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) The first filter only allows light that is polarized in a particular direction to pass through. Since the incident light is unpolarized, it contains an equal amount of light polarized in all directions. Therefore, when an unpolarized light passes through a polarizing filter, only 50% of the incident light will pass through the first filter.

(ii) When no sample is present, the second filter will block all the light that has been polarized by the first filter. Since the first filter only allowed 50% of the incident light through, no light will pass through the second filter. Therefore, 0% of the initial light will pass through the second filter when no sample is present.

(iii) If the intensity of light emerging from the second filter is 40% of the incident intensity, it means that the second filter is only allowing 40% of the polarized light from the first filter to pass through. Since the second filter is at a right angle to the first filter, the sample must have rotated the plane of polarization by an angle such that only 40% of the light can pass through.

(iv) If the second sample has half the sugar concentration of the first sample, it means it will rotate the plane of polarization by half the angle of the first sample. Since the intensity of light emerging from the second filter is proportional to the angle of rotation, the intensity of light emerging from the second filter will be half of the intensity when the first sample was present. Therefore, the intensity of light emerging from the second filter will be 40% / 2 = 20% of the incident intensity.

(i) To determine the percentage of incident light that will pass through the first filter, we need to consider that the first filter is at the entrance of the system, before any light has interacted with the sample.

The first filter is called an analyzer and is placed perpendicular to the plane of polarization of the incident light. In this configuration, the analyzer will block all light that is polarized perpendicular to its transmission axis, and only allow light that is polarized parallel to its transmission axis to pass through.

Since the incident light is unpolarized, it consists of an equal combination of all possible orientations of polarization. Therefore, when the unpolarized light passes through the first filter, half of the intensity will be blocked because it is perpendicular to the transmission axis, and the other half will pass through because it is parallel to the transmission axis.

So, the percentage of incident light that passes through the first filter is 50%.

(ii) If no sample is present, the second filter, called the polarizer, is positioned after the sample region. As the second filter is oriented at right angles to the first filter, it will block all light that passed through the first filter, since the first filter only allows light polarized parallel to its transmission axis to pass through.

Therefore, if no sample is present, none of the initial light will pass through the second filter, resulting in 0% transmission.

(iii) When the sample is placed between the two filters, the intensity of light emerging from the second filter is 40% of the incident intensity. This reduction in intensity suggests that the sample has rotated the plane of polarization of the incident light.

Since the intensity of transmitted light is proportional to the concentration of the molecule and the rotation angle, we can set up the following equation:

Intensity of transmitted light = Incident intensity * Concentration * Rotation angle

In this case, the intensity of light emerging from the second filter is 40% (or 0.40) of the incident intensity, which leads to the equation:

0.40 = 1.00 * Concentration * Rotation angle

Simplifying the equation, we find that the sample has rotated the plane of polarization by an angle equal to 0.40 / Concentration.

(iv) If a second sample with half the sugar concentration of the first sample is placed between the filters, we can determine the intensity of light emerging from the second filter using the same equation as before:

Intensity of transmitted light = Incident intensity * Concentration * Rotation angle

Let's denote the concentration of the second sample as C, which is half of the concentration of the first sample. So, the concentration of the second sample is 0.5 * Concentration.

Plugging this into the equation, we have:

Intensity of transmitted light = 1.00 * (0.5 * Concentration) * Rotation angle

Simplifying further, we find that the intensity of light emerging from the second filter in this case is 0.5 times the intensity of light when the first sample was present. Therefore, the intensity of light emerging from the second filter will be 0.5 * 40% = 20% of the incident intensity.