1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Behaviour of light thru a glass rod 'end to end'

  1. Nov 12, 2014 #1
    Pre-amble
    A glass rod with polished faces.
    Light goes in one end, and exits by the other end.

    We hear of light bouncing down fibre optics.
    What then happens with a glass tube, that has a rough ground inner and outer surface?
    Or worse..... when a part of the inner diameter is 'hot melt enlarged', creating a section with a larger OD, so that the light must now follow a bend in the glass, then up the tube walls and out of the end?

    Here is the Experiment:

    Take a capillary glass tube say 8mm OD x 3mm ID.
    Cut 4 off to size.... say 50mm
    Polish both ends (Face A & B)

    Sample 1
    The OD and ID are to a 'glass finish'.
    (as per supplied - a good quality borosilicate capillary tube)
    100% light in at Face A........ w % light out at Face B

    Sample 2
    The OD and ID are roughed up say 'sandblast finish'
    100% light in at Face A........ x % light out at Face B

    Sample 3
    At Face B...... the 3mm ID is hot formed to 5mm, for a depth of 15mm
    The enlarged section OD is now say 11mm
    The OD and ID are to a 'glass finish'.
    The light still enters by Face A..... but it now must pass through a reverse bottleneck.
    100% light in at Face A........ y % light out at Face B

    Sample 4
    At Face B...... the 3mm ID is hot formed to 5mm, for a depth of 15mm
    The enlarged section OD is now say 11mm
    The OD and ID are to a 'rough finish'.
    The light still enters by Face A..... but it now must pass through a reverse bottleneck.
    100% light in at Face A........ z % light out at Face B

    Notes:
    The area of Face A = area of Face B in all cases.

    Can we 'in principal' predict the loss in light transmission for all cases?

    But really..... what is happening to the light as it passes end to end, through this glass capillary tube?

    Or perhaps more precisely... through the walls of the tube?
     
    Last edited: Nov 12, 2014
  2. jcsd
  3. Nov 14, 2014 #2

    jedishrfu

    Staff: Mentor

    I think Snells law would apply in all cases with the question being what refractive and reflective indices to use to describe the media.
     
  4. Nov 14, 2014 #3
    Thanks for the input.
    I literally just this moment copied the post to General Physics, as I thought I was getting no replies.
    I'd posted here cos I thought lens makers might know.
    What to do.... I didn't know how to move the thread?

    Clearly I should have deleted this post before copying it to general.
    I've also posted the physical results (with the new post).

    I'll have a look at Snells Law. :)

    I don't know what to do about the posts, as there appears to be no delete once an answer has been posted
     
    Last edited: Nov 14, 2014
  5. Nov 14, 2014 #4

    Drakkith

    User Avatar

    Staff: Mentor

    I've moved your thread here to the classical physics subforum and deleted your other one. In the future, just contact a mentor or use the "Report" feature if you need a thread deleted or moved or if you have any questions.
     
  6. Nov 14, 2014 #5
    Results:
    It is very hard to judge.
    Most light comes out of Face B.

    When the rod surface is roughed up...... the rough area appears brighter, but perhaps this is merely how it is displaying the light that would otherwise have passed through the walls.

    For the reverse bottleneck:
    More light appears to be emitted at the bottleneck, AND subsequently from the larger diameter walls.

    Perhaps the bend in the glass walls was too much..... or perhaps any bend will allow light to escape.
     
  7. Nov 14, 2014 #6

    Drakkith

    User Avatar

    Staff: Mentor

    Well, first and foremost, a fiber optic is not like a glass capillary tube. The core of a fiber optic is solid, not hollow.

    Lots and lots of scattering of any light incident to either surface.

    Note that it will be very difficult to "bend" any significant amount of light through this tube since you are relying on normal reflection off a transparent surface and not relying on total internal reflection. Any time the light strikes the inner surface of the tube a non-trivial amount of it will pass through and be lost.
     
  8. Nov 15, 2014 #7
    "Well, first and foremost, a fiber optic is not like a glass capillary tube. The core of a fiber optic is solid, not hollow."

    Are you suggesting that the increased surface area of the tube, causes more light loss?
    I think this must be true.
    Or is there more significance to the lens having a hole in it?

    What then happens with a glass tube, that has a rough ground inner and outer surface......Lots and lots of scattering of any light incident to either surface.
    Yes, I witnessed this during physical testing.
    But do we think that more light is being lost?
    It seems logical.... instead of bouncing back, it flies out of the glass wall.
    I guess a sandblasted finish would cause the lens to glow, thereby losing a large amount of light.

