Apparent depth conceptual question

[Clack_Attack]

You know the proofs for finding the apparent depth of a swimming pool, or object submerged in one?

Well, it always assumes the object is directly above the original object. Does anyone know where the assumption comes from?
(see below)

 

[BvU]

Have you ever seen something under water touch the side of the pool when looking at an angle and actually be away from it when seen from above ?

 

[Clack_Attack]

Have you ever seen something under water touch the side of the pool when looking at an angle and actually be away from it when seen from above ?

I kind of see what you're saying. But that's for small incident angles, i.e. when looking straight down. I've seen the actual derivation for the apparent depth of a pool, You get an equation like:

(with d being the apparent depth, D being actual, n1 and n2 are the indeces of refraction and theta 1 being incident angle)

Anyway, then it goes on to say assuming we're looking at it from above, theta is so small and this reduces to:

But in the diagram I have, the fish isn't been seen from above, it's a much wider angle.

Is it like a fact of nature that it will appear at the same horizontal distance (from the observer's eyes) but just appear at a higher vertical distance? (assuming we're looking into a denser medium and all that)

 

[pixel]

To locate an image we need two rays. In your first diagram, let's also use the ray that starts at the same place on the fish but goes straight up, perpendicular to the surface. That ray doesn't get deflected (incident angle = refracted angle = 0). So we have two rays coming out of the water, the one you show and the second one that I describe. The image location is found by continuing those rays back to the point they appear to come from and that is directly above the point where the two rays started.

 

[Clack_Attack]

Ah, so if I were to surround the outside with a million eyeballs and a million rays emanating in all directions from one chosen point on the fish, the apparent images would all trace back to the same point above the fish?

 

[BvU]

I think they would all have different $\theta$, wouldn't they ?

 

[sophiecentaur]

You know the proofs for finding the apparent depth of a swimming pool, or object submerged in one?

Well, it always assumes the object is directly above the original object. Does anyone know where the assumption comes from?
(see below)

View attachment 98915

It would be very easy for you to draw your own ray diagram and establish, for yourself, what happens and where the image should appear. Three rays from the object (all you should need) will - or will not - result in three rays coming from the surface that intersect (produced backwards) in a point. If not, you can reckon that the image position will not be the same.
The change in direction at the surface will follow Snell's Law which we all know and love. You will need to dig out your old school protractor or print one off.
Have a go and don't rely on glubby pictures from Google.

 

[pixel]

Ah, so if I were to surround the outside with a million eyeballs and a million rays emanating in all directions from one chosen point on the fish, the apparent images would all trace back to the same point above the fish?

Thinking about this some more - what I wrote was true if you just used the vertical ray and a non-vertical (oblique) ray from the point on the fish, using a giant eye. If we just consider the oblique ray, as you increase the angle from the vertical the refracted ray, when followed back into the pool, will be higher and higher above the fish. So the answer to your question is no - they rays would not trace back to a single point above the fish when a large spread of rays from the fish are considered, which is what BvU may have been alluding to.

Realistically, as you look into the pool, rays with a single oblique angle and a small spread about that angle are entering your eye. You follow them back into the pool to locate the image of the point on the fish. As you move further away i.e larger ray angles to the vertical, the image moves up toward the surface of the pool. I've been playing with a spreadsheet that I created to calculate the location of the image and it does appear that it shifts to the right , as in your third diagram, as you move more to the right. Maybe someone else can confirm this.