You set up an experiment with two polarizer screens on an optical bench and coll

Question

You set up an experiment with two polarizer screens on an optical bench and collect data for the intensity of the light at the detector as you vary the relative angle between the transmission axes of the polarizers. You create a plot of light intensity (in lux) vs cos2 with your data and get a linear fit for your graph gives a slope of 120.2 ± 1.204 and a y-intercept of 15.45 ± 2.141

(a) What is intensity of the light source you used in your experiment?

(b) What is ambient light in the room as you took your data?

Explanation / Answer

a) The intensity of the light source can be determined from the point when the transmission axes are aligned (0 degrees). This corresponds to the y-intercept of a) The intensity of the light source can be determined from the point when the transmission axes are aligned (0 degrees). This corresponds to the y-intercept of 10.45 ± 1.276. Only half of the light gets through however, as the other half is reflected by the polarises. Therefore the intensity of the light source is120.2 ± 1.204

b) The ambient light in the room can be obtained when the transmission axes are perpendicular to each other. That is, when theta = 90 degrees (or pi radians). This is because in this scenario, no light from the light source can pass through the polarisors, thus only ambient light is recorded. As you have not given the units of slope, I can't know whether you have used lux/radian or lux/degree. Basically however use the slope to determine the intensity when theta = 90 degrees, and you'll have the ambient light in the room.. Only half of the light gets through however, as the other half is reflected by the polarises. Therefore the intensity of the light source is 120.2 ± 1.204

b) The ambient light in the room can be obtained when the transmission axes are perpendicular to each other. That is, when theta = 90 degrees (or pi radians). This is because in this scenario, no light from the light source can pass through the polarisors, thus only ambient light is recorded. As you have not given the units of slope, I can't know whether you have used lux/radian or lux/degree. Basically however use the slope to determine the intensity when theta = 90 degrees, and you'll have the ambient light in the room.

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