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.
(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.
(i) To determine the percentage of incident (unpolarized) light that will pass through the first filter, we need to consider that the first filter is at right angles to the polarized light. This means that the first filter will block all of the light that is polarized in the same plane as the filter. Since the incident light is unpolarized, it is made up of equal components of light polarized in all possible planes. Therefore, when the unpolarized light passes through the first filter, only the component that is perpendicular to the filter's plane will be transmitted.
This means that the first filter will allow 50% of the incident (unpolarized) light to pass through, as it filters out the other 50% that is polarized in the same plane as the filter.
(ii) If no sample is present between the two filters, the light that passed through the first filter will be polarized in a plane perpendicular to the first filter. When this polarized light reaches the second filter, which is at right angles to the first filter, it will be stopped completely. This is because the second filter will block all light that is polarized in the plane perpendicular to its own plane. Therefore, if no sample is present, 0% of the initial light will pass through the second filter.
(iii) When a particular sample is placed between the two filters, and the intensity of light emerging from the second filter is 40.0% of the incident intensity, we can determine the angle through which the sample rotates the plane of polarization. The intensity of transmitted light is directly proportional to the concentration and the angle through which the plane of polarization is rotated.
Assuming a linear relationship, we can write the following equation:
Transmitted intensity = Incident intensity * concentration * cos(angle)
Given that the transmitted intensity is 40% of the incident intensity, we have:
0.4 = concentration * cos(angle)
To solve for the angle, we rearrange the equation:
cos(angle) = 0.4 / concentration
Finally, we take the inverse cosine (arccos) of both sides to find the angle:
angle = arccos(0.4 / concentration)
(iv) If a second sample has half the sugar concentration of the first sample, we can use the same equation as in part (iii) to determine the intensity of light emerging from the second filter in this case.
Assuming the angle of rotation remains the same for the second sample, we can find the intensity by substituting the new concentration (half of the original concentration) into the equation:
Transmitted intensity = Incident intensity * concentration * cos(angle)
For the second sample, the intensity of light emerging from the second filter is equal to:
Transmitted intensity (second sample) = Incident intensity * (0.5) * cos(angle)
Therefore, the intensity of light emerging from the second filter in this case will be half of the intensity for the first sample.