflect the mirrors with respect to each other, like in the pic:

http://img200.imageshack.us/my.php?image=38102863.jpg

Now the centers of all the reflected mirrors are in a circle, so we just need to find where the yellow line first crosses a line outside the blue area.  With simple trig you can figure out the slope of the yellow line as

m=tan(90+d/2)

and since it goes through the point (r,0), it has equation

y=m(x-r)

The outside lines of the mirrors are in a circle which has equation

x^2+y^2 = (r+1/2)^2

If you solve these two equations together (which will probably be messy as hell), you'll find a point (x,y) defined in terms of d alone that gives the point at which the light breaks out of the circle defined by the outsides of the mirrors.

To get the distance the light travelled, find the distance of this point from the point where the light starts with the distance formula.  Then set that expression equal to the distance that light travels in .001 seconds and solve for d.  That will be the highest that the angle can be.

Maple sez about 0.0018274 radians.

c ~ 3e8 m/s

>>4
Clever; it's a lot better than the brute force attack I was thinking of.

>>4
Oh by the way, YOU'RE WELCOME! >:(

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>>9

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>>10-12
What is this faggotry doing on my /sci/?

>>13
Fuck off, this is MY /sci/.

>>14
Back to /b/,

>>15

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>>16
Back to /b/,

>>17


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>>15
>>17
Oh noes, he left off the "please".  He must be getting *really* pissed off now. :O

>>18
Back to /b/,

>>19
Meh, just being lazy about it now.

>>20

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Dammit, if you're gonna be an analdouchebag, you can't be fucking *lazy*.  lrn2 intertubes.

>>21
Back to /b/,

you can't be fucking *lazy*
lrn2
Haha.

>>22

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I said YOU can't be lazy. 

I can.

>>23
Back to /b/,

NO DOUBLE STANDARDS

>>24

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In that case, why should you be allowed to hang around when everyone else here has original thoughts and an IQ above 90?

>>25
Back to /b/,

>>26


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I rest my case.

>>27
Back to /b/,

  o           .'`/
      '      /  (
    O    .-'` ` `'-._      .')
       _/ (o)        '.  .' /
       )       )))     ><  <
       `\  |_\      _.'  '. \
         '-._  _ .-'       '.)
             `\__\

>>28
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>>30
Back to /b/,

Good.