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Show that if you have three polarizing filters, with the second at an angle of $45^{\circ}$ to the first and the third at an angle of $90.0^{\circ}$ to the first, the intensity of light passed by the first will be reduced to 25.0% of its value. (This is in contrast to having only the first and third, which reduces the intensityto zero, so that placing the second between them increases the intensity of the transmitted light.)
Therefore, the intensity of light after passing through all the three filters is 25$\%$ of theinitial intensity.
Physics 103
Chapter 27
Wave Optics
Simon Fraser University
Hope College
University of Sheffield
McMaster University
Lectures
02:51
In physics, wave optics is…
10:02
Interference is a phenomen…
01:28
Show that if you have thre…
00:28
The angle between the axes…
10:49
*Unpolarized light passes …
02:24
Unpolarized light with i…
01:52
Unpolarized light of inte…
04:40
Three polarizing filters a…
02:42
Three Polarizing Filters. …
01:55
00:56
03:01
$\bullet$ Three ideal pola…
02:46
02:55
Unpolarized light is incid…
01:49
$\bullet$ Two ideal polari…
01:04
You use a sequence of idea…
02:22
Unpolarized light of inten…
01:09
Prove that, if $I$ is the …
01:31
At the end of Example $27.…
01:39
Show that a diffraction gr…
00:41
If you have completely pol…
00:46
Additional ProblemsThr…
After the first filter, the intensity we have is i not then, after the second filter, we have intensity. I 1. After third, we have intensity. I 2 point we can write. I 1 is equal to. I not cos square theta, whereas i 2 is equal to. I 1 cos square theta 2 point. Then we can substitute the value of i 1. So this is theta 1 here, which is, i nod, cos square theta 1 times cos square theta 2 point so substituting the values for a that worn had to do, which is 45 degrees. I 2 will be. I nod times cos 45 degree square, cos. 45 degrees square, so simplifying this expression we get here, i 2 to be 0.25, not.
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