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Total internal reflection problem.

  1. Dec 30, 2003 #1
    If anyone could provide some insight into how I might go about solving this problem it would be greatly appreciated.

    Q: A ray enters the flat end of a long rectanglular block of glass that has refractive index n_2 as shown in the following figure. Show that all the entering rays can be totally internally reflected only if n_2 > 1.414

    Last edited: Dec 30, 2003
  2. jcsd
  3. Dec 31, 2003 #2
    Wow... I think that's going to be hard without any angles given...
  4. Dec 31, 2003 #3

    Well I was fooling around with the 1.414 number I realized that it was the sqrt of 2. I figured this must mean something so I did a search on the weeb and found some sites that discuss the rare 1.414 index.

    "for total internal reflection at 45 degrees, a dense to rare refractive index ratio of 1.414, the square root of two, is required."

    So I guess somehow I have to prove that the angle of light entering the glass will refract in such a way that its incident angle against the interior of the glass is at an angle of 45 degrees....
  5. Dec 31, 2003 #4

    Doc Al

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    Here's a hint. Figure out (in terms of n) the maximum angle Θ2. Then realize that the biggest Θ2 gives the smallest Θ3. Make that smallest Θ3 the critical angle and solve for n.
  6. Dec 31, 2003 #5
    Great man! Thanks for the tip! I'll give er a go and see what turns out.
  7. Feb 1, 2005 #6

    Hey all, totally bringing this up from the thread graveyard, but I too am having the same problems with this question. I understand where Doc Al is comming from but I still dont have a Θ1 to go from. Any help with this would be more then amazing.
  8. Feb 1, 2005 #7


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    Maybe you're supposed to get it from the picture?
    If the picture is given with the incoming light ray, just get your geotriangle (or whatever you call it) to measure the angle of incidence.
  9. Feb 2, 2005 #8

    Doc Al

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    Don't think in terms of a specific Θ1; instead, you must consider the light entering the block at any angle.

    This might help (since the diagram is no longer viewable): Light is entering one end of the block (at all angles of incidence) and must internally reflect from the sides of the block.
  10. May 24, 2005 #9

    hmmmm...I have the same problem and I'm still having trouble with it.

    the incident angle (Θ1) can range anywhere from 1 degree to 89 degrees as it hits the glass,

    which would make the max Θ2 = 89 degrees

    which would make the smallest Θ3 = 1 degree

    but a critical angle of 1 degree doesnt sound right.
    Any idea where I'm going wrong here?
  11. May 25, 2005 #10


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    I don't have the diagram, but I am guessing Θ2 is the angle between the ray and the normal inside the first surface of the glass, and that Θ3 is the angle between the ray in the glass and the side of the glass perpendicular to the entry surface. If that is correct, for total internal reflection Θ3 must be greater than the critical angle, when all the light hitting the second surface is reflected back into the glass.

    The geometry is really pretty simple. What has to be true of the sum Θ2 + Θ3 ? Light incident at 89.999 degrees will enter the glass at Θ2, and for any kind of glass that angle will be a lot less than 89 degrees. That light must be incident on the second surface at that same angle, or greater, to be totally reflected. Since the surfaces are perpendicular, the angle of refraction at the first surface has to be no more than 45 degrees, and the angle of incidence at the second surface has to be at least 45 degrees. So what must be the index of refraction?
  12. May 25, 2005 #11

    Doc Al

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    [itex]\theta_1[/itex], the angle of incidence at the end surface, ranges from 0 to 90 degress.

    [itex]\theta_2[/itex], the angle of refraction at the end surface, is constrained by Snell's law to be: [itex]n_1 \sin \theta_1 = n_2 \sin \theta_2[/itex]. Thus the maximum [itex]\theta_2[/itex] is given by [itex]\sin \theta_2 = (n_1/n_2)[/itex].

    As OlderDan points out, the relationship between [itex]\theta_2[/itex] and [itex]\theta_3[/itex] (the angle of incidence at the side surface) is a simple one. To ensure total internal reflection, that minimum [itex]\theta_3[/itex] must be at the critical angle, thus [itex]\sin \theta_3 = (n_1/n_2)[/itex]. A little trig (or geometry) will allow you to solve for [itex]n_2[/itex] assuming [itex]n_1 = 1[/itex].

    Here's another discussion of essentially the same problem, which you may find helpful: https://www.physicsforums.com/showthread.php?t=71377
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