Reflection/Refraction water tank question

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The discussion revolves around understanding visibility of marks in a water tank due to refraction. The initial question addresses whether marks can be seen from the top or bottom of the tank, with the conclusion that marks can be seen from the bottom coming up. Participants discuss the importance of sketching light rays from the marks to the observer's eye, emphasizing that the eye is in air, not water. Refraction at the air-water boundary is highlighted as a key factor affecting visibility. Overall, the conversation aims to clarify how light travels and refracts, impacting which marks can be seen.
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Homework Statement


[/B]
The 80-cm-tall, 65-cm-wide tank shown in the figure is completely filled with water. The tank has marks every 10 cm along one wall, and the 0 cm mark is barely submerged. As you stand beside the opposite wall, your eye is level with the top of the water.

Part A asked "Can you see the marks from the top of the tank (the 0 cm mark) going down, or from the bottom of the tank (the 80 cm mark) coming up?" I already answered this question: We can see the marks from the bottom of the tank coming up.

Which is the lowest or highest mark, depending on your answer to part a, that you can see?
d=__cm

I have attached an image I found online of the figure

Homework Equations



The only equation that I can think of would be Snell's Law n1sinθ1=n2sinθ2

The Attempt at a Solution



I am not sure how or where to start. If the 0 cm is at eye level, then wouldn't that be the highest mark you can see? Should I be sketching rays from each mark to the eye? If so, wouldn't they all reach the eye?
physics-drawin-png.25408.png
 
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Your idea to sketch rays from each mark to the eye is great! Just remember that your eye isn't in the water; your eye is in the air. If you'd like, you can imagine that your eye is very close to but not touching the edge (e.g. 1 mm or 1 cm). The wording in these sorts of problems can be a bit confusing at first.

Next time you're swimming in a pool, see if you can test this out.
 
Scott Redmond said:
Your idea to sketch rays from each mark to the eye is great! Just remember that your eye isn't in the water; your eye is in the air. If you'd like, you can imagine that your eye is very close to but not touching the edge (e.g. 1 mm or 1 cm). The wording in these sorts of problems can be a bit confusing at first.

Next time you're swimming in a pool, see if you can test this out.
Thank you for replying! So as the rays head toward the eye, does the incident ray hit the edge of the tank (air-water boundary?) and then refract in air before it hits the eye? Sorry, I'm still a bit lost. Wouldn't they all still hit the eye?
 
Yes, there is refraction at the air-water boundary on the horizontal surface of the water. The light then has to travel through some air -- even if it's less than 1 mm -- before it hits the eye.

Try it out with some rays, maybe starting 10 cm below the surface, and you'll see whether all of the rays hit the eye. You're correct to use Snell's law for this.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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