Help understanding refraction in water

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Homework Statement:
I think an old lecture note problem on refraction is wrong and I am right.
Relevant Equations:
none (I think) although the general solution would involve Snell's Law.
I recently had some trouble understanding refraction but after I finally understood where I went wrong (thanks to the people on this forum) I went back to my old lecture notes because I thought that what I recently learned didn't fit right with something that had previously been presented in my introductory to physics course. Here it is:

"You are standing waist deep in the sea on a calm day and spot something shiny that seems to be 1 metre ahead of you. Where actually is this shiny thing?"

Assumptions:
Calm water (smooth surface)
Sea floor is level
Height = 1.78m
Waist = half of height
n_air = 1
n_water = 1.33

Now my problem isn't in working out the answer because that whole process was presented in the lecture power point. My problem is that I thought that when refraction occurred in such a scenario as this, where the object appears to be should be directly above where it actually is. That is to say, there is no horizontal change in distance, only vertical due to refraction.

So, the image given was this:

lec1.png


But what I think it should have been is this:

lec2.png


So if I am right in this, then the question was wrong to begin with. The object should still be 1 meter away from you, or if it actually is closer to you, it should appear that way as well.

Am I correct in this? Thanks.
 

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  • #2
jbriggs444
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Homework Statement:: I think an old lecture note problem on refraction is wrong and I am right.
Relevant Equations:: none (I think) although the general solution would involve Snell's Law.

I recently had some trouble understanding refraction but after I finally understood where I went wrong (thanks to the people on this forum) I went back to my old lecture notes because I thought that what I recently learned didn't fit right with something that had previously been presented in my introductory to physics course. Here it is:

"You are standing waist deep in the sea on a calm day and spot something shiny that seems to be 1 metre ahead of you. Where actually is this shiny thing?"

Assumptions:
Calm water (smooth surface)
Sea floor is level
Height = 1.78m
Waist = half of height
n_air = 1
n_water = 1.33
In this case, the idea behind the question is that the target object is sitting on the flat bottom of the lake, pond, river or whatever. So the observer is not depending on binocular vision for his depth cues. He is, instead, projecting his line of sight to an ideal level bottom and estimating distance from the intersection of line of sight and sea floor.

There is a bit of an art to the process of examining a question and trying to guess at which physical principles the writer wishes to explicitly probe and which are being ignored for the moment. You will not always be able to read the writer's mind correctly.
 
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  • #3
DaveC426913
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Because the OP is looking at a more detailed answer than is required, it is fair game and relevant to point out that - due to that same refraction - the sea floor itself will not appear at the real depth - nor, in fact, will it appear level.

So the two apparently contradictory answers can actually be reconciled. The viewer will actually perceive this:
1633472078559.png
 
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  • #6
jbriggs444
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I believe the correct diagram should be like this
https://upload.wikimedia.org/wikipedia/commons/c/cc/Pencil_in_a_bowl_of_water.svg
(with the tip of the pencil equivalent to the shiny object).

I found the diagram here
https://physics.stackexchange.com/q...d-in-the-concept-of-apparent-depth-real-depth
and there is a related discussion.
That approach is using the virtual focus point (in the vertical plane) as the distance cue. That is not the only possible cue that could be used.

For instance, one can use the virtual focus point in the horizontal plane (i.e. binocular vision) as the distance cue and get a different answer.

When one learns about optics in first year physics, messy things like astigmatism are not mentioned. But they are real. You can have a optical apparatus where ray tracing in the vertical plane gives you one answer while ray tracing in the horizontal plane gives another. That is the situation here.

As I've suggested upthread, one might instead use the intersection of sea floor and projected line of sight as the distance cue.

Three possible approaches. Three possible diagrams. For me, the stipulation in the problem statement that the sea floor is level reveals the intent of the question setter.
 
  • #7
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In this case, the idea behind the question is that the target object is sitting on the flat bottom of the lake, pond, river or whatever. So the observer is not depending on binocular vision for his depth cues. He is, instead, projecting his line of sight to an ideal level bottom and estimating distance from the intersection of line of sight and sea floor.

There is a bit of an art to the process of examining a question and trying to guess at which physical principles the writer wishes to explicitly probe and which are being ignored for the moment. You will not always be able to read the writer's mind correctly.

Yeah, the following slides (that I've added at the end here) make it clear what they were asking for. I do get that. But this question was presented in the presentation just after describing refraction and just after describing Snell's Law so I'm having trouble reconciling the question itself and refraction especially after the last thread I made which also involved refraction.

So, am I correct in thinking that, yes, the apparent position of the object in this scenario SHOULD remain at the same distance away from the viewer but appear closer to the surface?

1.png
2.png
3.png
 
  • #8
haruspex
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So, am I correct in thinking that, yes, the apparent position of the object in this scenario SHOULD remain at the same distance away from the viewer but appear closer to the surface?
Assuming the viewer's head is upright and using binocular vision, yes. If it were tilted to one side, or with one eye closed, then the diagram in post #5 would be appropriate.
The diagrams in post #7 are wrong in all scenarios since they show the image as further away.

one might instead use the intersection of sea floor and projected line of sight as the distance cue.
Why would the sea floor not be subject to exactly the same change in apparent position? Or are you suggesting the viewer would subconsciously take the sea floor to be flat? That is not my experience of the phenomenon. I recall it appearing as though you are standing in a hollow, with the sea floor rising as you look further away (but asymptotically tangential to the surface, not as shown in post #5).

If looking down at an angle ##\alpha## to the horizontal and seeing sea floor at a true depth of h, the apparent depth would be ##h\frac{\sin(\alpha)}{\sqrt{n^2-\cos^2(\alpha)}}##.
 
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