Cont'd from "Left Side, Right Side, Bounded And Kinky" dated 10 Aug 2015,
For the case of μ2>μ1,
α2s<α+Δθ
and
α2p>α−Δθ
It is possible that α2s crosses below α2p when Δθ<0. α2s is not necessarily the left beam. Δθ can be adjusted to swing the beams over α.