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Optics Tricky Question

  1. Jan 26, 2008 #1
    Hey, on my test today on the bonus question was asking : if a light ray shines at 50 degrees into a diamond then enters air at what angle? n for diamond = 2.5
    n for air = 1.00.

    I know the answer is total internal reflection cause i guessed it and got it right, but how do you prove it and how do u knw if it is total internal reflection?
  2. jcsd
  3. Jan 26, 2008 #2
    also if you use the formula ni sin theta i= nr sin theta r you end up not being able to complete the formula
  4. Jan 26, 2008 #3


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    Well, what is the criterion for Total Internal Reflection? You should know this. It occurs when a light ray is moving from a more optically dense medium (n2 below) into a less optically dense medium (smaller index of refraction n1 below), and the angle of incidence is so large, that Snell's law breaks down...it produces a sine for the refracted angle that is greater than one. In fact, the critical angle occurs when the sine is equal to one. In other words:

    [tex] n_2 \sin \theta_2 = n_1 \sin \theta_1 [/tex]

    [tex] \frac{n_2}{n_1} \sin \theta_2 = \sin \theta_1 = 1 [/tex]

    [tex] \sin \theta_2 = \frac{n_1}{n_2} [/tex]

    [tex] \theta_2 = \theta_{\textrm{critical}} = \sin^{-1}\left(\frac{n_1}{n_2}\right) [/tex]

    This is the critical angle...for angles larger than this, the angle of the refracted ray is undefined...and there is no refracted ray. Instead, total internal reflection occurs.
    Last edited: Jan 26, 2008
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