Previous university exam problem on refraction (that I think is wrong)

[SaltMiner]

Homework Statement

You are spearfishing in waist-deep water when you spot a fish that appears to be 45° below the water and about 50cm underneath the surface. You recognize that the light coming from the fish to your eye has been refracted and you must therefore aim at some depth below where the fish appears to be. How far below where the fish appears to be should you aim?

Relevant Equations

n air = 1
n water = 1.35
angle 1 = 45 degrees
assume a flat interface

This is a question on a past exam at university. The answer was provided (for revision purposes and exam preparation) but I never understood it and it continues to frustrate me because even if I can't come up with the right answer to a problem, I'll at least 'get' (understand) the proper answer when it is given. But not this one.

There were three steps to solving this problem as presented:

1. First we find the angle of refraction using Snell's law which turns out to be 31.58 degrees. But then we take from 90 degrees this value to get 58.42 degrees.

2. We then find what the y value is by using some simple trig (tan(degrees) = O/A) and then solving for A which is our y value in the diagram. This turns out to be 50cm.

3. Lastly we use the same formula as in step 2 to find what the x value is. x = 50tan(58.42degrees).

Now I get each part of this answer, I just don't see how it is correct. This is because the y value here doesn't have to do with the actual fish's position, only its apparent position.

If we knew either how far down or away the fish actually was, then it would be easy to find the missing value because we also know the angle of refraction. But we know neither. In my university textbook, it even uses a similar diagram of a fish underwater showing that the fish is further AWAY and DOWN from its apparent position.

So even though its true that at 50cm away and at an angle of 58.42 degrees, the x value would be 81.3cm, this doesn't actually have to do with the fish's real position. What am I just not getting here? I hope this has made some sense.

 

[haruspex]

We then find what the y value is by using some simple trig

So far we have only found the angle the light ray made to the surface. To use trig to find a distance we must know a distance. What distance are they using here?

To calculate the depth, we have to consider two rays that start with a small angle between them. Then find the angle between those rays after they have emerged into the air. Tracing those lines back to where they meet under the water pinpoints the image.

 

[SaltMiner]

So far we have only found the angle the light ray made to the surface. To use trig to find a distance we must know a distance. What distance are they using here?

By using Snell's Law and knowing that nair =1 and nwater = 1.35 the angle of refraction was worked out to be 31.58 degrees.

The next step in the answer was to use this in the following way: 90 - 31.58 = 58.42 degrees. This is the angle below the surface of the water where the ray from the fish's actual position is coming from. That was understood.

The distance they then use is from the fish's apparent position which was worked out to be 50cm away from where the ray hits the water and 50cm down. But this is only for its apparent position. It doesn't give us anything about its real position which would obviously but further down and away because of refraction.

My confusion is trying to understand how you would use either of these apparent distances with the fish's real angle below the surface to find its death. Neither the fish's actual depth (which we have to work out) or how far it actually is, is given to us and I don't even know how one would ever work this out to begin with.

 

[haruspex]

By using Snell's Law and knowing that nair =1 and nwater = 1.35 the angle of refraction was worked out to be 31.58 degrees.

The next step in the answer was to use this in the following way: 90 - 31.58 = 58.42 degrees. This is the angle below the surface of the water where the ray from the fish's actual position is coming from. That was understood.

The distance they then use is from the fish's apparent position which was worked out to be 50cm away from where the ray hits the water and 50cm down. But this is only for its apparent position. It doesn't give us anything about its real position which would obviously but further down and away because of refraction.

My confusion is trying to understand how you would use either of these apparent distances with the fish's real angle below the surface to find its death. Neither the fish's actual depth (which we have to work out) or how far it actually is, is given to us and I don't even know how one would ever work this out to begin with.

As I posted, we need to consider two rays with a small angle between.
Suppose one ray from the fish has incident angle $\theta$ to the surface normal and the other $\theta+d\theta$. These come from depth y and hit the surface at horizontal displacements x and x+dx from the fish.
The rays then leave the surface at angles $\phi$, $\phi+d\phi$ to the normal, still dx apart horizontally.

You have an equation relating $n, \theta, \phi$.
Can you find equations relating:
$n, \theta, d\theta, \phi, d\phi$.
$y, dx, \theta, d\theta$
$y', dx, \phi, d\phi$
Where y' is the apparent depth?
From those, an equation relating $y, y', n, \theta, \phi$?

 

[jbriggs444]

Consider, for a moment the situation where the water has a refractive index of 1.0. No refraction. This is the imaginary situation that the viewer considers himself to be in when he judges that the fish is 50 cm below the water.

An imaginary ray from the man's right eye will strike the surface of the [non-refracting] water at a 45 degree angle and continue at a 45 degree angle, down 50 cm and forward 50 cm to the fish. The same for an imaginary ray from the man's left eye. The fish is located at the intersection of those two rays.

Now turn refraction back on. Give the water its refractive index of 1.35

Regardless of refraction, an imaginary ray from the man's right eye stays in the same vertical plane as before.
Regardless of refraction, an imaginary ray from the man's left eye stays in the same vertical plane as before.

So the actual horizontal position of the fish is still somewhere on the vertical line formed by the intersection of those two vertical planes. The original unrefracted lines of sight intersected on that vertical line. The new refracted lines of sight intersect on that vertical line. Therefore the true horizontal position of the fish is still 50 cm horizontal distance away from where the line of sight hits the water surface.