    Note that it will be very difficult to "bend" any significant amount of light through this tube since you are relying on normal reflection off a transparent surface and not relying on total internal reflection.

    I'm wondering if the degree of bending can go to a certain point, before the bending is such that the light can no longer rebound..... so leaves through the bend.
    Snells Law might provide the answer to that question.

    By 'total internal reflection' I guess you mean 'silvering' or some such reflective coating.
    I've just watched a couple of youtube videos....... it doesn't seem overly difficult.
     
  9. Nov 15, 2014 #8

    Drakkith

    User Avatar

    Staff: Mentor

    By total internal reflection I don't mean silvering the surface. I mean the actual effect of "total internal reflection", which occurs when a wave tries to pass from a medium of higher refractive index to a medium of lower refractive index at an angle shallow enough. Essentially all the light is reflected from the boundary and none passes through. Look up "total internal reflection" at wikipedia or Google or something. I'd provide the link myself, but I'm on my phone right now and it's an incredibly tedious task.
     
  10. Nov 15, 2014 #9

    Drakkith

    User Avatar

    Staff: Mentor

    Okay, I'm back on my PC so I can reply properly now.

    No. Light in a fiber optic travels through a solid core made of glass. Total internal reflection allows the light to bounce off the boundary between the core and the cladding (the material that surrounds the core) without losing any of the light.

    See here: http://en.wikipedia.org/wiki/Total_internal_reflection

    Exactly.

    If we assume that the inside of the tube is highly polished, then it's just a matter of figuring out the angle at which the light hits the inner surface of the tube and using that plus the reflectivity of the glass to figure out how much light is transmitted and how much is reflected.
     
  11. Nov 15, 2014 #10
    Thanks for that.
    The Wiki author hit the ground running, with a clear explanation:

    Total internal reflection is a phenomenon that happens when a propagating wave strikes a medium boundary at an angle larger than a particular critical angle with respect to the normal to the surface. If the refractive index is lower on the other side of the boundary and the incident angle is greater than the critical angle, the wave cannot pass through and is entirely reflected. The critical angle is the angle of incidence above which the total internal reflection occurs.

    So we can calculate light loss.
    And the slighter the bend, the better.

    Well the glass is borosilicate.
    The light source is not the sun, or laser..... so it's not parallel light.

    ..... but just looking at reflection:

    So we can be shocked....... most of the light seems to leave through Face B
    Because we are not dealing with parallel light, I presume that sharply angled light passes out through the walls, and the rest bounces its' way up the glass walls.

    So:
    If a large bend is created in the glass, an additional reflective jacket must be applied to the surface...... silvering.
    But because even parallel walls will lose light....... any bend will increase that light loss.
    Calculation is possible, but would involve a full understanding of the shape.
    Without a bend, it may be possible to calculate the light loss quite easily.

    The latter sounds a good route to take.
    Understanding how much light is being lost at the outset, would be good to know.
    Perhaps I could be helped through the calculations.

    I note that the refractive index for flint glass is (n)1.62
    I see snells law...... but how do we deal with the infinite angles of non-parallel light?
     
  12. Nov 15, 2014 #11

    Drakkith

    User Avatar

    Staff: Mentor

    Not exactly. Each ray will lose at least a small portion of light, with the amount of light lost increasing as the angle between the ray and the surface increase. So you'll lose the least light at a very, very shallow glancing angle, and you'll lose the most if the light strikes at a normal angle (90 degrees to the surface). For glass the amount of light reflected when light strikes at a normal angle is around 4-5% unless anti-reflective coatings are used. The rest is transmitted through the glass. For light striking at a very shallow glancing angle, the amount of light reflected instead of transmitted approaches 100% as the angle becomes closer to 0 degrees, but never quite reaches it. (Note that in optics the angle of the incident light is usually measured with respect to the normal angle. So light striking the surface at a perpendicular angle is said to have an angle of 0 degrees, and light parallel to the surface is said to be at 90 degrees, but I haven't used this convention in my explanation above.)

    If you mirror the surface, we have a mirror instead of a lens and it behaves differently. Mirrors have virtually identical reflectance for light striking at any angle. A simple method for determining how much light you'll get at the end of the mirrored tube is to count how many times the light bounces off of the inner surface and deduct a percentage of the incident light each time it reflects. A high quality optical mirror with an aluminum coating will reflect about 90% of the incident.

