012 (part 1 of 2) 10.0 points Unpolarized white light is incident on a po- lariz
ID: 2124404 • Letter: 0
Question
012 (part 1 of 2) 10.0 points
Unpolarized white light is incident on a po- larizer and then falls on an analyzer whose TA (Transmission Axis) makes an angle of theta = 33.8 degrees with the polarizer.
What is the fraction of the original light transmitted through both the polarizer and analyzer?
(decimal format)
013 (part 2 of 2) 10.0 points
If the analyzer is rotated until the transmit- ted intensity is 38.2 percent of the original light intensity, what is the new angle of the analyzer with respect to the polarizer?
Answer in units of (degrees)
Explanation / Answer
1) The intensity of unpolarized light is reduced to half of its intensity when it is passing through the polarizer. I1= Io/2 Apply the law of malus to the analyzer is I2 = I1 cos^2 %u03B8 = (Io/2)cos^2(33.8o ) = 0.345 Io So the fraction of the original light transmitted through both the polarizer and analyzer is I2/Io= 34.5 %. 2) Given that I/Io = 38.2 % The intensity of unpolarized light after passing through the analyser is I2 = (0.5) Io cos^2 %u03B8 from part (1) 0.382 = (0.5)cos^2 %u03B8 %u03B8 = 29.06o So the fraction of the original light transmitted through both the polarizer and analyzer is I2/Io= 34.5 %. 2) Given that I/Io = 38.2 % The intensity of unpolarized light after passing through the analyser is I2 = (0.5) Io cos^2 %u03B8 from part (1) 0.382 = (0.5)cos^2 %u03B8 %u03B8 = 29.06oRelated Questions
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