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Refraction of Incident Ray Through Water

  1. Jan 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Plot a graph of sinθ versus sinθ' where θ is the incident ray and θ' is the refracted ray and use the graph to determine the index of refraction of distilled water, the medium through which the incident ray passes. Draw the graph without including the origin as a data point. Comment on why the graph does not go through the origin.

    Use incident rays of 0, 10, 20, 30, 40, 50 and 60°.

    2. The attempt at a solution

    With respect to an acrylic medium that has an incident light ray approaching at 0°, the refracted light ray is also 0°. I don't understand why this would also not be the case for water. Is this difference due to the incident ray passing through the medium of the container holding the distilled water, then through the water, then through the medium again?
  2. jcsd
  3. Jan 15, 2012 #2

    Simon Bridge

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    Is this a practical experiment? How are you measuring the angles?
    Have you done it? Did the graph go through the origin?

    Not passing through the origin would indicate some sort of systematic error, though normal random uncertainty for so few data points could throw off the line.

    Anyway, you can test your idea on paper by tracing rays through a sample container with unreasonably thick walls and a very different refractive index.
  4. Jan 15, 2012 #3
    No, I haven't run this experiment yet. Based on the description of the lab, we're not supposed to have this plot run through the origin. It's confusing for me too, so I'm sorry if I'm not getting the point across. I want to understand the mechanics of why the line isn't supposed to run through the origin prior to doing the experiment because we have to hand in our reports at the end of class rather than at a later date.

    Yes, it's a practical experiment. We're measuring the angles using a protraction system built into a turn table upon which media of different refractive indexes and curvature will be placed.
    Last edited: Jan 15, 2012
  5. Jan 15, 2012 #4
    I think that Simon Bridges last point about "container with thick walls" may be part of the explanation because you are bound to have the water in a container which will 'shift' the light beam a little.
  6. Jan 15, 2012 #5
    Will this come into play if the incident ray is at 0° and perpendicular to the container's surface though? If the incident ray will pass through acrylic media, and the resulting refracted/transmitted ray will be at 0°, then why would the container affect the incident ray in such a strange way in the case of water?
  7. Jan 15, 2012 #6
    Anyone have any additional insight?
  8. Jan 15, 2012 #7

    Simon Bridge

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    But how is it used?

    Do you shine a narrow beam through the material while it sits on top of a protractor, and look through the material to read the angle off the protractor? Do you use pins in a board?

    No. How does it affect the other points though?

    When you plot the line, the value measured at 0 is only one of the data points which will determine the best fit.

    When you are doing the experiment, you want to look for other sources of systematic errors, also look at how standard errors are throwing the line off the theoretical one.
  9. Jan 15, 2012 #8
    The material (container + water in this case) sits on a circular turn table with a 360° protractor built onto its outer edge. The surface of the material that receives the incident beam is perpendicular to the normal marked at 0°. An incident beam, the result of an incandescent bulb's light passing through a 0.5mm vertical slit, passes through the material and is refracted. We set the value of the incident beam to 0° initially, and measure the resulting refracted/transmitted beam's angle based on the protractor on the opposite side. I expect it will read 180°, but apparently it's not supposed to according to the lab manual. We are NOT given a hint as to the shape or curvature of the container holding the water. I'm assuming it's cylindrical (a beaker).

    We aren't told how it will affect the other points. I guess I'll have to see when I conduct the experiment. I definitely keep track of anything I consider a source of random or systematic error, but I don't think that's what we're expected to look for here.
  10. Jan 15, 2012 #9

    Simon Bridge

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    It strikes me that if you put a ray in at 0deg and you measure a non-zero refraction, then it means the walls of the container are not parallel. (Either opposite walls with each other or two sides of the same wall).

    ... a beaker will not have sides of uniform thickness - but being cylindrical will throw everything way off. Sketch out how the ray would travel through a circular sample and you'll see.

    The curves glass will act like lenses on the beam.
    The incident beam has to be exactly radial to the center of the beaker to be normal incidence ... even a bit off with throw it out.
    The refracted ray is certain to hit a far wall that is not parallel to the one entered: it's a cylinder! The ray will trace a chord inside the beaker.
    So the exiting ray could go pretty much anywhere: the resulting graph won't be a line.

    I doubt you will be using a beaker.
    Possibly a specially constructed D-shape of perspex or a rectangular tank.

    In fact - from the description of the method, it's probably a D - the incident beam enters the tank in the middle of the flat side so the refracted ray will exit perpendicular through the curved side. If the incident beam is not dead-on center you'll get a systematic error.
    Last edited: Jan 15, 2012
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