Light of original intensity I0 passes through two ideal polarizing filters having their polarizing axes oriented as shown in Fig. E33.28. You want to adjust the angle f so that the intensity at point P is equal to I0/10. (a) If the original light is unpolarized, what should Φ be?

Video transcript

Welcome back, everyone. We are making observations about an un polarized beam of light. Now, we are told that it is incident in the x direction has an intensity of I incident right now crosses two linear sheet polarizes. Now, when it crosses the first one, the transmission axis is perpendicular to the X axis, while the transmission axis of the second sheet forms an angle fee with the first transmission axis. Now we are tasked with finding what is going to be if after leaving the second linear sheet polarizer, the second one. So I'll name this the second one and this the first one, we have an intensity of 1/6 of the original incident intensity. All right. Now, before getting started here, I do wish to acknowledge our multiple choice answers here on the left hand side of the screen, those are gonna be the values that we strive for. So without further ado let us begin. So we know that the intensity leaving the second linear sheet polarizer is 1/6 of the original intensity. But we can actually come up with another equality to the intensity leaving the second linear sheet polarizer after leaving the first one, we have an intensity of one half of the original intensity, which means if we are taking into account our angle fee, our intensity of our light passing through our second linear sheet polarizer is going to be our incident intensity divided by two times cosine squared of our angle fee. Since both of these equations equal the intensity of our light leaving the second linear sheet polarizer, we can set them equal to one another. So we have I ink divided by six is equal to I ink over two times cosine squared of P. Now, what I'm gonna do is I'm going to multiply both sides by two over I ink two over I ink. You see on the left hand side here, the I inks cancels out on the right hand side, the I inks cancels out as well as the two is in the denominator and numerator. Now what this gives us is that the square of the cosine of P is equal to 2/6 or one third. I'm gonna take the square roots of both sides to get rid of our exponent on the left hand side of our screen. And then in order to get rid of the trick function, all I have to do is take the inverse cosine of both sides of our equation. So then on the left hand side, the trick functions cancel out and we get that fee is equal to the inverse cosine of the square root of 1/3 which when we plug it into our calculator, we get that fee is 54.7 degrees corresponding to our final answer. Choice of a. Thank you all so much for watching. I hope this video helped. We will see you all in the next one.