Water Refraction and the depth of the water. I'm not sure if this is the right place to ask my question! But I hope I will find some help!.
Image distortion occurs by refraction of light at the boundary surface between air and water when a camera is set in air and the object underwater.  The distortion effects on measurements and shape. 
My question: Is the effects of the Water Refraction are related to the water depth (e.g. if we capture image for an object located underwater with 1 meter depth and we capture image for the same object with depth 2 meters). Is the effect of water Refraction on first image less than the second. 
I believe the answer is yes, but I don't have scientific evidence or justifications.
Any help!
 A: You need to look up something called Snell's law and critical angle.  If you are directly over your target, there is no refraction since the angle of incidence $\theta_i$, your divergence from perpendicular to the surface of the water, and the angle or refraction $\theta_r$, the object's divergence from perpendicular to the surface of the water, are both zero. When you are approaching your target, say in a boat, the relationship between $\theta_i$ and $\theta_r$ is Snell's Law:
$$
\frac{n_1}{n_2}=\frac{\sin{\theta_i}}{\sin{\theta_r}}
$$
where $n_1$ is the index of refraction of air (approximately equal to $1$) and $n_2$ is the index of refraction of water ($1.33$). These indices are actually measures of how much the speed of light slows down in the absence of a vacuum, such that $n=\frac{c}{v}$ where $c$ is the speed of light.

The difference between the angle of incidence and the angle of refraction makes an object under the water appear to be closer to the surface than it actually is. Suppose you are a spear-fisherman, your point of view is very close to the surface of the water, and your object is to skewer a large fish a meter or more underneath the water with an accurate throw.
Starting with your spear pointing straight down at the water, you would raise it an initial angle $\theta_i$ and then lower it by an amount $\theta_i-\sin^{-1}{(1.33\sin{\theta_i})}$ to hit your target. If you are looking at the water from a height which is within an order of magnitude to the depth of your target, the math gets much more complicated. It is somewhat indicative of the power of the human brain that fishmen in primitive cultures, with no formal knowledge of the effects of light refraction, are exceedingly good at this.
When the angle of refraction becomes very large, which happens as depth increases, the angle of incidence may exceed $90\,^\circ$. This is called the critical angle and it results in a total internal reflection of light under the water. Thus, as a target gets deeper into the water, the more nearly perpendicular your point of vantage needs to be.  Suppose you have two fish, one at a depth of $1\,\rm m$ and the other directly under it at a depth of $2\,\rm m$. As you approach, you will see the shallower fish first, and even though the water may be glassy smooth and crystal clear the fish at lower depths will still be invisible to you until you get closer.
Light attenuation is the amount of light left after it has passed through a certain distance of absorbing media. For a complete description, you will want to look up the Beer-Lambert law. I'll give you the highlights. Liquid water quickly absorbs most wavelengths of electromagnetic radiation, but readily admits light in the $360-750\,\rm nm$ range (visible light). As you go deeper, the amount of available light undergoes an exponential decay. Even so, after $200\,\rm m$ there is virtually no transmission of light.
The longer wavelengths (red) not only deviate the most under diffraction but also are attenuated quickest, while the shorter (blue-violet) wavelengths diffract the least and penetrate the most (this is why the deep ocean appears blue and deep-sea creatures are so often red-colored: makes them invisible).  The more light diffracts, the greater the distance it has to travel through a medium to reach your eye. This, combined with light intensity's exponential decay with respect to distance is what makes deep objects "disappear" so quickly.
So the answer to your question is YES, there is a relationship between refraction and attenuation. Although the two effects are separate, they work together to make red-colored items more difficult to see at great underwater distances. Also, you can see the deepest when you look straight down into the water, as previously noted.
Incidentally, if you've ever been scuba-diving or snorkeling, you may have noticed that there are "shafts of light" in the water. This is a good example of what you are talking about. Non-vertical light diffracts the most and is quickly attenuated, so the greatest penetration of light occurs in vertical shafts, an effect you also notice while looking up at moisture-laden rain clouds. (For me, understanding how this effect works only adds to their beauty.)
A: Refraction occurs due to the change in speed of light when entering the medium. Water has an optical density $\approx 1.33$ with respect to air. The angle of refraction is dependent on this optical density and is independent of the depth of water.
Please see Snell's Law
The greater distortion could be easily explained by realizing the that the base of the triangle formed by the extrapolated original path of light and the refracted ray increases with increase in depth,
