00:01
In this problem, we're going to be looking at an incident beam of light as it travels through three polarizers.
00:10
So here's our incident beam headed in this direction.
00:18
And we're given the intensity of the incident beam.
00:24
I'm going to call that si, which is 1 ,260 watts per meter squared.
00:33
What we're going to be looking for is the average intensity.
00:37
Of the transmitted light.
00:40
So i'll be calling that st.
00:43
That's what we're trying to solve for is we're going to have to look at the intensity as it changes through each of these polarizers until we get to the end.
00:52
So our polarizers, each one has a different angle measured relative to the vertical.
00:58
So here's our verticals.
01:01
This first one has an angle theta 1 equal to 19 degrees.
01:15
The second one has an angle theta 2 equal to 55 degrees.
01:24
And lastly, we have theta 3.
01:28
And that one we're told is 100 degrees.
01:33
So as it travels through each of these polarizers, the intensity of the light's going to change slightly and we're going to get at the other end a transmitted light that's a lower intensity than what we started with.
01:44
Now, since we have over here unpolarized light, to begin with, then we know that quite simply, after it goes through the first polarizer, the transmitted intensity is just one -half what we started with, no matter what angle theta is, because the light is starting off unpolarized, it's always going to end up being just one -half of that intensity after it goes through the first polarizer.
02:14
Because unpolarized light relative to a polarizer, half of it will get through the other half is perpendicular, and so you lose that half.
02:25
So through this first polarizer, we end up with an intensity, i'll call that s1 of 630 watts per square meter.
02:38
So this is the intensity of light.
02:41
Now that's going to hit the second polarizer.
02:43
So now that we have polarized light with a solarized light with a second.
02:46
For intensity...