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Apparent depth involving benzene on water

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data
    In a vessel, a layer of benzene (n=1.50) 6 cm deep floats on water (n=1.33) 4 cm deep. Determine the apparent distance of the bottom of the vessel below the upper surface of the benzene when viewed vertically through air. Ans 7 cm

    2. Relevant equations
    Snell's law n sin (theta)=n sin (theta)
    apparent depth/actual depth = n of air / n of material

    3. The attempt at a solution
    at first I thought I would solve the equation in parts so that for benzene:
    apparent depth = 1(6 cm)/1.5 = 4 cm
    and for the water
    apparent depth = 1.5 (4 cm)/1.33 = 4.5 cm
    adding them together you get 8.5 which is not the right answer. I don't know how you would use Snell's law with not having any theta.
  2. jcsd
  3. Dec 8, 2008 #2


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    This problem is simpler than you might think. What happens to the wavelength in each material? Actually, this reasoning may not be logically sound, but it works for apparent depth problems. To remember, just think depth and wavelength are both lengths, but don't forget to treat each material separately.

    It can be derived using Snell's law:
    1) Draw two rays that immenate from the object in the material.
    2) Find the ray paths outside of the material (Snell's law)
    3) Extrapolate those paths as virtual ray straight lines back into the material. The point where the virtual rays intersect is the image location.
  4. Dec 8, 2008 #3
    So as the rays go into the first material they are refracted. The ray that merges into the benzene from the air is refracted toward the normal because n of benzene > n of air. Then as that ray merges into the water is is refracted toward the normal because n of benzene > n of water.

    The path of the ray we see is imaginary which causes the bottom to look closer than it really is. So is there one set of imaginary rays that got through both the benzene and the water? Or is there a set of imaginary rays for each?

    And how do you find the ray paths with only knowing the depth? It seems that there is not enough information to use Snell's.
  5. Dec 8, 2008 #4


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    Since you are interested in the proof via Snell's law, I should specify that you can use the small angle approximation in this case. Do you know how to do this for a single layer of material, like finding the apparent depth of a coin at the bottom of a pool?


    Nope. You got it backwards. I'm assuming you are tracing the ray from the benzene into the water.

    I don't know what you mean by that. The depth that your brain reconstructs from the rays that enter your eye is imaginary in that, if you plunge a ruler into the container down to that depth, you will not hit the bottom, because it isn't the true depth. So, the point in space where your brain thinks that the rays intersect is not where they actually do. However, the rays that finally enter your eye after being refracted and reflected by whatever means are always real in any situation.

    Yes. I assume you mean that, for each physical ray in the set, you can trace it along a unique path from a point on the bottom to a point at the surface of the benzene.

    You can just choose a point on the bottom and consider the paths of two different rays that originate from that point out into the air. Draw a picture with three horizontal lines: the bottom line represents the bottom, the middle line represents that water-benzene boundary, and the top represents the surface of the benzene. Then, choose a point on the bottom line and draw from it two almost vertical lines, but slightly slanted. When these lines intersect the middle horizontal line, they become more vertical (they bend toward the normal from the water to the benzene). Continue these new segments until they hit the top horizontal line. At that point, the lines become more slanted (even more so than in the bottom layer, because n_air<n_water).

    You can write equations for the rays in the air, for instance in slope-intercept form, and then solve for the intersection of these two lines.
  6. Dec 8, 2008 #5
    How can your write an equation for in slope-intercept form when you have no other numbers besides depth? I'm not really following. I understand how the rays are to be drawn but not how to get the equations from that drawing.

    Attached Files:

  7. Dec 9, 2008 #6


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    You actually have more numbers than just depth, but you choose some of them. You have the indices of refraction, that are given, and you choose the angles of incidence to examine. But you shouldn't be plugging numbers in anyway; this is a derivation. Leave everything symbolic until the end.

    I can't see the png file that you attached. Anyway ...

    What I would do is assign the x-coordinate to the depth, and the y-coordinate to the horizontal (and turn the picture sideways if it makes you feel better). I would set the surface of the benzene to x=0. Then, from the point on the bottom to the first boundary (in the water), you can calculate a Delta y using trig and the fact that you know the depth of the water. You can calculate an additional Delta y in the benzene in the same way. The sum of these two Delta y's gives you the y-intercept of the ray in air (since x=0 there). You can find the slope of the ray in air using trig and Snell's law. Do this for both rays. Then find the value of x where the two lines intersect. This point will not be on the physical rays; the rays are extrapolated back into the material.

    OK, now I can see your image file. You have two bent ray paths represented by solid lines. Then, you have two dashed lines that look almost like extrapolations from the rays in the air, except that they look slightly bent at the boundary. Was this intentional? The dashed lines and the rays in the air should be colinear. Also, you should choose the same point on the bottom for both paths.
    Last edited: Dec 9, 2008
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