While studying about path of light rays while passing through a prism, I noticed that:
Although the prism is a 3d object, only a cross section of the prism is considered enough to talk about light rays. This cross section is the area of meeting of the prism's surface and the plane defined by incident light ray and the normal to the prism at the point of incidence.
It is known that for a given case of refraction, the incident ray , normal to the surface at point of incidence and refracted ray will be coplaner.
This implies that incident ray, first refracted ray and the respective normal lie in the same plane and so do the refracted ray(now incident ray for second refraction), emergent ray(second refracted ray) and their corresponding normal.
I wanted to ask if there was a way to prove that all the light rays and both the normals lie in the same plane.
Also, is this result unique to the triangular prism or it can be said to be true for any solid ( or maybe any prism)
Thanks alot.
The doubly-refracted path of the light ray is not in general planar.
Refraction effectively rotates the incident ray in the plane defined by the ray and surface normal. For the path of the light ray to be planar, the two rotation axes at the refractive interfaces must be parallel.
For simplicity, let’s say that the first refracting surface is the $x$-$y$ plane and that the incident ray $\mathbf r = (r_x,r_y,r_z)$ strikes it at the origin. Let’s also assume that $\mathbf r$ is not parallel to the $z$-axis, since otherwise no refraction occurs at the first surface and the path is trivially planar. The ray is rotated about the axis $\mathbf k_1 = \mathbf r_1\times(0,0,1)=(r_y,-r_x,0)$ and continues on in some direction that’s a linear combination of the original direction and the surface normal, $\mathbf r' = (\lambda r_x,\lambda r_y,\lambda r_z+\mu)$.
Suppose the second refractive surface is parallel to the first. The corresponding rotation axis has direction $\mathbf r'\times(0,0,1)=(\lambda r_y,-\lambda r_x,0)$, which is obviously parallel to the first rotation axis. So, if the two refractive surfaces are parallel, the light’s path is always planar.
Now let’s look at the classic triangular prism. If we extrude an isosceles triangle in the $x$-$z$ plane parallel to the $y$-axis, the surface normals of its slanted faces are $\left(\pm\sqrt3/2,0,1/2\right)$. Taking the “back” face, the rotation axis is $$\mathbf k_2 = \mathbf r'\times\left(-\frac{\sqrt3}2,0,\frac12\right) = \left(\frac12\lambda r_y, -\frac12\left(\lambda r_x+\sqrt3(\lambda r_z+\mu)\right), \frac{\sqrt3}2\lambda r_y \right)$$ and for this to be parallel to $\mathbf k_1$, we must have $$\mathbf k_2\times\mathbf k_1 = \frac{\sqrt3}2 r_y\left(\lambda r_x, \lambda r_y, \lambda r_z+\mu \right) = \frac{\sqrt3}2 r_y \mathbf r' = 0.$$ We know that $\mathbf r'\ne0$, so this condition is only satisfied when $r_y=0$, i.e., when the incident ray is parallel to the triangular ends of the prism (alternatively, when it’s perpendicular to the prism’s axis). By symmetry, a refraction out of the “front” face yields a similar result, so if the light ray comes in “obliquely”, its path through and out of the triangular prism will not be planar.
This can be generalized to any pair of refractive surfaces by using an arbitrary surface normal $\mathbf n=(n_x,n_y,n_z)$ for the second refraction. The condition for the two rotation axes to be parallel becomes $$(r_x n_y - r_y n_x)(\lambda r_x,\lambda r_y, \lambda r_z+\mu) = (r_x n_y - r_y n_x) \mathbf r' = 0.$$ The parenthesized term is the third coordinate of $\mathbf r\times\mathbf n$, so the condition for planarity of the light’s path is that the incident ray and the second surface normal lie on a plane that’s perpendicular to the first refracting surface.
There is one other possibility to consider: if $\mathbf r'$ is parallel to $\mathbf n$, then there is no second refraction and the path is trivially planar. For this to happen, $\mathbf r$, $\mathbf n$ and $(0,0,1)$ (the first surface normal) must be coplanar, and this occurs when their triple scalar product vanishes. We must therefore have $$\det\pmatrix{r_x&r_y&r_z\\n_x&n_y&n_z\\0&0&1} = r_xn_y-r_yn_x = 0,$$ but this is exactly the more general condition derived above. From this, we get a fairly simple statement of the condition for a planar path: the incident ray and the two surface normals must be coplanar.
I found it interesting that indices of refraction didn’t really enter into the above considerations. They contribute to the coefficients $\lambda$ and $\mu$, but the exact values of those coefficients turned out to be irrelevant. However, if you try to generalize this to non-planar interfaces, then the surface normals will in general depend on the specific path taken by the light ray, which will in turn depend on the indices of refraction of the media involved.