1Mu2 * 2Mu3 * 3Mu1 = 1
See the diagram enclosed.
Suppose ray of light travels from medium 1 to 2 to 3.
1Mu2 = refractive index of medium 2 wrt 1 = sin i / sin r --- eq 1
2Mu3 = refractive index ray goes from 2 to 3 = sin r2 / sin i3
Now as the interface surfaces are parallel in the medium 2, the angles r and r2 are same.
2Mu3 = sin r / sin i3 - eq 2
substitute these two values on the LHS of given question
sin i/sin r * sin r/sin i3 * 3Mu1 = sin i/sin i3 * sin i3 / sin i = 1
To know 1Mu3 is sin i3/ sini , let us imagine that there is no medium 3. Then the ray on emerging from medium2 goes out parallel to incident ray in medium1. Simple, as the reverse of the refraction at 1st interface happens at interface2. So angle of refraction would be i. So when medium 3 is present the angle of refraction is i3.
1Mu3 = sin i / sin i3
But 1Mu3 = 1 / 3M1 Hence the result.
without going into refraction diagrams, we can do this simply as
1Mu2 = Mu2 / Mu1 Mu1 = refractive index wrt vacuum
2Mu3 = Mu3 / Mu2
3Mu1 = Mu1 / Mu3
Multiply all them you get 1