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. (b) If the original light is linearly polarized in the same direction as the polarizing axis of the first polarizer the light reaches, what should Φ be?

Video transcript

Welcome back, everyone. We are making observations about a polarized light of a laser diode that passes through to polarize. Now, for polarizer P, one, we have that its transmission is along the y axis. While the transmission axis of the second polarizer makes this angle fee with the y axis. Now the incident light has an initial intensity of I initial, right? And we are tasked with finding what is going to be P if the intensity at point M is equal to 1/8 of our initial intensity. Now, before getting started here, I do wish to acknowledge our multiple choice answers on the left hand side of the screen here, those are the values in which we want to strive for. So without further ado let us begin. All right. Well, we actually have an equation for this where you're going to have that the intensity at any point. And then in this case, we'll take it at point M is going to be equal to the initial intensity times the cosine squared of P. Now dividing both sides. Well, actually, before I divide both sides, I'm going to sub in our value for our IM we know that this is I, our initial intensity divided by eight is equal to our initial intensity cosine squared of B dividing both sides by I eight. You see that the I eight or sorry, just the initial intensity, the initial intensities cancel out on both sides. And what we get is that cosine squared of P is equal to 1/8. In order to get rid of that exponent, I'm going to take the square root of both sides. And then in order to get rid of the cosine function, I'm gonna take the inverse cosine of both sides. You'll see on the left hand side, all we are left with now is P and we have that P is equal to the inverse cosine of the square root of 1/8 which plugging this into our calculator gives us 69.3 degrees corresponding to our final answer. Choice of B. Thank you all so much for watching. I hope this for your help. We will see you all in the next one.