Refraction - A fish under water

[rishch]

Refraction -- A fish under water

Homework Statement

1 A fish under water obliquely sees a fisher-man standing on the bank of a lake. How does he appear?
2 A parallel sided glass plate is introduced in the path of a converging beam. What happens to the point of convergence of the beam?

The Attempt at a Solution

It's not that I don't know how to solve these questions it's just that my answer differs from the one the textbook gives.

1 The textbook answer is that he appears taller than he actually is. Why? I thought that he would appear a bit higher than he actually is due to normal shift?

2 Textbook answer is that the point of convergence shifts towards the glass plate. But when I visualize, I think that it moves further away from the glass plate.

 

[Ibix]

Sketch! It's always the answer to ray optics problems. It's what ray-tracing packages do, ultimately.

1. Draw a ray from the man's feet to the fish, and from his head to the fish. What can you say about the ray deviations?

2. Depends on the refractive index of the medium that the glass block is sitting in. Assuming that's air, I agree with you. The rays are refracted closer to the optical axis inside the block, so travel further than they would without the block.

When you think the book answer is wrong, always double check that you read the question right, though!

 

[rishch]

Okay, I'll try to draw ray diagrams. As for the second one, yes the medium is air, but both the answers are correct. Their answer is correct if you assume that the point of covergence was in the glass block in which case it would move forwards towards the edge, but I was imagining that the point of convergence was past the block, in which case it moves away. It's like a car going from. If there's a wall in front of it, it moves towards it, but if the wall is behind it, it moves away from it.

 

[rishch]

I tried a bit, but I still don't get it because I can't get the precise angles. In the book they demonstrate normal shift, and I thought that they just move upwards by a fixed amount with the image being a exact replica of the object.

 

[rishch]

Can anyone explain...?

 

[Ibix]

Take a simple case. Put his feet at the same level as the lake surface, and the ray to the fish is an easy one. Then draw a ray from the fish, refract it, and then draw the man. Then you can just measure the angle subtended. See what you get.

I'm not sure the answer to (1) is ever correct. The focus always moves downrange, whether or not it's in the block, I think.

 

[rishch]

Yes it moves downrange but it depends on how you look at it. See the car and wall analogy.

 

[rishch]

Could someone please post a diagram. Because when I draw it, it seems that he becomes shorter as the rays from his lower body are incident at a greater angle and refract moer, while the rays from his upper body are incident at a comparatively less angle so are refracted less, making him shorter :/

 

[ehild]

The fish sees a point of the man at the extension of the ray incident to its eyes. So the ray A starting from the head of the man is refracted and makes a smaller angle with the vertical in the water as makes in the air.

ehild

 

[ehild]

Here are the drawing showing the plate and convergent rays (assuming the refractive index of the plate is higher than that of its surroundings). You are right, the rays are shifted parallel with the original one, and the upper ray shifts upward, the normally incident keeps its direction, so the point of convergence moves away from the plate.

ehild

 

[Ibix]

The fish sees a point of the man at the extension of the ray incident to its eyes. So the ray A starting from the head of the man is refracted and makes a smaller angle with the vertical in the water as makes in the air.

I must say I agree with rishch - the image of the man has a smaller angular subtense than the man, so he appears shorter.

A qualitative argument for this is that the whole "dome of the sky" above the fish is compressed into a viewing angle of ±θ_c from the vertical - why would the image of the man grow when the rest if the above-water world shrinks?

My criticism of your diagram, ehild, is that I don't see why you have moved the image to the plane where the ray B enters the water. Where would the image of a toy boat floating just in front of the man appear? I think a better diagram is my attached. Note that the image is smaller, although this is hardly conclusive as I did not measure anything.

 

[rishch]

You can't be sure where exactly the image is formed, unless you draw more than one ray from the same point and there aren't any convenient rules here so that we can predict how they'll get refracted, as is possible in the case of convex mirrors, so for an accurate diagram you'll need accurate measurements :/ Although I do find what Ibix said, how the dome of they sky above the fish is compressed a very good point.

 

[ehild]

I must say I agree with rishch - the image of the man has a smaller angular subtense than the man, so he appears shorter.

A qualitative argument for this is that the whole "dome of the sky" above the fish is compressed into a viewing angle of ±θ_c from the vertical - why would the image of the man grow when the rest if the above-water world shrinks?

You are right, the feet of the man appears at higher position, too, what I did not take into account. The whole world above the water appears inside some smaller viewing angle, everything shrinks, so must the man, but to know the image exactly, we have to draw at least two rays from every point.
I could have been right if the man stood on the water surface :tongue2: so the fish would see his feet directly and his head high above, but in that case, the part of body in and out of water would be separated.

ehild