tan(θ2−α2s)=μ2μ1tan(θ1−α)
when
θ1−α<0
ie incident angle α large,
tan(θ2−α2s)=−μ2μ1tan(α−θ1)
tan(θ2−α2s)=μ2μ1tan(180o−α+θ1)
When μ2<μ1,
θ2−α2s<180o−α+θ1
α2s>Δθ−180o+α
But if α2s is to make a turn of −180o,
θ2=θ1
Δθ=0
And so,
α2s>−180o+α
Also consider when μ2>μ1,
θ2−α2s>180o−α+θ1
α2s<Δθ−180o+α
But if α2s is to make a turn of −180o,
θ2=θ1
Δθ=0
And so,
α2s<−180o+α
So, when μ2=μ1
α2s=−180o+α
α2s is reflected back along α, as α2s is measured anticlockwise positive. In all cases, α2s is reflected back into in medium μ1.
tan(θ2+α2p)=μ2μ1tan(θ1+α)
for large incident angle α, such that
θ1+α>180o
This happens with EMW where θ is measured towards the positive E direction.
Let x+180o=θ1+α then
tan(θ1+α)=tan(x+180o)=tan(x)
So,
tan(θ2+α2p)=μ2μ1tan(θ1+α−180o)
tan(180o+θ2+α2p)=μ2μ1tan(θ1+α−180o)
When μ2<μ1,
180o+θ2+α2p<θ1+α−180o
α2p<−360o−Δθ+α
180o+θ2+α2p>θ1+α−180o
α2p>−360o−Δθ+α
In these cases, α2p is not in the same medium.
The two beams α2s and α2p behave differently for large incident angle α. α2s is reflected back along α, the incident ray and α2p is displaced from the extrapolated path of α by −Δθ.
Total internal reflection due to velocity changes as photons pass through the two mediums is a distinct phenomenon apart from these.