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.
Physics
To answer these questions, we need to understand the principles of optical activity and the setup described in the question.
(i) To find the percentage of incident (unpolarized) light passing through the first filter, we consider that the first filter is polarized. When unpolarized light passes through a polarizing filter, only the component of light aligned with the transmission axis of the filter can pass through.
In this case, the first filter is placed at the initial position with no sample present. Since the initial light is unpolarized, it consists of equal intensities of light oscillating in all possible directions. When this light passes through the first filter, which is at right angles to the initial light direction, only the component of light aligned with the transmission axis of the first filter will pass through.
Therefore, the percentage of incident light passing through the first filter is 0% since all the light is blocked by the first filter when there is no sample present.
(ii) Similarly, when no sample is present, but the second filter is in place, the transmitted light intensity will depend on the angle between the two filters. In this case, the two filters are at right angles to each other. When unpolarized light passes through two filters at right angles, no light will pass through the second filter. Therefore, the percentage of the initial light passing through the second filter in this scenario is 0%.
(iii) When a sample is placed between the two filters, the intensity of light emerging from the second filter is stated to be 40.0% of the incident intensity. We need to determine the angle by which the sample rotates the plane of polarization.
Since the sample is a sugar solution, an optically active substance, it rotates the plane of polarization of light passing through it. The angle of rotation is proportional to the thickness of the sample and its concentration.
In this case, the intensity of light emerging from the second filter is reduced to 40.0% of the incident intensity. This reduction in intensity suggests that the sample rotates the plane of polarization by an angle that causes only 40.0% of the initially transmitted polarized light to align with the second filter's transmission axis.
To find the angle of rotation, we can use the equation:
Intensity transmitted = intensity incident * cos^2(angle of rotation)
Let x be the angle of rotation. Then, we have:
0.4 (incident intensity) = incident intensity * cos^2(x)
Dividing both sides by the incident intensity:
0.4 = cos^2(x)
Taking the square root of both sides:
cos(x) = sqrt(0.4)
Taking the inverse cosine (arccos) of both sides (note: inverse cosine gives the principal value between 0 and π radians):
x = arccos(sqrt(0.4))
So, by calculating the arccosine of the square root of 0.4, we can find the angle of rotation.
(iv) For the second sample, with half the sugar concentration of the first sample, we need to find the intensity of light emerging from the second filter.
Since the sugar concentration is halved, the angle of rotation caused by the sample will also be halved. Therefore, we can use the equation from part (iii) with half the angle of rotation.
Let y be the angle of rotation for the second sample. Using the same equation as before, but with y instead of x:
transmitted intensity for the second sample = incident intensity * cos^2(y)
Since y is half of x (the angle of rotation for the first sample), and cos^2 is a symmetric function (cos^2(θ) = cos^2(-θ)), we know that cos^2(y) = cos^2(x/2).
So, the transmitted intensity for the second sample can be calculated as:
transmitted intensity for the second sample = (incident intensity) * cos^2(x/2)
By substituting the value of x/2 from part (iii) into this equation, we can find the intensity of light emerging from the second filter for the second sample.