Struggling to understand the concept of this question (ice cube in water optics question)

[murshiddreamengineer]

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook.

Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water.

I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an idea, or do we need more information to approach this question?

 

[.Scott]

I believe the author intends for the observer to be looking down into the water from above the water.

There is enough information provided in the problem statement.
You should have been given an equation for computing the actual depth given the apparent depth when viewing down through a transparent material.

Here is a link: Apparent Depth

 

[murshiddreamengineer]

I believe the author intends for the observer to be looking down into the water from above the water.

There is enough information provided in the problem statement.
You should have been given an equation for computing the actual depth given the apparent depth when viewing down through a transparent material.

Here is a link: Apparent Depth

I'm sorry, the question says that the ice cube is partially immersed (floating). I mistakenly say immersed

 

[.Scott]

Are you quoting the problem statement precisely?

If so, we are supposed to presume that this ice cube is a real cube - and therefore, 10 cm high.
Once again, the phrase "from the water" is very ambiguous - it could mean from the surface of the water or from beneath the water surface or from over the water.

I take it to mean from "over the water", because anything else would require more explanation.

You will still need that apparent depth formula.
You might want to start by calculating what that 10 cm height would look like if the entire cube was submerged.

Then, it will just be a matter of determining what portion is actually submerged.

 

[kuruman]

I believe the author intends for the observer to be looking down into the water from above the water.

In that case how does one interpret "from the water" as in "An observer observes the ice cube from the water"? Is the observer immersed in the water or is something lost in translation?

To @murshiddreamengineer: Is there a drawing or diagram that accompanies this problem?

On edit: I see that @.Scott has the same question.

 

[haruspex]

Is the observer immersed in the water

Wouldn’t that make the height seem greater?

 

[murshiddreamengineer]

In that case how does one interpret "from the water" as in "An observer observes the ice cube from the water"? Is the observer immersed in the water or is something lost in translation?

To @murshiddreamengineer: Is there a drawing or diagram that accompanies this problem?

On edit: I see that @.Scott has the same question.

yes, he is in the water and look the ice cube

 

[kuruman]

Wouldn’t that make the height seem greater?

Yes it would make the height seem greater. I assume that "height" means the dimension perpendicular to the surface of the water. Here, we are told that the "length" of 10 cm appears to the observer as 7.75 cm. Presumably "length" is perpendicular to "height" hence parallel to the surface. I am trying to visualize the line of sight of the observer.

I cannot resist adding a thought I just had. This could be trick question the answer to which is, "An ice cube with a length of 10 cm has a height also of 10 cm because it's a cube, no?"

On Edit: OK the observer is in the water.

 

[murshiddreamengineer]

yes, he is in the water, and look at the ice cube

Actually, it is a cuboid.

 

[kuruman]

View attachment 366074Actually, it is a cuboid.

This clears it up. Thank you. Look up "Apparent depth at near normal incidence" and use the same reasoning but from below the surface.

 

[haruspex]

This clears it up. Thank you. Look up "Apparent depth at near normal incidence" and use the same reasoning but from below the surface.

I still fail to see how it could appear to be less than the true height when viewed from the denser medium.

 

[.Scott]

View attachment 366074Actually, it is a cuboid.

The reason I didn't think the problem would be set up this way is that now you are using a lens system intended for air and putting it in direct contact with the water.

This drawing also clears up another issue - "long" means "high".

Do we get to assert that any reasonable human observer would be wearing swimming goggles?
If so, you can use the same equations that would be used for the view from above.

If not, then I believe the problem is understated. I think that you would need to consider the curvature of the lens surface in contact with the water - which may vary based on how the viewer is attempting to focus,

 

[jbriggs444]

yes, he is in the water and look the ice cube

Anyone with eyes that function underwater and using stereoscopic binocular vision for distance detection will correctly determine underwater distances. It is simple triangulation irrespective of how one brings the two images into focus.

I agree with @.Scott about the manner in which being underwater could affect range detection by focal distance.

 

[kuruman]

I still fail to see how it could appear to be less than the true height when viewed from the denser medium.

So do I. That is why I asked for clarification about the point of view and suggested that OP considers the "apparent depth" from the other side.

 

[murshiddreamengineer]

The reason I didn't think the problem would be set up this way is that now you are using a lens system intended for air and putting it in direct contact with the water.

This drawing also clears up another issue - "long" means "high".

Do we get to assert that any reasonable human observer would be wearing swimming goggles?
If so, you can use the same equations that would be used for the view from above.

If not, then I believe the problem is understated. I think that you would need to consider the curvature of the lens surface in contact with the water - which may vary based on how the viewer is attempting to focus,

Even if we assume he is wearing goggles. How does the apparent height depend on the height of ice cuboid immersed in the water

 

[kuruman]

Even if we assume he is wearing goggles. How does the apparent height depend on the height of ice cuboid immersed in the water

Look up the derivation for apparent depth. Adapt it for the case where the “depth” is viewed from the other side.

