The sugar concentration in a solution (e.g., in a urine specimen) can be measured conveniently by using the optical activity of sugar and 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, as shown in the figure. The intensity of light transmitted through the two filters can be compared with a calibration chart to determine the concentration.

Question

(i) What percentage of the incident (unpolarized) light will pass through the first filter?
(ii) If no sample is present, what percentage of the initial light will pass through the second filter?
(iii) 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?
(iv) A second sample has half the sugar concentration of the first sample. Find the intensity of light emerging from the second filter in this case.

Answer 1

(i) half (it arrives randomly polarized. The first filter takes out all the x components for example)

(ii) None. The x components go out with the first filter. The second filter takes out the y components :)

(iii)http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polcross.html

50 % got through the first filter
now it is turned through angle theta by the sugar
so angle to the second filter is 90-theta
.4 original = x (.5 original)
x = .8 gets through
.8 = cos^2(90-theta)
90-theta = 26.6
theta = 63.4

iv)
intensity will be lower because not turned so much by the sugar
theta = (1/2)(63.4) = 31.7
x = cos^2(90-theta) =.276
x(original) = .276*.5 = .138 of original gets through

Answer 2

thank you soooo much.