Paralel light conditions after passing from a sphere

Question

Is it possible to get such result that light will be parallel after passing from the sphere?

what is the total condition for such result if possible?

Thanks for answers

Answer 1

Absolutely. The angle under which the beam comes out if there is a single reflection inside the sphere can be worked out to be $\pi + 2\alpha - 4\beta$ where $\alpha$ is the angle of incidence and $\beta$ is the angle of refraction. $\alpha$ and $\beta$ are related by Snell's law: $\sin\alpha / \sin\beta = c$ is constant (the constant depends on the material and the wavelength of the light). In this case we must have $\alpha = 2\beta$, and we get that $2\cos\beta = c$.

This happens in the rainbow, only you cannot see it: 1) light going to the centre of the rainbow is blocked by the shadow of your head 2) all visible light reflects in all directions sufficiently close to parallel, so it will be a uniform white and not stand out much like the boundary, which is the actual rainbow.

Answer 2

Well the diagram you have drawn is physically incorrect. The reflected ray you have drawn, is NOT reflected by TIR; total internal reflection, as the ray could not have entered the sphere and the angle of reflection is always the negative of the angle of incidence (effective refractive index is -1).

So an optical path as you have shown it is not possible; the reflected ray has to be the same path length as the incident ray inside the presumed perfect sphere.

Even so, there might be one unique angle of incidence, on the sphere, where the emerging ray is parallel.

Note that the angle of incidence for the once reflected ray must be identical to that at the point of reflection. From that we can deduce that there can be no emerging ray if there actually is TIR at the foldback point.

Ergo the angle of incidence has to be smaller than the 90 degree incidence that gives TIR, so there must also be a Fresnel partially transmitted ray, and a partially reflected ray at both points A and C.