Unpolarized light of intensity I₀ is incident on a stack of 7 ...

Question

Unpolarized light of intensity I₀ is incident on a stack of 7 polarizing filters, each with its axis rotated 15° cw with respect to the previous filter. What light intensity emerges from the last filter?

Accepted Answer

Hello, everyone. Let's go through this practice problem. The intensity of light emitted by an incandescent light bulb is I sub B. The emitted light is directed towards a system consisting of four polarizer. The first polarizer axis is vertical. Whereas the 2nd, 3rd and 4th polarizer axes are 45 degrees, 90 degrees and 100 35 degrees counterclockwise from the vertical respect. Calculate the intensity of the light that emerges from the system. Option A 5.0 multiplied by 10 to the power of negative one I sub B. Option B 2.5 multiplied by 10 to the power of negative two I sub B. Option C 1.3 multiplied by 10 to the power of negative two I sub B and option D 6.3 multiplied by 10 to the power of negative three I sub B. So there's a lot of information given in this problem of four different three different angles for four different polarizer that we have to work with. But fortunately, despite all the information and even all the steps that we're gonna have to go through to solve this problem. It is ultimately a fairly simple application of polarizing laws. So let's go through each polarizer, each of the four polarizer and discuss how the intensity I sub B will be affected by each polarizer. So the first thing I'm going to do is set up a formula for the intensity after it's passed the first polarizer, which I'm going to call is sub one. Now, when the light first emit, when the light is first emitted from the incandescent light bulb, we say that the light is un polarized because it hasn't reached any polarizer yet. Fortunately, for us, the math on this is very simple. If any un polarized light passes through any polarizer, no matter what the angle is, the intensity is split in half. That is an important law to recall. So after the light is passed, the first polarizer, I sub B is divided into two. So that is what the new intensity is after the light is passed through the first polarizer. Now let's come up with a formula for the intensity after the light has passed through the second polarizer, which I'm going to call is sub two. So now that we're talking about polarized light passing through more polarizer, this is where the formula starts to get a little bit more complicated re remember something called Malice's law. Malice's law is what is going to apply to the later polarizer. And Malice's law states that if polarized light passes through a polarizer, then the new intensity is equal to the previous intensity before it reached the polarizer. So I of one, in our case multiplied by the square of the cosine of the difference of angles between the two polarizer. So the angle difference between the first two polarizer is represented by this angle theta. So for our purposes, I sub one is represented by I sub B divided by two multiplied by the cosine squared of the angle difference. The which is given in the problem as 45 degrees because the second polarizer is 45 degrees, whereas the first polarizer is at an angle of zero degrees. So there's a 45 degree difference there. Now let's talk about the intensity after the light has passed through the third polarizer or I sub three. Now, for the rest of the problem, we're just going to keep using malice's law which states that the intensity after it passes through the polarizer is equal to the intensity just before it hits that polarizer multiplied by the square of the cosine of the angle difference between that polarizer. And the previous one, we're just gonna keep using the same method. So for polarizer three, for intensity three, it's basically just going to be intensity two multiplied by the cosine squared of that angle difference. And as we just discussed, this is I I sub two is, is sub B divided by two multiplied by the cosine squared of theta. But now we're multiplying all of this by the cosine squared of the new theta. The one thing that's very nice about this problem is that the value of the actually isn't changing for us because remember this, the doesn't represent the angle, it represents the angle difference between the current polarizer and the previous one. So even though the second polarizer or the third polarizer is at an angle 90 degrees, that theta is still only represented that still only equals 45 degrees because 90 degrees is 45 degrees higher than 45 degrees, which was the angle of the previous polarizer. So these data are going to stay the same because they both represent a 45 degree difference. So let's simplify this a bit by writing by putting those cosine squared together, this is I sub B divided by two multiplied by cosine squared of theta which is also squared. So finally, we now need to discuss I sub four, the intensity after the light has emerged from the fourth polarizer. Once again, we're using the exact same concept of malice's law. So we're going to take our expression for I sub three. So I sub B divided by two multiplied by the cosine squared of theta all squared. And then multiply it by again, the cosine squared of the new angle difference theta which is represented by a change from 90 degrees up to 100 35 degrees, which you might notice is another 45 degree shift. So we can use the exact sa we can keep using the exact same theta. And in its most simplified form, this is just equal to I sub B divided by two, multiplied by the cosine squared of theta all cubed. Now, so this is a fairly simple formula at this point. All we got to do is plug in 45 degrees for theta. So we're taking I sub B dividing by two and multiplying by the cosine squared of 45 degrees cube. And if we put this into a calculator, then we find a value for the constant for the coefficient of 6.3 multiplied by 10 to the power of negative two multiplied by I sub B. That is our answer to the problem. And we can see this corresponds to option D but that is it for this problem. I hope this video helped you out. And if it did, please consider checking out our other tutoring videos which will give you more experience with these types of problems. That's all for now. And I hope you all have a lovely day. Bye bye.

Unpolarized light of intensity I₀ is incident on a stack of 7 polarizing filters, each with its axis rotated 15° cw with respect to the previous filter. What light intensity emerges from the last filter?

Key Concepts

Malus's Law

Polarization of Light

Cumulative Effect of Multiple Polarizers

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