Derivation of: n = 1 / Sin C

Posted on 2006-04-25
Last Modified: 2012-08-14
The refractive index of a material can be found using:

   n = Sin i / Sin r

Which is also equal to:

   n = 1 / Sin C

Where C is the critical angle. But I don't get why.

Because, when i=C, r=90, therefore:

   n = Sin C / Sin 90 = Sin C / 0


So, where'd the 1/Sin C come from?

Question by:InteractiveMind
    LVL 37

    Expert Comment

    by:Harisha M G
    Sin 90 = 1
    LVL 25

    Author Comment

    Oops, yer.

    In that case:

       n = Sin C / Sin 90 = Sin C

    Where does the 1 over come from?
    LVL 37

    Accepted Solution

    Critical angle is the incident angle for which the refracted ray will move along the surface, the refracted angle measured from the normal.

    From Snell's law,

    n1 sin(t1) = n2 sin(t2)

    So, considering that the relative refractive index n = n1/n2 where n1 > n2, then t1 will be the incident angle, t2 will be the refracted angle (Note that TIR occurs only when the light travels from a denser medium to a rarer medium). So,

    n = sin t1/sin t2 = sin(90)/sin C = 1/sin C

    Isn't it ?
    LVL 25

    Author Comment

    I still don't get why the incident ray (in the case of TIR, the critical angle C) is on the bottom of the fraction ...
    LVL 37

    Expert Comment

    by:Harisha M G
    You are using the formula, wherein, "n" represents the refractive index of the material compared to air (or vacuum). Then, the denser medium will be the medium having "n" and rarer medium will be air. You can't achieve TIR in that case. So, it should be dervied using Snell's law only

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