B Why do reflections in water sometimes appear different from what we see?

[kuruman]

For quite some time I was under the mistaken impression that the specular reflection of objects from the calm surface of still water forms an image that is the same as the image I see except it's upside down. That is not the case. In the picture below^* I have circled in white a couple of features that are just missing from the reflection. There are other discrepancies which the interested reader may wish to search for.

What's going on? Two ideas need to come together. First, objects in the back are partially hidden by objects in the front; how much is hidden depends on one's point of view. Second, what one sees (or photographs) in images like this is a superposition of two points of view: (a) the "erect" POV formed by rays reaching the eye directly and (b) the "inverted" POV formed by rays that are reflected off the surface before continuing on to the eye. The diagram below shows schematically how an object can be missing from a reflection. What one sees in the reflection is not the upside-down image of what the eye sees but the upside-down image of what the "reflected" eye sees.

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^* https://www.befunky.com/learn/mirror-photography/

 

[tech99]

This effect is used when planning microwave communications over water. The water reflection is a cause of problems because it varies in phase as the tide comes up and down, causing fading. So it is usual to try to hide the transmiter behind some object to block the water-reflected path.

 

[kuruman]

This effect is used when planning microwave communications over water. The water reflection is a cause of problems because it varies in phase as the tide comes up and down, causing fading. So it is usual to try to hide the transmiter behind some object to block the water-reflected path.

I would assume that the phase-dependent fading you describe is due to interference between the direct and reflected signals, correct?

 

[A.T.]

For quite some time I was under the mistaken impression that the specular reflection of objects from the calm surface of still water forms an image that is the same as the image I see except it's upside down.

That's how fake reflections made in Photoshop look like.Interesting related question that came up in a previous thread:

If you have a shiny sphere with a highlight above the water, where will the highlight be on the reflected image of the sphere?

https://www.physicsforums.com/threads/optical-versus-digital-reflections.1002553/

 

[tech99]

The fading caused to microwave paths over water is caused by several phenomena. As you suggest, the rise and fall of tide will alter the phase of the reflected wave and produce very large changes in received power. Water also has a tendency to have stratified atmosphere above it which causes problems. For instance, we often see an evaporation duct over the water which can trap or guide waves and either enhance or diminish the received signal. For these reasons, it is desirable to avoid water paths or at least to try to avoid a water reflection.

 

[kuruman]

If you have a shiny sphere with a highlight above the water, where will the highlight be on the reflected image of the sphere?

I did some scribbling and this is what I came up with. I assumed a Cartesian coordinate system with its origin at the center of the shiny sphere of radius $R$. Angle $\alpha$ is measured from the 12 o'clock position at the pole. I assumed that parallel rays from a distant source are incident on the sphere at $\alpha =0$ (parallel to the y-axis). Thus, the scattering angle is $2\alpha.$

An observer's eye is now placed at coordinates $\{x_0,~y_0\}$. My scribblings show that the ray from the center of the highlight that reaches this eye is incident on the sphere at angle $\alpha$ obtained by solving $$\tan(2\alpha)=\frac {x_0/R-\cos\alpha}{y_0/R-\sin\alpha}.$$I suspect that this equation can only be solved numerically or graphically. The figure below shows the geometry.

The "reflected" eye is below the surface, at coordinates $\{x_0,~-(h+y_0)\}$ which would be the new input in the equation. Clearly, the ray that reaches the "reflected" eye is incident at a larger angle $\alpha$ than the direct ray. This then would be the picture to invert in order to simulate the reflection of the shiny sphere and its highlight. Finally, I note that there should always be a solution as long as the reflected eye is not in the shadow of the shiny sphere, i.e. as long as $x_0/R>1$.

 

[DaveC426913]

Interesting related question that came up in a previous thread:

If you have a shiny sphere with a highlight above the water, where will the highlight be on the reflected image of the sphere?

https://www.physicsforums.com/threads/optical-versus-digital-reflections.1002553/

Indeed - not merely related - that thread contains an analysis of the very phenomenon that the OP refers to in demonstrating that the original photo was real. (Skip down to post #9.)

You can see a slight rotation caused by flipping the real image and its reflection. (The axis of rotation parallel to the plane of the pic, from side to side.)

 

[kuruman]

Nice work with the .gif.

