Scientists are testing a transparent material whose index of refraction for visible light varies with wavelength as n = 30.0 nm1/2/λ1/2 , where λ is in nm. If a 295-nm-thick coating is placed on glass (n=1.50) for what visible wavelengths will the reflected light have maximum constructive interference?

Video transcript

Chiropractic index that is equal to a divided by Lambda to the 0.5 or the square root of Lambda where LAMBDA is the wavelength of light and nanometers and A is a constant equal to 37.6 with units of the square root of a nanometer. A 200 nanometer thick layer of the material is deposited on a glass substrate that has a refracted index of 1.55. We're asked to determine the wavelength of visible light that exhibits constructive interference between reflected waves from the top and bottom interfaces of this thin layer. Our multiple choice answers in units of nanometer are a 30 sorry 384 and 609 B 425 114 C 526 109 or D 486 120. So the equation that they give us in this problem N of Lambda is equal to a divided by the square root of Lambda. We can combine that with the equation that we know for our constructive interference, which is LAMBDA of C is equal to two multiplied by N multiplied by D divided by N where M for constructive interferences are positive integers. So M will be 123. And so, and so our, to, in this equation, N is again our refractive index and then D is the thickness of our leg. And so when we combine these two, solving for our um constructive wavelength, we have two multiplied by a multiplied by D divided by the square root of Lambda. Sub C multiplied by M sub C. And so we can pull the land of seas together on the same side. And we'll have Lambda C is equal to two multiplied by a multiplied by D divided by and sub C raised to the two thirds cap. And from here, it's simply a loving chug using our known values for the constant A and our thickness D and then varying our N. So lambda C where M is equal to one will be two multiplied by 37.6. And those are units of the square root of the nanometer multiplied by our thickness, 200 nanometer divided by M. This is M one all that raised the two thirds power now because we're working in nanometers. And our answer is in nanometers and we're not using any other units, we can keep everything in nanometers and we don't have to move it to our standard unit of a meter. And so we plug that into our calculator and we get 609 nanometers. We'll do the same thing for our constructive wavelength when lambda or sorry, when M is equal to two and we get 384 SS. OK. Let's keep going see what happens when and is equal to three, we get 293 nanometers. OK. So now we are outside of our visible wavelengths which we can recall the human eye can see wavelengths from around 380 nanometers to 700 nanometers. And so these two answers are the answers that are in that visi visible spectrum. And so when we look at our multiple choice answers, we can see that this aligns with answer choice A so a is the correct answer for this problem. That's all we have for this one, we'll see you in the next video.