Horizontally polarized light is incident on a polarization filter. The initial intensity of the light is 0.55 W/m 2 , but is then reduced to 0.40W/m 2 after passing through the filter. Calculate the angle of the transmission axis of the polarization filter with respect to the horizontal.

everyone. So, hopefully, you got a chance to try this practice problem on your own. Let's work it out together. So we're told that initially horizontally polarized lights passes through a polarization filter. It's not unpolarized. Alright? The initial intensity of this light is 0.55. After it passes through the filter, it gets reduced to 0.4. We wanna calculate the angle of the transmission axis of this polarization filter with respect to the horizontal. Alright. So first things first, let's just draw a quick little sketch of what's going on. So we've got my little axis like this. We've got lights that is initially horizontally polarized. Alright? So don't make the mistake of drawing all these lines because it's not, unpolarized light. So, basically, what I'm gonna do here is I'm gonna draw this light kind of like this, and I'm actually just gonna say that this is 0 degrees. Okay? So this is 0 degrees, and I'm gonna label this my i0, and this is gonna equal, 0.55. Alright? And now what happens is it passes through a polarization filter, and that polarization filter is angled at some angle, and that's basically what I'm trying to find here. Alright? So if this is like my horizontal, what I'm really trying to find here is what is theta. Alright? So that's the theta that I'm trying to find, such that when it comes out the other side, it's gonna be polarized now in the new direction like this, a little bit different. Right? It's gonna be polarized at some angle and the intensity of this light, I'm gonna call this I one, it gets gets reduced down to 0.4. Alright? So this is a little bit different than some of the other problems we've seen because in the other problems we're usually calculating what this final intensity is. Now we have the final intensity, and we're actually looking for the angle, so we're doing something else. But still, we're just gonna go through our steps here. We're gonna use, our steps for finding, the for, you know, solving for the nth polarizer. There's really only one here, so there's only one equation that we have to use. So if we take a look at this, what happens is, remember, this light here is gonna be polarized initially. So when it comes time to use our equations, which one do you think we're gonna use? Are we gonna use the one half rule, or are we gonna use the cosine square rule? Well, hopefully, we realize because it's initially polarized already and we're not gonna use the one half rule, we're actually just gonna go straight into using our cosine squared rule. Right? So with this first filter, sometimes that light will be unpolarized and sometimes it will be initially polarized. Alright? So don't always assume that it's always going to be the one half rule at the beginning because that might not always be true. Alright? So this basically says that, I one is going to equal I zero, right, I one, and then you reduce it, you use the light from before. This is going to be cosine squared, and this is going to be of theta over here. So this is really what I'm trying to find here. What's my theta, all right? So I've got actually both of these numbers. I have i1 and i0, so I could just slowly start to solve for theta. If I move this to the other side, what I get is that i1 divided by i0 is equal to cosine squared theta. All right? Now I've got a cosine squared, but I wanna find the theta. So what I have to do is I have to take the square roots. So I have to take the square root of both sides, and then basically what happens is I get that, cosine of theta is equal to the square roots of I one over I naught. Now you could start plugging some numbers in, that's totally fine, but there's actually just one last step to do, which is that we have to take the inverse cosine to get rid of this cosine over here. So what happens is when you take the inverse cosine, that goes away, and then this becomes the inverse cosine of the square root of I 1, which is 0.4, divided by I naught, which is 0.55. Now make sure that your calculator is gonna be in degrees mode. So this I'm gonna put here degrees mode here. Alright. So make sure your calculation's in degrees mode. Otherwise, you're gonna get radians. You're gonna get something weird. And then what your answer should be is your answer should be 31.48 degrees, and that means that that is the angle with respect to the horizontal of that transmission axis. Alright? So, basically, this angle over here with respect to 0 is about 31 degrees. So this is 31.8. Anyway, folks, so that's it for this one. Let me know if you have any questions. Let's move on.

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Horizontally polarized light is incident on a polarization filter. The initial intensity of the light is 0.55 W/m2, but is then reduced to 0.40W/m2 after passing through the filter. Calculate the angle of the transmission axis of the polarization filter with respect to the horizontal.

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Horizontally polarized light is incident on a polarization filter. The initial intensity of the light is 0.55 W/m², but is then reduced to 0.40W/m² after passing through the filter. Calculate the angle of the transmission axis of the polarization filter with respect to the horizontal.