I agree, the squirrel picture passes this test. The picture shows a section of the squirrel's rump. Look at the the leaf seen nearly edge-on near the center of the top (direct) picture. It is entirely surrounded by fur. In the bottom picture the leaf appears near the edge of the fur. This is consistent with the lowered view of the "reflected" eye below the surface of the water.

 

[A.T.]

View attachment 294172
The "reflected" eye is below the surface, at coordinates $\{x_0,~-(h+y_0)\}$ which would be the new input in the equation. Clearly, the ray that reaches the "reflected" eye is incident at a larger angle $\alpha$ than the direct ray.

Yes, larger angle α but for large x₀ and α ≈ 45° (distant horizontal line of sight) not much larger. So the highlight in the reflection would still be on the side of the sphere further away from the surface, like is the case for the eye of the squirrel. This is was what raised suspicion in the other thread, but is correct.

 

[Charles Link]

I have a scene of the Chicago skyline on my Facebook cover page, and it appears to be taken during a sunrise, where the sun rises in the east over Lake Michigan. In a number of places, the reflection of the sun off the buildings is much brighter in the reflected image in the water than it is when viewing it directly. @kuruman 's post offers the explanation to this. Let me see if I can upload the photo. Yes, click on it, and you get a much better view of it.

 

[Charles Link]

and a follow-on to the above: When I first spotted these anomalies in the photo a year or two ago, I had to question whether the photo was for real. @kuruman 's diagram in the OP post above would explain it, where the specular reflection from the sun is such that it reaches the viewer from a couple of the building faces only off the reflection from the water.

 

[kuruman]

and a follow-on to the above: When I first spotted these anomalies in the photo a year or two ago, I had to question whether the photo was for real. @kuruman 's diagram in the OP post above would explain it, where the specular reflection from the sun is such that it reaches the viewer from a couple of the building faces only off the reflection from the water.

I am not so sure about this. The inverted reflected light view is the same as the direct view except it is seen as coming from a different angle. Look at the bright red reflection near the middle of your picture. There is nothing like it in the direct view. I suspect that the bright lines are reflections of a few street lights that have not been turned off yet. Something like the photo below.

 

[A.T.]

I am not so sure about this. The inverted reflected light view is the same as the direct view except it is seen as coming from a different angle.

That different angle could be the one that sees the reflection of the rising sun, while the direct line of sight doesn't

Look at the bright red reflection near the middle of your picture. There is nothing like it in the direct view. I suspect that the bright lines are reflections of a few street lights that have not been turned off yet.

Some of it maybe. But some of it matches the buildings too well.

 

[DaveC426913]

I had to question whether the photo was for real.

My opinion is that - whether or not the pic has been enhanced - the reflection is certainly real.

You can see a line midway to the shore that divides disturbed water from glassy water. This is quite common near shores adjacent to skylines populated by tall buildings (I live in Toronto) - the buildings provide a wind shadow that keeps the water glassy smooth out to a certain distance from shore.

It defies belief that someone would go to the effort to fake that in.

As for the sun's reflection, some of the buildings might be reflecting the morning sun, but many are surely just reflected streetlights:

This one is unusual though, and difficulter to 'splain away:

 

[Charles Link]

Very nice analysis @DaveC426913 . :) In any case, regardless of the exact nature of a number of the objects, the reflected image is considerably different than the straight-on view of the objects, and this goes along with the ideas of @kuruman in the OP of this thread.

 

[PeterO]

My opinion is that - whether or not the pic has been enhanced - the reflection is certainly real.

You can see a line midway to the shore that divides disturbed water from glassy water. This is quite common near shores adjacent to skylines populated by tall buildings (I live in Toronto) - the buildings provide a wind shadow that keeps the water glassy smooth out to a certain distance from shore.

It defies belief that someone would go to the effort to fake that in.

View attachment 294216As for the sun's reflection, some of the buildings might be reflecting the morning sun, but many are surely just reflected streetlights:
View attachment 294217

This one is unusual though, and difficulter to 'splain away:
View attachment 294218

The rough water provides a series of crests which will include a section that reflects the light in the appropriate direction that is why the reflected light is so spread out. That bright area you have the red arrow pointing to is just the spread reflection of the bright section on the roof of the building. Notice how it does not go "all the way to the shore" where there is calm water. The reflections of the lower light sources do, however, appear to go all the way to the shore.