To add comments or start new threads please go to the full version of: Seeing A Fish
PhysForum Science, Physics and Technology Discussion Forums > General Sci-Tech Discussions > Homework Help

longAgoStudent
6th January 2009 - 05:06 PM
Hello All,

This isn't really a homework problem, I've just been frittering about trying to re-grasp some basic physics concepts and I don't know where else to turn.

My question is:

If I know Snell's Law of Refraction and
I know the refraction index for air (1.00) and
I know the refraction index for water (1.33) and
I know that a point on a fish is 5' below the surface of still clear water and
I know that my eye is 4' vertically above that surface and
I know that my eye is 12' horizontally away from the point on the fish...

Is there a way to determine where within the critical angle I need to look (place the normal) so that I am viewing that point on the fish? (In order to keep it really simple, let's say dead i.e. motionless fish!)

This has been driving me nuts & it has been a long time since studying this stuff in school. I suspect simple (not so simple?) trig will give me the answer... Can someone please help?

Thank you in advance for your assistance in this matter. I appreciate it.

j.
boit
8th January 2009 - 04:34 PM
QUOTE (longAgoStudent+Jan 6 2009, 05:06 PM)
If I know Snell's Law of Refraction and
I know the refraction index for air (1.00) and
I know the refraction index for water (1.33) and

First assume that the fish is magically suspended in a dry pond, what will be the angle of depression of your gaze? When filled with water you will elevate your gaze a little bit.
Chalnoth
8th January 2009 - 09:02 PM
Yeah, it is pretty simple. The easiest way to do this sort of problem is to draw it out on a piece of paper, identify the unknown variables, and use Snell's law to relate them. Then just solve for the unknowns (there should only be one), and you're done.
longAgoStudent
9th January 2009 - 01:04 PM
Thank you both for responding to my question.

QUOTE (boit+Jan 8 2009, 04:34 PM)
you will elevate your gaze a little bit.


It is precisely that "little bit" that I'm trying to figure out!

QUOTE (Chalnoth+Jan 8 2009, 09:02 PM)
draw it out on a piece of paper, identify the unknown variables, and use Snell's law to relate them


After drawing it out and pretending it is a floating fish in a dry pond, I am able to figure out that the angle of depression from the viewer to the fish is 36.87°, but I was not able to progress further than that. Since 1/1.33 = .719, I tried multiplying 36.87° by .719 to get 26.51°. Taking that angle from the viewer to the boundary gave me a refracted angle of 63.49°. Using Snell's I then got an incident angle of 40.05°. This brought me close to the center of the fish, but not exactly to the point I was seeking. Is there a better way?

Thanks again, I really appreciate your assistance in this goofy exercise (it is driving me nuts, though!)

j.
boit
12th January 2009 - 01:48 PM
QUOTE (longAgoStudent+Jan 9 2009, 01:04 PM)
Thank you both for responding to my question.
It is precisely that "little bit" that I'm trying to figure out!

You can actually get better results with little mathematics. Just draw two lines from that part of the fish you want to see' one line say 30 degrees and the other 45 degrees (avoid angles more than 48 degrees). Directly up is 0 degrees, avoid that too. Now bend those line away from the normal the moment they hit the surface according to the index of refraction. Extrapolate those lines back under the surface and you will have the apparent depth of (that part of) the fish. You can then draw any line from that point to your eyes.

Tips:1) First deal with sines then consult the trigonometric tables to get the angles.
2)sine of incident ray angle/sine of refracted ray angle=Constant
longAgoStudent
12th January 2009 - 05:16 PM
Thank you boit (and/or others) for your patient help as I learn to work this through...

Following your instructions, I drew two rays from a point on a fish, one at 45° and one at 30°. The 45° incident ray then refracts at 70.13° at the water-air boundary, and the 30° incident ray then refracts at 41.68° at the water-air boundary. When I extrapolate those lines back into the water, they converge on a point 1.13' below the surface (my fish point is at 5' below the surface), and 1.88' horizontally away from the fish point (the x axis). I drew a line from the convergent point back to my vision point which crossed the water-air boundary at 3.64' horizontally away from the fish point on the x axis, and created a refracted angle of 57.33° at it's normal. By using Snell's again, I figured that the incident angle should be 39.27°. But when I carry a line on that angle from that normal back to the fish point, it doesn't converge on the initial fish point. What am I doing wrong??

