# How would we see the earth if refraction of light was significantly stronger?

I'm assuming the easiest way to minimally change the laws of physics is to make the effective speed of light more dependent of the medium it travels through. Unless this is incredibly far from the truth, I don't wish you to discuss that in your answer in much detail.

Instead I want to focus on the following: what apparent shape would the earth and (large) objects on it have as observed from various heights relative to the surface (I'm assuming things get interesting when you get close to the surface) and what would different wavelengths of light breaking at different rates entail?

I'm inclined to believe that the world would look like one part of a two sheeted hyperboloid.

• From how far away? The Earth's atmosphere is like a thin skin over a large sphere, so from any appreciable distance the index of refraction of the atmosphere is irrelevant. I don't think the comments on your linked reference show a lot of understanding of optics. It'd take not only a high index but some serious layering of different indices to have the effect they claim. – Carl Witthoft Oct 7 '14 at 12:36
• @CarlWitthoft I noticed the comments were making wild claims and my (limited) knowledge of optics didn't entirely agree with them. I'm not entirely convinced that a photon will ever curve (not reflect) back down in our atmospehere regardless of the refraction rate, since the refraction will be continuous, not discrete. I'm not sure though, which is why the question is: what would the earth look like from on the surface if the refraction index was higher? – overactor Oct 7 '14 at 12:53
• @CarlWitthoft don't mirages require abrupt rather than gradual changes in density? – overactor Oct 7 '14 at 14:08
• Not necessarily. See en.wikipedia.org/wiki/Gradient-index_optics , where it's explicitly stated that some mirages occur due to GRIN-lensing. – Carl Witthoft Oct 7 '14 at 15:59
• You could simulate that easily with a raytracing software like Povray. Simply define materials with very high refractive index and see what happens. I don't think it will look very different from a pond, except that the surface will be mostly reflective and there will only be very small angles under which one can look "inside" the atmosphere. – CuriousOne Oct 7 '14 at 16:46

Let's pick apart the problem from various vantage points to see if an answer becomes interesting:

1. Standing on the ground Looking around, the world will not look much different in a high refractive index material. Think about swimming in a pool. A high index makes objects appear closer than a low index (the optical path is shorter). Therefore, objects would appear closer, but only in comparison to a world of low refractive index.
2. A satellite observing the ground Here things start to get interesting. Like most physics, the real fun happens at the interface between two different media. In a geo-stationary orbit, looking down on the planet's surface, all the features would appear to be at the surface in almost a flat manner. The high index material has no optical path compared to the physical distance. Therefore, everything appears painted on the surface of the interface between the atmosphere and vacuum. Even depth perception becomes difficult. Very high features appear flat.
3. Approaching the planet from a significant distance When you approach the planet from afar and you are close enough to perceive it as a disc in space, there are two dominant effects that come to mind: refraction and Fresnel reflection.

A. Refraction The refraction is extremely strong. Therefore, looking at the atmospheric limb, it will appear as a very strong lens, bending light straight down towards the surface. Again, the features on the surface will appear printed at the interface between the atmosphere and vacuum. Even at grazing incidence, at the limb, light rays will refract straight down to the planet's surface.

B. Fresnel reflectance This is the big bummer. The larger the index of refraction difference between two materials, the larger the impedance mismatch. Therefore, energy (light) does not couple through the interface very well. The surface becomes highly reflective.

$$R={{(n_1 - n_2)^2} \over {(n_1 + n_2)^2}}$$

So, the surface will appear very reflective. It the atmosphere is very dense, then the interface between vacuum and atmosphere will transition over a short distance, enhancing this reflectivity. If on the other hand, the transition is occurs over a large range, then the reflectivity will be much less and the surface will appear as I've described, "painted" on the shell of the atmosphere.

If you want to see this in real life, take a look at Calcite.