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The first polarizing filter polarizes the light. along its axis: The second polarizing filter is rotated and the polarization axis angle between the first filter an...

Question

The first polarizing filter polarizes the light. along its axis: The second polarizing filter is rotated and the polarization axis angle between the first filter and the second filter is 60e. Find the intensity ratio of the light after the first filter (L) and the second filter (Lz). Give LPE Lfh = 0 L/li = 0.25 Lfh = 0.5 Iz/hi = 0.75 Lli = 1prnmIngPdarizing Iliet

The first polarizing filter polarizes the light. along its axis: The second polarizing filter is rotated and the polarization axis angle between the first filter and the second filter is 60e. Find the intensity ratio of the light after the first filter (L) and the second filter (Lz). Give LPE Lfh = 0 L/li = 0.25 Lfh = 0.5 Iz/hi = 0.75 Lli = 1 prnmIng Pdarizing Iliet



Answers

Three polarizing filters are stacked with the polarizing axes of the second and third at 45.0$^\circ$ and 90.0$^\circ$, respectively, with that of the first. (a) If unpolarized light of intensity $I_0$ is incident on the stack, find the intensity and state of polarization of light emerging from each filter. (b) If the second filter is removed, what is the intensity of the light emerging from each remaining filter?

After the first polarizer, the intensity will go down so we can write. Intercity will be. I know. Let's say it was I note after a fastball riser, which is making 60 degrees. So then it will go off course 60 square go 60 square. So if we make a square off a ghost 60 then there. The answer we get here is course sixties by not is half and court for So after first we get Ah, the intensity quarter off the initial. And with the 2nd 1 which is horizontal, then I know. Or four or four course off 30 goes off 30 square. So course off thirties. So they're here. We have I know Lord over too. We're four. Sorry times cause off. 30 is a square root off too. Books 33 over two square. So you were two whole square. And excuse us. Excuse us. I know. Or, um, full force of 16. Some 16 3 So I know there were 16. This will be our intensity after second polarizer. End off the problem. Thank you for

How's drank? Slash style door discussion. Suppose you have a war rising for just okay, this is everything okay? No. The light off in density iron ore is incident on the first. Polarizing. Okay, you have to man pain this fi as such, you will get a final intensity off. I know, but okay. So firstly, suppose you have unpasteurized light and this is your polarizer one and this is a polarizer to Okay, so after first polarizer, when this intensity is this un polarized light is passes you will get the intensity I note by two because plane polarizer will convert on polarized on pullers on were like to polarise so intensity will reduce to house So it will be I know this I know two will fall on second polarizer and you have to just five so that you will get the intensity off. I know it by 10 So we have the formula Icicles toe. I know cause where tip in this case, we want to get eyes. I know by 10 on I notice I know it by two because after first polarizer, we will have I note by two intensity anti ties oversight in this case. Now, After solving this, this is one by five. So, Joe, point a little course in worse, you will get the value of five. And if it sold this, we will get an angle of 63.4 degree approximately. Okay, 63.40. Now suppose instead off an polarized light. If there is the plane polarized like that plane polarized simply some first polarise unchained remains painful rush. So our intensity instead off I note by two will be, I note. So I know. Because where Tito intensity will not change in case off polarized light. Okay, so it is. I know, right? Then I note coast where Tita, I know it will cancel out. So from this fire from here, you will get the value of five goes inward. 0.1 under through. And that is 71.5 degrees. Okay, well, this is all for me for this video. I hope you will like that with you. Thank you.

In the first part of this problem, we have finally really off I for richer the human condition holes. I saw that intense to off light when the bosses with the forcible rising scrutinize it, my prime Michels to whatever to off I know what this I don't is that intensity often pull right light. When the life passes through the second polarizer its intensity will be Eichel's too. I pride uh, cause I squared off five No, inciting the well off I and I probably weaken white this equation, as I know d wanted by 10 equals two. I know you wanted by two because I m squared off five. Well, now sold in this question Far five. We can write It is PFI calls to our cause I off one divided by Sequera Little fight five So this will give us a value for fires. Michael's too 63 point for three degree In part B of this problem, we have to kill Clear the value off the five. We told us that you want condition so there will be no change in the intensity or flight after passing through the possible riser So we can white I prime equals toe. I don't now. They did. Still flight of passing through the second polarize it will be. I equals two I pride. And then because I square off five. Well, when taking the weather's into this equation, weaken right. I have the well off I know you wanted by 10. And that is I note, and then we have co sign square off. I well, falling this question for five weaken white equals two are co sign off one divided by a squared off death. So this is your value for fires. Michael's too. So 21 point five. So indignity. So this is the required answer. Thank you.

Question for the six states that, um, incident light with intensity. I not is it happens bond. Three polarized filters I shown here for the 1st 1 is at zero degrees. The next one is rotated 45 degrees in the next one is already stated 45 degrees after that. In this problem, we wouldn't find what I three is based on this information. So we can see that after the light passes to this first filter at angle of zero degrees, we confined simply that this intensity is just I not over to I to it happens, passes through filter at 45 degrees. So that intensity goes as co signed squared of the angle co sine squared 45. Well, I'm not sorry I want is just I not over to second replace that cause I'm 45 as a route to so co sign Square 45 is just 1/2 sorry one over route to coast 45. So the coastline square 45 is just 1/2 times 1/2 is I not over for And similarly, for I three, this final intensity of our late it's I to times again coastline scored 45 because the filter three is rotated by 45 degrees from filter, too. I could replace I two with what I found in the line before. Which is this incident intensity over four times again. 1/2 to find that the final intensity of the light after passing through these three filters is I not over eight. There you go.


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