A) Answer: 50% B) Answer: 0% C) When a particular sample is placed between the t
ID: 1284090 • Letter: A
Question
A) Answer: 50%
B) Answer: 0%
C) When a particular sample is placed between the two filters, the intensity of light emerging from the second filter is 35.3% of the incident intensity. Through what angle did the sample rotate the plane of polarization?
Answer: 57.2 degrees
D) A second sample has half the sugar concentration of the first sample. What percentage of the initial light will pass through the second filter in this case?
Answer: __________%
A) Answer: 50% B) Answer: 0% C) When a particular sample is placed between the two filters, the intensity of light emerging from the second filter is 35.3% of the incident intensity. Through what angle did the sample rotate the plane of polarization? Answer: 57.2 degrees D) A second sample has half the sugar concentration of the first sample. What percentage of the initial light will pass through the second filter in this case? Answer: __________%Explanation / Answer
Law of Malus
I= Io cos2 (theta)
.23= .5 cos2 (theta)
.46 =cos2 (theta)
sqroot .46 = cos (theta)
.678 = cos (theta)
arccos .678 = theta
47.3 = theta
And again, wrong angle so minus it from 90
90 -47.3 = 42.7
then we multiply them all together for each intensity
.5 * cos2 (42.7) * cos2 (90-42.7) = 12.4 %.
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