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Thursday, August 13, 2015

More Bending Of Light

From the post "Split Cannot Mend" dated 10 Aug 2015,

tan(θ2α2s)=μ2μ1tan(θ1α)

when

θ1α<0

ie incident angle α large,

tan(θ2α2s)=μ2μ1tan(αθ1)

tan(θ2α2s)=μ2μ1tan(180oα+θ1)

When μ2<μ1,

θ2α2s<180oα+θ1

α2s>Δθ180o+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,

θ2α2s>180oα+θ1

α2s<Δθ180o+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.


When we consider,

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+α180o)

tan(180o+θ2+α2p)=μ2μ1tan(θ1+α180o)

When μ2<μ1,

180o+θ2+α2p<θ1+α180o

α2p<360oΔθ+α

and when μ2>μ1,

180o+θ2+α2p>θ1+α180o

α2p>360oΔθ+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.