A single optical coating reduces reflection to zero for λ = 55...

Question

A single optical coating reduces reflection to zero for λ = 550 nm. By what factor is the intensity reduced by the coating for λ = 430 nm and λ = 670 nm as compared to no coating? Assume normal incidence.

Accepted Answer

Welcome back. Everyone. In this problem. A glass surface is coated with a special material that reduces the reflection of light to zero for a wavelength of Lambda equals 560 nanometers for Lambda equals 510 nanometers and lambda equals 600 nanometers. By what factor is the intensity of the reflected light reduced compared to no coating assume normal incidents. A hint is that we can use the intensity reduction formula that says I divided by IO equals two, multiplied by one plus cost four PI ND divided by Lambda where D is the thickness of the coating with zero reflection at Lambda equals 560 nanometers A says that at 510 nanometers, the factor is 9.41% and at 600 it's 4.37%. B says they are 9.41 and 5.83% respectively. C 10.2 and 4.37% and D 10.2 and 5.83%. Now, how can we figure out by what factor? OK. The intensity of the reflected light is reduced compared to no coating. Well, for starters we'll, we'll need to figure out or we'll need to use our information, use what we know when we're comparing it to new coating. That's, that's our baseline. OK. Now, we already know based on our problem that the reflection of light is reduced to zero. OK. For a wavelength of 560 nanometers. In other words, at LAMBDA equals 560 nanometers. OK. Then our factor I divided by not is gonna be equal to zero. So if we put this into our formula, OK, then we should be able to solve for the thickness D and then we will be able to use that value to help us figure out the factor for different wavelengths. OK. So let's see if we can go ahead and do that. Now this tells us then that I divided by a knot is equal to two multiplied by one plus the cosine of four pi nd divided by pi. OK. And we know at this point it's going to be equal to zero. Since it's equal to zero, we can divide both sides by two which tells us that one plus a cosine of four pi and D divided by Lambda. Oops should have put lambda here. My apologies is going to be equal to zero. And thus that means then that the cosine of our expression is going to be equal to negative one. No, for the cosine function to equal negative one, the argument must be an odd multiple of pi that is, that is, and that is four pi nd divided by Lambda must be equal to two M plus one pi OK. And odd multiple of pi where M is an integer. And now at the minimum thickness OK. At the minimum thickness, I'll get rid of this line in a minute. But at the minimum thickness OK. Then from our expression, we can say that M is going to be equal to zero. Thus, this tells us then 0.4 PND divided by Lambda is going to be equal to two times zero plus one which is one multiplied by pi or just pi. Notice here we can cancel pi from both sides of our equation and know when we solve for D. That means our thickness D is going to be equal to Lambda divided by four N. We know our value of Lambda here is 560 nanometers. So we can even write that here as 560 nanometers divided by four N. OK. Know that we have an expression OK? For our value of D or thickness, then we can use it OK to substitute it into the intensity reduction formula for our varying wavelengths because no, that tells us. Then let me put this in red here that at Lambda, for example, let's call it Lambda one as 510 nanometers. OK. Then that means or reduction factors is going to be equal to two multiplied by one plus the cosine. OK. Remember our formula tells us it's gonna be a cosine of four pi ND. OK. So that's gonna be four pi N multiplied by our value for D Lambda. Sorry, 560 nanometers divided by four N. OK? And now we know all of this is going to be divided by this value for Lambda that we have. OK. This new value for Lambda is 510 nanometers. I'm sorry, I apologize for my um my mark here, my writer, I'm not sure why it's just skipping a bit but just ask you to be a bit a little patient with me. OK. Now here in this case, notice that we can cancel four N from both terms. OK? And now when we simplify at Lambda being 510 nanometers, then our factor is going to be two multiplied by one plus the cosine of 560 nanometers pi OK. Divided by our value for a Lambda which is 510 nanometers. OK? And no, when we substitute here and solve, OK, then we should get our reduction factor to be 0.09412 which is approximately equal to 9.41% to three significant figures. OK. So when, in other words, we're saying when the wavelength is 510 nanometers, it has a reduction factor of 9.41%. Now let's just basically do this again to calculate the intensity reduction at 600 nanometers because no uh let's call this lambda two lambda two equals 600 nanometers. Then our factor we can basically just use this same formula that we've simplified it to. So our factor is going to be equal to two multiplied by one plus the cosine OK. Of 560 nanometers pi divided by 600 nanometers. OK. And again, when we go ahead and solve, we should get this to be 0.04370 or approximately 4.37%. Thus, when our wavelength is 510 nanometers, then our factor is 9.41%. And when it is 600 nanometers, our factor is 4.37%. If we look back at our answer choices that tells us a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

A single optical coating reduces reflection to zero for λ = 550 nm. By what factor is the intensity reduced by the coating for λ = 430 nm and λ = 670 nm as compared to no coating? Assume normal incidence.

Key Concepts

Optical Coating

Interference of Light

Wavelength Dependence

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