Why does the region outside the circular cone appears dark and the region inside the cone appears bright to a fish? I can't understand why as Im thinking that light rays from underwater objects are still reflected by the water surface into the fish's eyes, then why is it dark outside the cone? Can anyone please answer me?
[Physics] Fish eye view problem
opticsreflectionrefractionvisible-light
Related Solutions
Dear Rootosaurus, when you're looking at an image of a chair behind you in a flat mirror, then you're observing the so-called virtual image of the chair. If the mirror's surface is located in the $x=0$ plane and the coordinate of the real chair is at $(x,y,z)$, then the virtual image of the chair is at $(-x,y,z)$.
However, the light rays coming from the real chair that are reflected by the mirror and that ultimately end up in your eyes have exactly the same directions as the light rays of a hypothetical chair that would be actually located at the point $(-x,y,z)$ behind the mirror. So geometrically, you can't really distinguish a reflection of a chair that seems to be located at $(-x,y,z)$ from a real chair that is located at $(-x,y,z)$ and that you're observing through a window without a mirror. The geometry of the light rays is identical. That's why the concept of images is so useful.
In particular, myopia means that one has some trouble to observe distant objects. Distant objects - imagine a distant point - have the property that they emit light rays that are nearly parallel. The further an object is, the more parallel its light rays look when they arrive to your eye. However, the lenses in your eyes have to convert these parallel mirrors into converging mirrors - so that all the light rays coming from the distant star end up at one point of the retina.
Myopic eyes are good in converting "steeply divergent" light rays from nearby objects to converging ones, but they're doing too much of a good thing. When you get too parallel light rays, myopic eyes make them converge too much, too early - the intersection will be inside the liquid in your eye. Hyperopia is the opposite disorder in which eyes make the light converge less than is needed. But what's relevant for your question is that the virtual image of the chair at $(-x,y,z)$ "emits" the same excessively parallel rays as a real chair at the same point, so a myopic eye will have the same trouble seeing it. After all, it shouldn't be paradoxical: the total distance that the light rays have traveled includes the distance of you from the mirror as well as the distance of the mirror from the real chair - because the colorful photons ultimately came from the chair and they were just reflected, not created, at the plane of the mirror.
Gah. Once again, I see this question crop up because these textbooks/etc draw inaccurate diagrams.
With ONE ray, your eye can never determine where the object is. Note that in your diagram, the eye can deduce the line along which the apparent image is, but to make a point, we need two lines! Who told the eye that the apparent object was directly above the real object? So there has to be a second ray of light.
The actual diagram should be something like this:
Ok, in this diagram, we have a stick(long object) instead of a fish, and the angle $\theta$ is not small enough for the apparent image to be directly above the object, but my pount here is that in taking RD/AD , we have to take two rays. Taking one ray works only while calculating it, but it is not the reality.
Second thing: To judge depth, we need two eyes. With one eye, we can only judge depth if we know the size of the object beforehand. The second eye can be thought of as picking up the 'second ray' in the above diagram. Now there will always be some angle between alteast one eye and the normal, so refraction will still happen.
To try this out, close one eye. Now hold your arms straight out, nearly stretched, but not completely stretched. Now extend your index fingers (there should be about one inch between them). Now try touching them together. Try this a few times, then repeat with both eyes open. You'll realise how necessary the second eye is for judging depth.
While considering optical lengths and RD/AD, we always consider near-normal viewing. This means that the angle $\theta$ is very small, but not exactly 0. This allows us to draw diagrams like the one in your question to find apparent depth by considering only one ray.
Another way to look at it is to limit $\theta$ to zero in your formula for apparent depth. Surprise! No change, since there is no $\theta$ term in the final formula anyways.
So, summing up:
- One ray is not enough to judge depth
- Neither is one eye
- So perfectly perspendicular viewing makes no sense for a two-eyed depth-judging being.
- So we can easily apply the formula for real depth apparent depth when the observer is on the normal.
Oh and the answer to the question is 8 m/s, as you get $v_{app,\perp}=\frac{v_{real,\perp}}{n}$ from differentiating the apparent depth formula.
Best Answer
I think you are referring to Snell's Window :
The area outside the cone is a reflection of light travelling upwards from underwater objects. If there are no objects below the surface which emit or reflect light, or if they are so far from the surface that such light is dim when it reaches the eye, the area seen outside the cone will be dark.
Light reaching the eye inside the cone comes from the sky. It is compressed from an angle of 180 degrees into an angle of 96 degrees, so it appears brighter even than it does when you are not under the water. This enhances the contrast with the dim reflections from under the water.
In addition, ripples on the surface of the water make it possible for light outside of the cone to reach the eye more directly, at points where it is refracted towards instead of away from the eye. So the area outside of the circle is brighter when the water surface is rough - as in the sea or a crowded swimming pool - and darker when it is perfectly still - as in a slow-flowing river or a pond.