(II) Two polarizers A and B are aligned so that their transmis...

Question

(II) Two polarizers A and B are aligned so that their transmission axes are vertical and horizontal, respectively. A third polarizer is placed between these two with its axis aligned at angle θ with respect to the vertical. Assuming vertically polarized light of intensity I₀ is incident upon polarizer A, find an expression for the light intensity I transmitted through this three-polarizer sequence. Calculate the derivative dI / dθ ; then use it to find the angle θ that maximizes I.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with Malice's law. So in this problem, a student is conducting an experiment in her physics lab, she uses a beam of vertically polarized light with intensity not and it passes through three polarizing filters. The first filter P one is aligned vertically. The second filter P two is placed at an angle of theta with respect to the vertical. And the third filter P three is set at an angle of 60 degrees relative to the vertical. She wants to know how much light intensity I is transmitted. After passing through all three filters, there are two parts to this question. Part one, we need to find an expression for the transmitted light intensity I. And for part two, using the derivative of D I by D theta determine the angle theta that max bs as the transmitted intensity. For her experiment, we're also given four possible choices as our answers. For choice. A for part one, the intensity is I is equal to I knot multiplied by cosine squared theta multiplied by cosine squared of the quantity of 60 degrees minus the. And for part two, the intensity is maximized at theta equal to 30 degrees. For choice B for part one, the intensity is I is equal to I not multiplied by cosine squared. The multiplied by cosine squared of the quantity of 90 degrees minus theta. And for part two, the intensity is maximized at theta equal to 60 degrees. For choice C for part one, the intensity is I is equal to I not multiplied by cosine squared. The multiplied by cosine squared of the quantity of 60 degrees minus the. And for part two, the intensity is maximized at 30 equal to 45 degrees. And for choice D for part one, the intensity is I is equal to I not multiplied by cosine squared, the multiplied by cosine squared of the quantity of 90 degrees minus theta. And for part two, the intensity is maximized at theta equal to 90 degrees. Now, for part one, we need to find the intensity after the light passes through all three polarizer. And so what we're going to do is use malice's law to figure out the intensity after each polarizer. So starting with our first um filter here. P one recall that from Alice's Law, the intensity after the polarizer which we'll call I one is equal to the incoming intensity, which is I not multiplied by the cosine squared of the angle between the direction that the light is polarized. And the direction of the filter. In this case, we have both vertically um polarized light and a vertically aligned filter. So that means the angle between them is actually zero degrees. So the cosine squared of zero degrees. So that means that I one is just equal to I not, we'll do the same thing for the second filter P two. So we'll have its intensity after the filter called I two. And this could be equal to the incoming uh intensity I one multiplied by the cosine squared of the angle between the direction of the light polarization and the direction of the filter. So in this case, after the first filter, we still have the l uh light being polarized vertically. And the second filter is placed at an angle theta with respect to the vertical. So the angle between those two is just going to be equal to theta. So we can replace I one with I not. And so that means we'll have I two is equal to I kno multiplied by cosine squared. The, we'll do the same thing for the third filter except we'll call its intensity after that. Um After it passed through the filter as I, since that's gonna be the intensity after all three filters and that's gonna be equal to the incoming intensity, which is I two multiplied by the cosine squared of the angle between the polarized light and the angle of the filter. In this case, our polarized light is coming in at an angle of theta and our filter is set at an angle of 60 degrees. Both of those are measured relative to the vertical. So we'll have the angle of the filter which is 60 degrees minus the angle of the light, which is at this case at an angle of theta. The difference between the two is just gonna be 60 minus the, we can go ahead and plug in for I two and we'll have I is equal to I kno multiplied by cosine squared of data multiplied by cosine squared of the quantity of 60 degrees minus the. And that could be your answer for part one. Now, for part two, we need to find the angle that maximizes the transmission here. So the way we'll do that is to take a derivative our, of our expression of I with respect to theta and then set it equal to