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## Main Question or Discussion Point

i need a equation that discrip the movement of laser ray in the tube in a way that it wont get out

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i need a equation that discrip the movement of laser ray in the tube in a way that it wont get out

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Simon Bridge

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google for "total internal reflection".

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the tube is open from the both ends

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Simon Bridge

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Sounds like some sort of pipe that is hollow if it can be described as "open at both ends".

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Simon Bridge

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At each reflection, some is absorbed, some transmitted, and some reflected.

Depending on the tube material and the color of the laser, the absorption could be 100% at the first contact to almost nothing. iirc: it does not depend on the angle of the incident beam so you are only concerned to get enough reflections to absorb a significant percentage of the photons in the length of the tube.

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- #8

Simon Bridge

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Its the same as reflecting off two mirrors.

Lets say you want to reflect N times, in a tube length L and diameter D.

Put your laser on-axis, but angled towards one side.

If N=1, you need to aim the laser so it just grazes the edge of the entrance and hits half way down. If N=2, it's the same but aimed a third of the way down and so on. So you are making N+1 isosceles triangles, with height D and base 2L/(N+1). The angle of attack is therefore given by:

[itex]\tan\theta = L/(N+1)D[/itex]

If you aim the beam well off-axis though, you can get more reflections spiraling around the outside. So you have quite a lot of freedom... the same sort of argument above can be used to find out how many reflections you get off-axis.

The secret is to draw the picture.

Lets say you want to reflect N times, in a tube length L and diameter D.

Put your laser on-axis, but angled towards one side.

If N=1, you need to aim the laser so it just grazes the edge of the entrance and hits half way down. If N=2, it's the same but aimed a third of the way down and so on. So you are making N+1 isosceles triangles, with height D and base 2L/(N+1). The angle of attack is therefore given by:

[itex]\tan\theta = L/(N+1)D[/itex]

If you aim the beam well off-axis though, you can get more reflections spiraling around the outside. So you have quite a lot of freedom... the same sort of argument above can be used to find out how many reflections you get off-axis.

The secret is to draw the picture.

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Simon Bridge

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Math is a language - you use almost the same way... picture what you want, draw it, turn it over in your mind and when you can feel it, describe it in math.

Turning words into a picture is usually the slow part.

Look how long it took to get the right picture of what you wanted.

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