Well, in other words, because you don't want to accept that space can get curved, you postulate that there are "real lens" around objects such as the Sun and the galaxies etc.
Good for you.
If you think about Eddington's - or any other - experiment, it's clear that the characteristic thickness of the "glass" - let me call it "glass" - surrounding the Sun has to be comparable to the solar radius. By the thickness, I mean the thickness of the region where the refractive index's deviation from one is at least 50% of its maximal deviation from one.
Shape of the lens around the Sun
For Eddington's solar setup, the thickness can't be much smaller because that would mean that the stars away from the Sun - visually further than one apparent radius - are not affected but they are affected. And the thickness can't be much greater because of the same reason: too distant stars from the Sun would be affected.
Fine. The bending observed by Eddington - or at least others who did the same thing more honestly - is 1.37 arc seconds which is 6.6 microradians. By Snell's law, the deviation of $n_{\rm max}$ from one has to be comparable to this angle, so you need a "glass" with something like
$$ n \approx 1+ 10^{-5} $$
The air has $1+3\times 10^{-4}$, so if you just dilute air by a factor of 30, you get exactly the "glass" that you need for your lens around the Sun.
Now, the starlight has to go through this "glass" and we may ask what is the optical thickness. Well, the solar radius is about 700,000 kilometers so the starlight would have to penetrate through 700,000 kilometers of a material whose density is similar to 1/30 of the air. That's equivalent to a 20,000 kilometers thick layer of air. I assure you that for any conceivable material, no light would survive if it had to go through a 20,000 kilometers thick air-like wall. Note that even the atmosphere which is just 10 kilometers thick (effectively) absorbs a big portion of the light.
Looking through dozens of miles of water
Alternatively, you may prefer water with $n=1.33$. If you diluted it about 30,000 times, it would get to $n=1+10^{-5}$. So 700,000 kilometers of your "glass" are equivalent to 20+ kilometers of water. If this "glass" were surrounding the Sun, the stars around the Sun would look similar like if you observe them from the Mariana Trench (times 2-5) - and I assure you that not much starlight can be seen in the Mariana Trench. ;-)
This example with the Sun was far too modest. You would have to reproduce the same results for the huge galactic gravitational lens. They bend the light by similar angles but the physical size of the "glass" would have to be many light years. The light in your model would have to penetrate through many light years thick wall of air. Nothing would be left. Moreover, if the density of your "glass" around galaxies resembled the air (over 30), the "glass" would be far more massive than the galaxy itself. ;-)
The latter point may be made spectacular if we estimate the average density of a galaxy. The Milky Way's mass is $6\times 10^{42}$ kg and its radius is $5\times 10^{20}$ meters so $M/R^3$ is $5\times 10^{-20}$ kilograms per cubic meter or so. Compare with the $0.04$ density of the "air over thirty" material. The "glass" you would need to surround the galaxy would be $10^{18}$ times denser than the actual average density of the galaxy.
Do you understand why your model is safely dead (by dozens of orders of magnitude) by now? The Universe is - and has to be - remarkably empty. This fact is also being underestimated by proponents of various spin networks and spin foam - they're incompatible with the observations for similar reasons as your "glass".
Designing a better glass
You could try to invent a "better material" for your "glass" - one that doesn't absorb much and one that has the same index of refraction for all colors - an important condition needed to avoid a rainbow-like color pattern of the stars, as explained in another answer by Ted Bunn.
If you were working hard to obey all the required conditions, you would find out that the required material looks nothing like the atomic matter containing electric charges (e.g. electrons and protons). It would look like a sea of virtual gravitons that affect all kinds of light in the same way, and only affect its direction and not its intensity.
At the very end, if you were doing it right, you would rediscover the "material" of the virtual gravitons in a quantized version of general relativity. ;-) This "material" acts on all photons in the same way (because of the equivalence principle); it maximizes the bending effect relatively to the extra mass it adds to space (which is really zero - the curved spacetime may be fully attributed to the Sun's mass at the center); and it doesn't absorb light.
