Sin 90 = 1

Solved

Posted on 2006-04-25

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

Erm.

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

Thanks

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

Erm.

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

Thanks

5 Comments

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 ?

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