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Homework Help: Sin i / sin r graph from ray through glass block

  1. Jan 23, 2010 #1
    1. The problem statement, all variables and given/known data
    I have done the light ray through a glass block experiment at different angles.

    I've taken the measurement for i1 which is from the incident ray. and i2 which is from the emergent ray. Then I have to take the average of i1 and i2 and then plot a graph using the average value.

    I have to of course do this for various incident angles and their corresponding emergent angles. So it'll be a sin i over sin r graph.

    However, if I measure an incident angle i1 to be 15 degrees and the corresponding i2 emergent to be 16, then that means the average will be 15.5 degrees. I have to plot the graph using average values.

    However the protractor only measures to an accuracy of 1 degree, so would I be correct in plotting using the value for sin16 as opposed to plotting for sin15.5 ?

  2. jcsd
  3. Jan 24, 2010 #2
    I don't understand why you are taking an average of the incident and emergent angles.
    The graph uses a list of values of i and plots them against the corresponding values of r.
    Are you sure the instruction was not "average values of i" plotted against "average values of r", where each value of i and r is the result of an average taken of a number of measurements.
  4. Jan 24, 2010 #3
    sorry I probably haven't worded my question very well.

    The experiment consisted of shining a light ray throught the glass block at various incident angles.

    Each different i1 angle would be measured and recorded and its corresponding i2 angle measured and recorded. The i2 being the emergent angle with the normal. Then I would take the average of these two (both would of course be very similar) and plot the average value.

    The sin of these would be plotted against the sin of the average refractive angles. That is, the sin of the average of the refractive angle at the incident and the refractive angle at the emergent.

    So its just a standard sin i over sin r graph.

    Though when finding the average sin i values there is the following problem.

    I measure an incident i1 angle of, for example, 40 degrees and then measure the i2 angle for the same light ray. i2 being its emergent ray angle aginst the normal. Though this i2 value is, say 41 degrees.

    So I take an average of the two and I have to plot the sin value of the average on a graph. Though the average is 40.5 degrees. I'm thinking for this I should plot using a value which is the sin of 41 degrees (rounding up) instead of plotting using a value of sin 40.5 degrees.

    Since the protractor cannot measure to an accuracy of 0.5 degrees.

    There wouldn't be this problem if, for example, the I1 value was 50 and the I2 52 since the average is 51. That is, a whole number which the protractor can measure to an accuracy of.
  5. Jan 24, 2010 #4
    I see.
    The reason for taking the average of say, 40 and 41 is to improve the overall accuracy of the measurements. Even though the protractor only has a scale with one degree graduations, you can use the value 40.5 in your sin and graph. There would be no point in taking the average of the 2 values if you then rounded up or down to the nearest scale value.
    In addition, by taking readings for a number of different values of i and r, and drawing a graph of sin i against sin r, you are also using these multiple readings to increase the accuracy when you measure the gradient.
    You can look at the overall probable accuracy of your value at the end by examining the "best" and "worst" values of the gradient.
  6. Jan 24, 2010 #5

    That's right, yes.

    And it would be done for various incident angles 20, 30, 40 etc and taking the corresponding i angle at the emergent, and then taking the individual averages. The intention being that each i value and each r value is more accurate.

    That's what was puzzling me as we previously covered in class how various measuring instruments are only accurate to various amounts. Since the protractor is only accurate to one degree I wasn't sure if it was appropriate to take the sin of a degree which included a half degree.


    Thanks, yes that's part of the homework too. Thanks again Stonebridge. It's very much appreciated! :smile:
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