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) 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
thank you soooo much.
(i) To determine the percentage of incident (unpolarized) light that passes through the first filter, we need to consider that the first filter is perpendicular to the polarization direction of the incident light.
Since the first filter is at right angles to the polarization direction, it blocks all the light that is polarized in that direction. Therefore, none of the polarized incident light will pass through the first filter. Only the unpolarized light component will pass through.
Hence, the percentage of incident (unpolarized) light that will pass through the first filter is 100%.
(ii) If no sample is present, the intensity of light passing through the second filter will depend on the orientation of the second filter with respect to the polarization direction of the incident light.
Since no sample is present, the light passing through the first filter will still be unpolarized. If the second filter is aligned with the polarization direction of the unpolarized light, it will allow all the light to pass through. Thus, the percentage of initial light that will pass through the second filter is 100%.
(iii) The intensity of light emerging from the second filter is 40.0% of the incident intensity. This indicates that the sample in between the two filters is rotating the plane of polarization by a certain angle.
To determine the angle of rotation, we can use the relationship between the angle of rotation and the intensity of transmitted light:
Intensity of transmitted light = Initial intensity * cos²(θ)
Where θ is the angle of rotation.
Given that the transmitted intensity is 40.0% of the initial intensity, we can set up the equation:
0.4 = 1 * cos²(θ)
Solving for θ:
cos²(θ) = 0.4
cos(θ) = √(0.4)
θ = arccos(√0.4)
θ ≈ 66.42 degrees
Therefore, the sample rotates the plane of polarization by approximately 66.42 degrees.
(iv) If the second sample has half the sugar concentration of the first sample, the rotation of the plane of polarization will be proportional to the concentration.
Since the concentration of the second sample is half of the first sample, the rotation angle will also be half of the previous angle.
Therefore, the angle of rotation for the second sample will be approximately 66.42 degrees divided by 2, which is approximately 33.21 degrees.
To find the intensity of light emerging from the second filter in this case, we can use the same equation as before:
Intensity of transmitted light = Initial intensity * cos²(θ)
Substituting the new angle of rotation, we have:
Intensity of transmitted light = 1 * cos²(33.21)
Calculating the value, we find that the intensity of light emerging from the second filter in this case is approximately 89.78% of the initial intensity.