[QUOTE] "and you will have the apparent depth of (that part of) the fish"

I thought that the apparent depth changed as the viewer's relationship to the water changed (i.e. the closer to the surface the viewer was, the closer the surface the fish appeared. That's what's driving me crazy, I can't figure out how to determine where a viewer (from a given position) sees a fish (from a given position) and at what apparent depth.

Wish I could furnish a sketch to help explain my problem...

Any suggestions? Thanks in advance.

j.
boit
13th January 2009 - 05:42 AM
QUOTE (longAgoStudent+Jan 12 2009, 05:16 PM)
I drew a line from the convergent point back to my vision point which crossed the water-air boundary at 3.64' horizontally away from the fish point on the x axis, and created a refracted angle of 57.33° at it's normal.  By using Snell's again, I figured that the incident angle should be 39.27°. 

I thought that the apparent depth changed as the viewer's relationship to the water changed (i.e. the closer to the surface the viewer was, the closer the surface the fish appeared.


Now I see where you are going astray. Once you get the apparent (image) part of the fish just assume again the pond is dry and just draw a line straight to any observer's eyes. Don't refract it again. If you do that it is like you have placed another fish in the pond. That is why the line can't go bark to your original fish.

The apparent depth changes not only with the viewer's proximity to the surface but also with the object's (fish). But you will notice the angle is gradually elevated naturally when the observer moves away horizontally. Same thing if the observer descends (or to a shorter viewer). So you are actually right in your second statement.

Should you get stack I'll email you a diagram. Cheers.
longAgoStudent
13th January 2009 - 12:38 PM
QUOTE (boit+Jan 13 2009, 05:42 AM)
Now I see where you are going astray. Once you get the apparent (image) part of the fish just assume again the pond is dry and just draw a line straight to any observer's eyes. Don't refract it again. If you do that it is like you have placed another fish in the pond. That is why the line can't go bark to your original fish.

Curiouser & Curiouser...

Isn't there no such thing as a light ray traveling from water to air without refraction? (Excepting a perpendicular.) The line has to go back to the original fish doesn't it?

The two angle method outlined above gives me different "apparent image points" depending on the angles selected. And drawing a straight line from those apparent image points to a viewer at a certain position doesn't seem (to me anyway) to demonstrate a refracted ray.

Maybe I am confusing the issue too much. Let's forget about viewer for a moment. Take a point on a fish, draw a ray from that point to the water-air boundary (within the critical angle), refract it according to Snell, and continue it a certain distance through the air. Now put a viewing point at the end of that ray. Presumably this shows how the ray of light travels in two straight lines: from the fish to the boundary, and from the boundary to the viewer. Now erase the ray lines (leaving the beginning fish point and the end viewer point.) Is there a way (either mathematically or geometrically) to determine those same two straight lines we just drew by using the positions of the viewer and the fish?

Thanks to any and all who can help me out with this.

j.
boit
13th January 2009 - 04:24 PM
QUOTE (longAgoStudent+Jan 13 2009, 12:38 PM)
Curiouser & Curiouser...

Isn't there no such thing as a light ray traveling from water to air without refraction? (Excepting a perpendicular.)  The line has to go back to the original fish doesn't it? 

The two angle method outlined above gives me different "apparent image points" depending on the angles selected.  And drawing a straight line from those apparent image points to a viewer at a certain position doesn't seem (to me anyway) to demonstrate a refracted ray.

Maybe I am confusing the issue too much.  Let's forget about viewer for a moment.  Take a point on a fish, draw a ray from that point to the water-air boundary (within the critical angle), refract it according to Snell, and continue it a certain distance through the air.  Now put a viewing point at the end of that ray.  Presumably this shows how the ray of light travels in two straight lines: from the fish to the boundary, and from the boundary to the viewer.

There trully has to be refraction for angles within the critical excluding perpendicular. The line that goes back to the fish has to be the refacted one.

I need to really see how two apparent image points come to be.

You have explained perfectly in the last paragraph(edited version) how to work out the problem. As for drawing straight line from those apparent image to a viewer not seem to demonstrate a refracted ray I'll direct you to my teacher Doc Al and I quote him below

"Sure. Make yourself a diagram of the fish at a distance D under the water surface. Draw a vertical line from the fish upward. Draw a ray of light from the fish (assume it's a point source) that makes some angle with the vertical (not just straight up). Show how the ray refracts as it traverses the boundary. Use Snell's law to find its new angle in the air. Then trace back the refracted ray until it hits the vertical line. Use a bit of trig to figure out at what distance below the surface it hits the line. That's the apparent depth of the fish as seen from the air. (If you did this with several rays at different angles, they would all hit that vertical line at the same point.)"