Given the horizontal position, the depth is a simple trig calculation away.

 

[Doc Al]

So the actual horizontal position of the fish is still somewhere on the vertical line formed by the intersection of those two vertical planes. The original unrefracted lines of sight intersected on that vertical line. The new refracted lines of sight intersect on that vertical line. Therefore the true horizontal position of the fish is still 50 cm away from where the line of sight hits the water surface.

I don't think that's true. The apparent position depends upon the angle of viewing. Only when viewed from (nearly) above will the real and apparent positions be in the same vertical line. (This is the case usually considered since it's much easier to figure out.)

I think the path to the correct solution is as outlined by @haruspex above. (If only I wasn't so lazy...)

 

[jbriggs444]

I don't think that's true. The apparent position depends upon the angle of viewing. Only when viewed from (nearly) above will the real and apparent positions be in the same vertical line. (This is the case usually considered since it's much easier to figure out.)

We are given a viewing angle of 45 degrees from the horizontal. We can assume that the eyes are level with each other. We can assume that depth perception is based on binocular vision cues.

I do not believe that the apparent horizontal position is a function of viewing angle. The argument I've given seems sound.

If we were laying flat on an air mattress with our eyes in a vertical line then the binocular distance cues would indeed be affected by the refraction. But that is not the situation that I imagine. (Spearfishing and laying flat on a mattress are not normally done concurrently).

 

[Doc Al]

We are given a viewing angle of 45 degrees from the horizontal. We can assume that the eyes are level with each other. We can assume that depth perception is based on binocular vision cues.

Interesting point! I'll have to ponder that one a bit. (Not sure I buy it.)

I'm imagining small cones of rays emanating in all directions from the object; in particular, one centered on that 45-degree angle of view. And the task is to figure out where that small cone appears to originate.

Offhand, I'd say that the apparent position and depth both depend on the angle of view, but I could be wrong. (Never considered that tilting your head made a difference!)

 

[jbriggs444]

I'm imagining small cones of rays emanating in all directions from the object

That is a nice picture. I am concerned that a narrow cone like that will no longer be a right circular cone once refraction has done its work. Binocular vision only gets to work on one axis across the cone.

 

[haruspex]

That is a nice picture. I am concerned that a narrow cone like that will no longer be a right circular cone once refraction has done its work. Binocular vision only gets to work on one axis across the cone.

I agree that the question setter probably intends depth based on binocular vision, with the eyes in the same horizontal plane. That indeed makes it much simpler.
But with the eyes in the same vertical plane, the answer is different.

With monocular vision, judging distance from focus, it should be rather confusing because, as you say, a small cone of rays of circular section gets turned into one of somewhat elliptical section.

 

[Doc Al]

Good stuff. Thanks to both of you, @jbriggs444 and @haruspex.

 

[jbriggs444]

With monocular vision, judging distance from focus

For what it is worth (irrelevant to the discussion in this thread), depth perception based on monocular focus is very much suppressed relative to depth perception based on binocular vision. I've run tests both in my youth (pegboard in my bedroom) and in the past few months (chain link fence at the neighborhood tennis courts) where it is possible to use the regular pattern to get ones eyes to alias two different picture elements as if they were one. This decouples focus-based depth perception from binocular-based depth perception.

Binocular wins handily.

You can also watch while your brain works the focus adjustment. It looks exactly like an autofocus camera hunting. Extremely cool the first time you see it.

By default, the autofocus algorithm tracks well with the intersection point of your binocular vision. It takes a certain amount of practice to hold binocular vision in place on an aliased pair of fuzzy chain link fence wires while autofocus gets aggressive enough to bring those wires both into sharp focus. Definitely a weird appearance once the resulting stereoscopic image has firmed up.

 

[SaltMiner]

Thanks for the replies but I'm still not understanding it. I've gone to the trouble of presenting the answer given as follows:

Here is a picture I found on the net with some added stuff on my behalf to help visualize the problem. It lists all of the current knowns.

In this first step to solving the problem we used Snell's Law as follows:

n₁ sinθ₁ = n₂ sinθ₂

sinθ_2 = (sinθ_1)/n_water

sinθ_2 = (sin(45))/1.35

θ_2 = 31.58 degrees

remaining angle = 90 - 31.58 = 58.42 degrees

In step 2 we find the y-value:

tan45 = 50cm/y

y = 50cm/(tan45)

y = 50cm

In the final step we combine what we found in steps 1 and 2 to come up with:

tan(58.42) = x/50cm

x = 50tan(58.42)

x = 81.3cm

Now my problem with this answer given is that this is true if and only if the fish was directly below its appeared image. That is to say, it actually was 50cm away. However, is this the case? Some images I've seen showing such a scenario show the fish or other object ahead (closer to us horizontally, as is the case with these drawings), directly below, or (as in the case my textbook) further away (from us horizontally). So which is it?

I DO understand that because of refraction objects appear closer to the surface than they really are but how about horizontally speaking? I've even watched contradicting videos on YouTube about this.

Sorry for my lack on not 'getting it'.

 

[haruspex]

my problem with this answer given is that this is true if and only if the fish was directly below its appeared image

Consider the paths of one ray to each eye. Viewing the paths from above, they will appear as straight lines. As each ray passes through the surface of the water it stays in the same vertical plane. So the image is indeed directly above the fish. The diagrams are misleading in that respect,

 

[SaltMiner]

Thank you very much! @haruspex and @jbriggs444. I understand it now.