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!

Homework Help: Max. n for TIR - air-medium interface

  1. May 24, 2012 #1
    1. The problem statement, all variables and given/known data

    A light ray in air is obliquely incident on a parallel-sided slab of refractive index 'n'.
    The light ray refracts and hits the left side of the slab and undergoes total internal refraction.
    What is the maximum possible 'n' value?

    NOTE - This is a theoretical problem that I created upon observation of a numerical problem that asks for range of incident angles. I do not know if maximum 'n' value can exist for any given circumstances. Also, I have neglected the fact that at critical angle incidence, T.I.R. doesn't occur. However, a small dθ after c should cause T.I.R, so I have left out that negligible amount. Please point out any clarification and/or errors in my assumptions.

    2. Relevant equations

    n * sin θ = constant
    sin c = 1 / n

    3. The attempt at a solution

    Now, assume the ray is incident at angle i on the top side.
    Upon refraction, we can say → sin i = n * sin r.
    Also, since maximum value of sin() is 1; sin i < 1 → n * sin r < 1
    ( we disregard i = 90°, as the ray is obliquely incident on, and not grazing, the slab)

    Now, for air-slab interface, of critical angle c, we know that sin c = 1 / n.
    Also, when we draw a diagram, the normals to the incident ray and refracted, T.I.R-ing ray form a right triangle, with c and r as the other two angles → c + r = 90° → c = 90° - r
    Taking sin() on both sides, we get sin c = sin (90° - r) = cos r
    OR, cos r = 1 / n → sin r = √(1 - ( 1 / n^2 ) ) → n * sin r = √( n^2 - 1)
    OR, √( n^2 - 1) < 1 → n^2 < 2 → n < √2

    Therefore, the maximum value for 'n', for ray at air-slab interface to undergo T.I.R, is √2.

    Attached Files:

  2. jcsd
  3. May 25, 2012 #2
  4. May 25, 2012 #3
    Your diagram is wrong angle r should be less than angle i
  5. May 26, 2012 #4
    @truesearch -

    My apologies! I rushed through while making the diagram on Paint.NET.

    Regardless, what do you think of the proof?
  6. May 26, 2012 #5


    User Avatar
    Homework Helper

    The result is correct, but the proof is not mathematically rigorous.
    So sinr≤1/n, cos(r)≤1/n...

  7. May 29, 2012 #6
    What exactly does that mean?

    What are the implications of this proof and your counter-arguments?

    I'm a bit confused...
  8. May 29, 2012 #7


    User Avatar
    Homework Helper

    For total reflection, 90-r, the angle of incidence at the side of the slab must be greater than or equal to the critical angle θc even with the greatest r, which is equal to θc

    90°-r≥θc,→ r≤90°-θc even for the greatest r:

    r ≤θc≤90°-θc→ θc≤45°

    The sine function is monotonously increasing between 0°and 90° so

    sinθc=1/n≤1/√2 →n≥√2:

    the refractive index can not be smaller than √2. √2 is the smallest, the minimum refractive index that ensures total reflection. There is no maximum refractive index. It can be arbitrary greater than √2.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook