- #1

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1)1 2)3/2 3)2/3 4)4/3

Solution: 1

Method ive tried: I really dont know how to solve this problem! i know if there is only one liquid. but there are two liquids so iam little confused.

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- Thread starter sylwesh98
- Start date

- #1

- 42

- 0

1)1 2)3/2 3)2/3 4)4/3

Solution: 1

Method ive tried: I really dont know how to solve this problem! i know if there is only one liquid. but there are two liquids so iam little confused.

- #2

lightgrav

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Ray 1 from "coin" at vessel bottom center, straight along the Normal all the way into air.

Ray 2 from coin is tilted at angle θ

"virtual ray 2" traces back straight along Ray 2 in air, but does not refract at either surface.

- #3

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- #4

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That is the ray 1 and its correct . Make one more ray .ray is not clear i dont get it! is it the ray?

- #5

DEvens

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- Trace a ray from the coin to the top. But have it at some angle so that it will refract at each boundary.

- Follow the ray all the way from the coin to the top. And project this ray to an observer. Don't forget to do this as a function of the value of A.

- Trace another ray straight up. (You have that.)

- Now you have a vertical, and an apparent angle at the observer.

- Work out from this where the coin looks like it is sitting, but do it as a function of the value A.

- Now you were told where the coin looks like it is sitting.

- #6

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- #7

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please can anyone provide me with a rough ray diagram so that i can proceed?!

- #8

DEvens

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please can anyone provide me with a rough ray diagram so that i can proceed?!

I will need to see you make some effort before I will help you more. I think I have given you plenty to solve the problem if you will just make the effort.

Review your notes on what happens when light passes from a medium with one refractive index to a medium with a different refractive index. Look for Snell's law.

- #9

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- #10

DEvens

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So, now, what formula applies at each interface? When a ray moves from a higher index media to a lower index media, does it get closer to the normal or farther away? And by how much? Don't forget that you have two places where the ray changes material.

To figure out the apparent depth of the tank you will need to be comparing to a situation with no fluid in the way that produces the same situation for an observer. So when you figure out where the rays are above the tank you will want to be projecting them back as though the liquid was not there, and see where it makes it look like the object at the bottom of the tank is.

- #11

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when the ray 2 leaves the medium one it will refract towards normal as it is passing from rarer to denser medium, after that from medium 2 to air it will be passing away for the normal as it is passing from denser to rared medium from that ray when it is seen from the eye it appears that the ray is coming from the middle of the vessel. but how can we solve if there are 2 liquids.

- #12

DEvens

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What formula applies at each interface? Write down Snell's law. You have a formula that relates the index of refraction at the red-to-grey interface. And now you can write down some formulas that give you the apparent depth as a function of the index of refraction. And you know the depths. And you know how deep the coin appears to be. Start writing down the formulas, what you know, and what you need to know.

- #13

lightgrav

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(I would also draw a "construction line" vertically from where ray 2 enters the top fluid)

The 2 "apparent depths" will show up along the 2 vertical lines.

- #14

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i dont know how to proceed from there! any clue please

- #15

lightgrav

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How steep is the (blue) angle in air, in terms of the (red) angle in the bottom layer?

You're supposed to find the apparent depth (top left triangle + middle top triangle),

compared to the actual depth (big \| in gray + little \| in red).

each of these individual depth comparisons is just a ratio, whose formula you prob'ly know.

- #16

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i think only snells law is applicable in these cases.

- #17

lightgrav

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The small-angle approximation to Snell's law is :

n

- #18

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but how it is applicable there?

- #19

lightgrav

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(generally) relate the apparent depth to the actual depth and the refractive index "A", and solve for A.

Snell relates the various θ , and tan θ ≈ the triangle width / height ... label the dimensions you know (or want).

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