 

[murshiddreamengineer]

Look up the derivation for apparent depth. Adapt it for the case where the “depth” is viewed from the other side.

I do not understand what you mean by "viewed from the other side"

 

[.Scott]

I do not understand what you mean by "viewed from the other side"

What is significant about looking down from above is that you are looking at the water through a horizontal surface between the air and the water. Looking up through such a horizontal surface will produce the same effects.

 

[haruspex]

Even if we assume he is wearing goggles. How does the apparent height depend on the height of ice cuboid immersed in the water

The apparent height of the whole block is the sum of the apparent height of the submerged portion and the apparent height of the exposed portion. Ignoring issues to do with the eye/water interface, the apparent height of the submerged portion would be the true height while that of the exposed portion would be altered by refraction at the air/water interface.
Thus, the change in apparent height of the whole block depends on what fraction is submerged.

 

[murshiddreamengineer]

I got it. Thanks.

 

[kuruman]

I do not understand what you mean by "viewed from the other side"

I mean that the eye, instead of being in air looking down to the bottom of the cube which is under water, is in the water looking up to the top of the cube which is in air. The situation is shown in the picture on the right. The red star is a point at the top of the cube (not shown) emitting rays of light in all directions. Only two of these rays are shown.

If one extends the refracted rays above the surface into the air (dashed lines), the apparent position of the top of the cube will be at their intersection. Note that the apparent height above the surface is greater than the true height. What does that imply about the apparent total height of the ice? Is it going to be greater or smaller than the true height of 10 cm?

 

[Steve4Physics]

It’s possible that there is a mistake with the question’s wording (Post #1) and a mistake in the diagram (Post #9).

If the viewer's eye is in the air directly above the ice, looking down, then a calculation with the given values shows that 9 cm of the ice is below the water surface.

This is what you would expect for floating ice, since the density of ice is roughly 0.9 times the density of water. This seems too much of a coincidence.

 

[.Scott]

When dealing with Physics assignments, I am willing to assume everything is frictionless, there are no aerodynamics, and things like levers and ropes are massless.

But putting an eye or camera underwater is a different matter. We need to be explicit about what constitutes a "vanilla" camera - the one we store on the same shelf as our massless "vanilla" rope and our frictionless "vanilla" slide.

In most cases our vanilla camera will be a pin hole camera with a flat "retina". And although the pin hole has a diameter of zero, there will never be a problem with exposure time or resolution.

When it comes to putting this "vanilla" camera under water, there are two flavors of "vanilla".
Flavor 1 is leak-proof and keeps the optical properties of the camera exactly as it was.
Flavor 2 is well vented and will fill the distance between the pin hole and the retina with the same transparent media that is found outside the camera.

Drawing the set up with the observer depicted as a disembodied eye, strongly suggests the Flavor-1, leakless observer. And, of course, there are other optical issues with the underwater use of a severely non-flat-field lens system such as the eye.

In one of my posts above, I suggested using an eye / camera encapsulated within an air pocket (swimming goggles).

Since the problem description has not addressed this issue, I suggest posters in this thread describe the "flavor" of their observers.

 

[kuruman]

It’s possible that there is a mistake with the question’s wording (Post #1) and a mistake in the diagram (Post #9).

If the viewer's eye is in the air directly above the ice, looking down, then a calculation with the given values shows that 9 cm of the ice is below the water surface.

This is what you would expect for floating ice, since the density of ice is roughly 0.9 times the density of water. This seems too much of a coincidence.

I was about to suggest rephrasing the problem along the lines that you suggested, but now I won't.

 

[haruspex]

In most cases our vanilla camera will be a pin hole camera with a flat "retina".

But that cannot apply here since it would provide no way to judge distances.

Without goggles, a real eye cannot focus because the lens cannot be made fat enough to stop the image being behind the retina. That should mean the object appears nearer than the near point, no matter how far away it is. I would think that means the brain gives up on judging distance entirely.
That leaves real eye + goggles, or two eyes as pinholes, as the only reasonable models for an underwater observer; but there is still the snag that it should make the perceived height greater, not less.

 

[murshiddreamengineer]

It’s possible that there is a mistake with the question’s wording (Post #1) and a mistake in the diagram (Post #9).

If the viewer's eye is in the air directly above the ice, looking down, then a calculation with the given values shows that 9 cm of the ice is below the water surface.

This is what you would expect for floating ice, since the density of ice is roughly 0.9 times the density of water. This seems too much of a coincidence.

Yes, I misunderstood the question; he has to look at the ice from air to water. They did not provide a diagram; I uploaded the one I had understood earlier. Sorry for the confusion.