zero, define the extremo. So we'll have the derivative of I with respect to theta. That's gonna be equal to the derivative with respect to the, of the quarantine here will plug in our formula for I. That's gonna be I not multiplied by cosine squared of theta multiplied by cosine squared of the quantity of 60 degrees minus theta. So all we need to do is take a derivative of this expression. So we'll have our derivative of I with respect to theta, gonna be equal to the I knots constant. So I can pass through the derivative. So I'll have I not multiplying the quantity. And here I need to use our product rule to take a derivative of both cosine squared terms. So taking a derivative of the first term, I'll get minus two cosine data multiplied by sign data. And that's multiplied by the second term which is cosine squared of the quantity of 60 degrees minus the. So just recall that the derivative of cosine is equal to minus sine. So that's where the negative sign shows up. We'll have to do the same thing for the derivative of the second term, the second cosine squared term. And so we'll have plus two multiplied by the cosine of 60 degrees minus the multiplied by the sine of 60 degrees minus the. And this could be multiplied by cosine squared, the which is our first term and all that has to be equal to zero in order to find our ex drema. So for the second term, um recall that again, our derivative of cosine is equal to minus sine, but you then have to take the derivative of the angle inside the cosine term. And that's where the second negative term comes in since I have minus the there, so two negative signs cancel. So that's why that second term is positive. So if I look at both of these terms inside of our brackets here, we have a factor of two that we can factor out and then we can divide both sides by two I not. And so that basically gets rid of those terms since I was setting the right hand side to zero. What I can do now is factor out a cosine data and cosine of 60 degrees minus data out of both of the, both of our remaining terms. And so I'll have the quantity of cosine data multiplied by cosine of the quantity of 60 degrees minus the. And this could be multiplying what's left, which is going to be minus sine theta multiplied by cosine of 60 degrees minus data. They don't have plus sign of 60 degrees minus data multiplied by cosine theta. That product has to be equal to zero in order for us to find the xtreme. Now, if I look at the first term here, the cosine multiplied by co uh the cosine data multiplied by the cosine of the quantity of 60 degrees minus the, if either of those two terms are zero, that means that it will have blocked the transmission of light out of our second polarizer or our third polarizer. So if the is 90 degrees, that means that cosine of the here is equal to zero. So that means that we'd have no in light passing through the second polarizer according to malice's law. Likewise, if the cosine of 60 degrees minus data is equal to zero, that means we'll have no light being transmitted through the third polarizer. So setting that first term equal to zero is um going to give us the minimum for that intensity. But the question is asking for the maximum. So we're going to ignore that first term and only set the second term in our product here equal to zero. And so when I do that, I can move the negative term over to the right hand side of our equation. And so I'll have the sign of 60 degrees minus the multiplied by cosine data is equal to sine theta multiplied by cosine of 60 degrees minus the. So I need to solve this for the, and one way I can do that is divide both sides by cosine theta and then divide both sides again by cosine of 60 degrees minus the. That will give me sine divided by cosine which gives me a tangent. So I have the tangent of the quantity of 60 degrees minus data is equal to the tangent of data. If I take the arc tangent of both sides, I'll have 60 degrees minus theta is equal to theta. And so we can easily solve this for theta by adding data to both sides and then divided by two. And that gives me data is equal to 30 degrees. So that could be our answer for part two. So if I look at our answer choices, this actually corresponds to answer choice. A so just a quick recap of what we've done for part one, we use malice's law to calculate the intensity after each polarizer. And that allows to calculate the total intensity as being transmitted through all three filters for part two in order to find the maximum angle uh the light uh sorry, the angle that maximizes the intensity, we first need to take a derivative of the intensity with respect to theta and set that equal to zero in order to find our ex extreme. And then we just um solve what was left there and that gave us an angle of theta equal to 30 degrees. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Malus's Law

Intensity Transmission through Multiple Polarizers

Optimization and Derivatives

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