    I'd simplify it into a finite number of rays and do the math from there.
     
  13. Nov 16, 2014 #12
    :)

    Questions answered, and so many more opened........

    normal angle = 90 deg (thank you for that quick lesson, cos I'd been wondering what.....). ;)
    I understood your convention,.that this is not how this angle is calculated (should be 0 deg).

    But what is a 'ray'?
    I had this image of a 'ray' not rebounding, but passing through the glass wall (or rebounding).
    Now you're telling me that only part of the 'ray' is given up.

    So is a 'ray', one photon wide....... is the ray a stream of photons, passing through the air in a wave?
    ..... and some, by chance don't reflect?

    It was easy to imagine a 'ray' at the wrong angle...... "sorry mate, but your out"!
    Now, it is more like a bouncer......... "you lot are in...... but not you; you're out"!

    It's another layer of complexity.... thanks for introducing me to that :)

    Re the silvering mirror.... I was first shocked at the 90% figure, until I realised that this must deal only with the light that would otherwise have left through the un-mirrored walls.
    That's right no?

    I note that the refractive index for flint glass is (n)1.62
    I see snells law...... but how do we deal with the infinite angles of non-parallel light?
    Okay; I think I see what you're getting at here.

    If we assume a light source, leaving the emitter, is evenly spread..... we have 180 deg of light.
    Each degree accounts for 0.555% of the total.
    Therefore we could do 90 equations @ 1 degree instance, to discover the light loss.

    Is that it?
     
  14. Nov 16, 2014 #13

    Drakkith

    User Avatar

    Staff: Mentor

    A ray is just a simplification that allows us to make the math easier. Instead of dealing with waves and interference and a lot of complex math, we can think of the light as straight little lines that will bounce around according to simple rules like Snell's law. If you divide the incoming wavefront of light into equal sections, you can assign each one its own ray and do the calculations for each one. When a ray is reflected, a portion of the light "assigned" to it is lost, being either absorbed or transmitted depending on the medium its interacting with. Think of a ray as the path that light takes, not as the light itself.

    No, I mean that only 90% of ALL the light incident on the mirror will reflect. The rest is absorbed. The next reflection will reflect 90% of the remaining light, and the third reflection will reflect 90% of this remaining light. In my telescope, light bounces off of the mirrors twice, so I lose 10% of the initial light and then 10% of the remaining light, leading to only 81% of the initial light making it to the eyepiece (with further losses in the glass of the eyepiece).

    That looks good to me, though I can't say I have any practical experience in calculating these things.
     
  15. Nov 16, 2014 #14
    Got the ray thing....... useful as a device....... and succinctly explained. :)

    The mirror revelation is then still shocking.
    But..... yours is a scope dealing with parallel light, bouncing back in the 0 deg - 20 deg region...... effectively the worst sector for losing light.

    If the light was being mirror reflected at 70 deg - 90 deg, surely almost no light would be lost..... is that right?

    Calculation?
    I'm up for it.

    D1 = outside diameter of tube
    D2 = inside diameter of tube
    L1 = length of tube
    n = 1.62 (refractive index)
    F1 = D1 sq - D2 sq
    F2 = D1 sq - D2 sq

    If non-parallel light enters through F1...... how much light is emitted at F2?

    Well.... framing the calculation is the first step :)

    I'll look at it more tomorrow.
    In the meantime, any ideas would be welcomed, on how this calculation might be made.
     
  16. Nov 16, 2014 #15

    Drakkith

    User Avatar

    Staff: Mentor

    I don't know how the angle of incidence effects the amount of light reflected from a mirror. I was under the impression that, unlike reflection from a transparent surface, the amount of light reflected from a mirror was essentially the same over any angle.
     
  17. Nov 17, 2014 #16
    Oh...
    I assumed that the silvering must absorb more light, if the light hit it at the normal angle.
    A bit like a bullet glancing off metal at an angle, or going straight through it at perpendicular.
    But it was just an assumption.

    Re the calculation.
    Having thought about the problem, it seems quite complex.
    With the light being non parallel..... 90 calculations would only cover the first bounce off the walls.

    Each degree of light would rebound differently up the walls of the tube..
    @ 1 degree there would be very many rebounds.
    @ 90 degrees there would be no rebounds.

    Perhaps it would need a computer program, to calculate light loss.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Behaviour of light thru a glass rod 'end to end'
Loading...