Just google Optics exam tomorrow and follow the first link. You will get the full thread. Cheers!

--------------------------------------------------------------------------------

longAgoStudent
14th January 2009 - 03:06 PM
Thank you again for seeing me through this, I really appreciate it.

QUOTE (boit+Jan 13 2009, 04:24 PM)
(If you did this with several rays at different angles, they would all hit that vertical line at the same point.)"


I'm not sure if this is true...

Look on p. 181 of this book at fig. 47:

http://books.google.com/books?id=g2JEAAAAIAAJ

Though the drawing is from 1905, these angles are very much like the ones I'm generating on my CAD program. Projecting the refracted rays back into the water, does not converge them on a single point on the vertical line. In fact they all hit the line at different places. Looking at this drawing leads me to believe that the apparent image is based on where the viewer is, and instead of always being at some point along a vertical line from an object, must follow a sort of elliptical path from slightly above the object near the vertical (when viewed from directly above), to slightly below the surface near the intersection of the critical angle (when viewed from near the surface).

This is what's giving me fits. Where is that apparent image & how to determine it?

Any optical specialists out there who can explain it to me like I'm a six-year-old??

Thanks again for your continued interest and assistance in this matter.

j.
Ron
14th January 2009 - 04:29 PM
Hi All,
Hey Long,
check out this hyperphysics link. It may help.
Peace,
Ron

http://hyperphysics.phy-astr.gsu.edu/hbase...pt/refr.html#c1
boit
14th January 2009 - 04:49 PM
QUOTE (longAgoStudent+Jan 14 2009, 03:06 PM)
Projecting the refracted rays back into the water, does not converge them on a single point on the vertical line.  In fact they all hit the line at different places.

The rays are intersecting the vertical at different points because of conversion of sines to angles, accuracy is pretty difficult. If you could draw a bigger diagram of say ratio 1:4 you will find there are less errors. Or if you were to use a block of glass with a pin glued underneath you could have a practical exercise to compare with your accuracy level (take note of its refractive index though). As I pointed earlier do not multiply refractive index by the angles but by their sines then check for the angles.
Now for some more irrelevant "information" bellow

The ray that you are actually looking for is the one that will pass through the optical center (which happens to be the geometrical centre) of the eye's lens.

One ray, and only one ray from that point of the fish will pass through that centre.

Since your eye is only 9 inches (actually less thanks to refraction) the will be no ray from that point parallel to your ray, which is coinciding with the principal axis now that you have elevated your gaze.

The lens will bulge to accommodate the image, so that it does not overshoot the retina.

For an observer far enough rays are said to be coming from infinity hence parallel to your ray and image will form on the focal point.
boit
14th January 2009 - 05:53 PM
QUOTE (boit+Jan 14 2009, 04:49 PM)
Since your eye is only 9 inches (actually less thanks to refraction) the will be no ray from that point parallel to your ray, which is coinciding with the principal axis now that you have elevated your gaze.


Sorry, I meant less than 15 inches. While am at this sine 30 degrees is 0.5 and sine 45 degrees is 0.7071
I tried drawing the refracted ray and got disappointing results too. I'll try to work on them too. I did not have a protractor though. I'll have to get one.
longAgoStudent
14th January 2009 - 10:18 PM
Ron,

Great site! However, it frightens me a bit because as I try to answer my silly problem, I drill down to calculus functions & derivatives which I cannot comprehend.

I'm beginning to think that the answer to my question may have to do with Fermat's... When I look at that drawing accompanying the equations on the Fermat's Refraction page, I can see that what I need to find are a & b when I know x & d & a+b. I can grasp that the principal has to do with the hypotenuses of the incident and refracted right triangles over the velocities of light in those respective mediums, but I can't make the leap to how to determine a & b as they relate to my above questions.

Thanks to all for any assistance you can provide as I continue my lesson...

j. (a bit grey himself!)
boit
15th January 2009 - 04:57 AM
QUOTE (longAgoStudent+Jan 14 2009, 10:18 PM)
Ron,

Great site!  However, it frightens me a bit because as I try to!)

Actual depth (4') divide by refractive index of water which is 1.33 (or 4/3) will give me an apparent depth of 3'. Trust me and assume the pond is dry again. This is the easiest way out. Actual experiment will confirm this. Naturally if you move to infinity the fish will be horizontal to your eyes (zero angle of depression) whether the pond is filled or not.
longAgoStudent
15th January 2009 - 02:45 PM
Hello All,

Thanks to all for your assistance (and to boit for his perseverance), but I think it might be time for me to give up. I've worked it over and over again and can't figure out.

