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Optical activity of sugar

  1. Nov 18, 2008 #1
    The sugar concentration in a solution can be measured conveniently by using the optical activity of sugar nad other asymmetric molecules. In general, an optically active molecule like sugar will rotate the plane of polarization through an angle that is proportional to the thickness of the sample and to the concentration of the molecule. To measure the concentration of a given solution, a sample of known thickness is placed between two polarizing filters that are at right angles to each other. The intensity of light transmitted through the two filters can be compared with a calibration chart to determine the concentration.

    a. What percentage of the incident (unpolarized) light will pass through the first filter?

    b. If no sample is present, what percentage of the initial light will pass through the second filter

    c. When a particular sample is placed between the two filters, the intensity of light emerging from the second filter is 40.0% of the incident intensity. Through what angle did the sample rotate the plane of polarization?

    d. A second sample has half the sugar concentration of the first sample. Find the intensity of the light emerging from the second filter in this case.

    For a my guess would be that the percentage would be 50% of half of the light will pass through the first filter.
    For b my guess is that it would be 0% because then it would be at a 90 degree angle and cosine of 90 = 0
    I am not quite sure how to do c and d either

    Any help would be appreciated.
     
  2. jcsd
  3. Nov 19, 2008 #2

    Redbelly98

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    a and b are correct answers. However, I'll point out that transmitted intensity depends on cos2θ. You still get 0 for a 90 degree angle, so it makes no difference in part b.

    For c and d, use the fact that the transmitted intensity is
    cos2θ
     
  4. Nov 29, 2010 #3
    Use the Law of Malus!
    I = I_o cos ^2 (theta)
    so for part c it will be
    .4 = .5 cos^2 (theta)
    But you have to do 90-theta to get the real value. You can draw a diagram and you can see that doing arccos will give you the wrong angle value.
    Part d you can do the law of malus again!
    I2= ½(I0)cos2((90- (answer to part c/2)
    You are solving for I2/I0. That should help.
     
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