**Snell's
Law **

Consider the ray
going through this composite block.

- Angle of refraction
at boundary AB (beta) is the angle of incidence at boundary BC (corresponding
angles)
- (Same for the
'gammas' in the next medium)
- The ray is speeded
up by the same amount it is slowed down (it has to be travelling at
its original speed by the time it gets iback into the original medium!).
Therefore the 'alpha' angles are the same.

Now by Snell's Law
n = sin i / sin r, so

_{A}n_{B}
= sin a /sin b

and

_{B}n_{C}
= sin b / sin g

and

_{C}n_{A}
= sin g / sin a

So we can 'jiggle'
these equations around to get the refraction at the interior boundary
in terms of the refractive indices of the two exterior ones...

_{B}n_{C}
= sin b / sin g
= (sin a
/_{A}n_{B }) / (sin a / _{C}n_{A})
= _{C}n_{A}/_{A}n_{B}

So.....
_{B}n_{C}= _{C}n_{A}/_{A}n_{B}

This
is given on your data sheet as