QUOTE (boit+Jan 15 2009, 04:57 AM)
This is the easiest way out.


I'm not looking for an easy way out! I just want to know if it can be done (too many variables?).

This is what I want to know:

Is it possible to determine the point of refraction on a boundary by knowing?:
a) The beginning position point of the ray (from a light reflecting object i.e. fish)
b) The ending position point of the ray (to a light receiving object i.e. eye)
c) The position of the boundary plane
d) Snell's & indices of refraction

I'm able to create a big right triangle based on the start and end points, and use simple trig to determine the point on the hypotenuse where it intersects the boundary (creating complementary angles). What I can't figure out is the distance to "shift" that point along the boundary to land at the point where refraction of the ray will take place (non-complementary incident and refracted and unknown angles).

So can someone please answer: "No, dummy, you cannot determine that point without knowing at least one of the angles at that point." (my suspicion) or "Yes, dummy, it's easy to determine that point, here's how."

Sorry to be so exasperated, but I've devoted too much time to this exercise already, and despite all of the kind hints and encouragements received here, am frustrated that I still cannot grasp it.

Thank you in advance for your yes or no answer.

j.
boit
15th January 2009 - 05:03 PM
QUOTE (longAgoStudent+Jan 15 2009, 02:45 PM)
"Yes, dummy, it's easy to determine that point, here's how.

Yes You Can! But as for how I may be a fairly good student but a horrible teacher. I wonder why Doc Al hasn't joined us yet. Any way by trig. I got sine 53.3333 which translates to 32 degrees( the 12' by 9' triangle sized down to 12' by 8')

Sorry buddy. I guess I know how it feels. Am also one to get stuck on a thing till it is resolved to my satisfaction but sometimes been forced to reluctantly move on which fortunately is not often.

Good bye and good luck.
Lasand
16th January 2009 - 09:19 PM
Maybe something from this site will help:

http://www.glenbrook.k12.il.us/GBSSCI/PHYS...frn/u14l1b.html

Anyway, it's interesting that the Archer fish can do it in reverse.
longAgoStudent
17th January 2009 - 11:47 AM
QUOTE (Lasand+Jan 16 2009, 09:19 PM)
Maybe something from this site will help:

Anyway, it's interesting that the Archer fish can do it in reverse.

Maybe the Archer fish can answer, "Yes, I use calculus to do it." or "Just lucky I guess!"
Edward 3
17th January 2009 - 12:53 PM
Sounds a bit like Trout to me !!
showboat
19th January 2009 - 11:06 PM
good point.

But a spook fish [deep sea. has a divided eye that uses mirrors.

http://www.telegraph.co.uk/earth/wildlife/...-sea-gloom.html

[The brownsnout spookfish, which lives at a depth of more than 3,000ft]

And the ever favorite,mudskipper.
Which does something similiar, but better.

There was a PBS TV show, it was pretty amazing.

Looked like a fish/salamander fighting and biting territory intruders on its tiny patch of mudflat at low tide, no water except stored in the mouth to breath and giving signals by raising the coloured back fin..

.

http://www.britannica.com/EBchecked/topic/.../four-eyed-fish

http://www.answersingenesis.org/creation/v18/i1/fish.asp

longAgoStudent
21st January 2009 - 10:31 PM
The instinct and biology of sea creatures is most interesting, I'm sure, but I came here because I'm interested in getting some assistance with finding an answer to my question...

Check out this approach. Image placement seems to conflict with both the 1905 text drawing I quoted earlier, and boit's apparent depth method.
Apparent Thickness

I would be most grateful and appreciative to anyone who can instruct me as to whether or not I can determine that refraction point only knowing coordinates as previously described, and additionally how to determine where on the sight-line the image falls.

In an attempt to discourage "clever fish discourse," today let's refer to the object as an underwater coin (numismatists need not reply).

Thanks in advance to the educators.

j.
PhysOrg scientific forums are totally dedicated to science, physics, and technology. Besides topical forums such as nanotechnology, quantum physics, silicon and III-V technology, applied physics, materials, space and others, you can also join our news and publications discussions. We also provide an off-topic forum category. If you need specific help on a scientific problem or have a question related to physics or technology, visit the PhysOrg Forums. Here you’ll find experts from various fields online every day.
To quit out of "lo-fi" mode and return to the regular forums, please click here.