A thin oil slick (n_o = 1.50) floats on water (n

Question

A thin oil slick (n_o = 1.50) floats on water (n_w = 1.33). When a beam of white light strikes this film at normal incidence from air, the only enhanced reflected colors are red (653 nm) and violet (392 nm). From this information, deduce the (minimum) thickness t of the oil slick.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem, a thin film of soap that has a refractive index of N equals 1.40 forms on a bubble which is floating in the air. When the sunlight hits the bubble, the most prominent reflected colors are yellow. Lambda equals 580 nanometers and blue. Lambda equals 460 nanometers. Based on this observation, what is the minimum thickness tea of the soap film responsible for these colors? So that's our angle. Our final answer that we're ultimately trying to solve for is we're trying to figure out what the minimum thickness T is for this soap film in order to produce these reflected colors of yellow and blue. So now that we know that we're ultimately trying to figure out what the value of T is, we're trying to figure out what the minimum thickness value is. Let's read off our multiple choice sensors to see what our final answer might be noting that all of our multiple choice sensors are in the same units of meters. So A is 2.4 multiplied by 10 to the power of negative six. B is 5.3 multiplied by 10 to the power of negative six C is 6.7 multiplied by 10 to the power of negative six. And finally, D is 8.9 multiplied by 10 to the power of negative six. OK. So first off, we need to recall and use our equations to help us. So for the constructive interference, so we need to recall a note for this particular problem. When constructive interference occurs, we need to consider the equations for the thickness tea for both the yellow and blue light conditions separately. So for example, when we, as we should recall, we should write that the equation for T the thickness for the first case which will say T equals M one plus one half multiplied by lambda one F. So I'm gonna just say one F instead of saying one comma F. So this is lambda wavelength one F. So this is the first wavelength film condition or the wavelength of the films for the first condition I should say. And this is all divided by two. And we also need to recall, we have to consider our second condition which T equals M two plus one half multiplied by lambda two F all divided by two. So to keep things simple, let's make a quick little note here, let us note that we're calling LAMBDA one is going to be equal to yellow light, which is 580 nanometers. And LAMBDA two is gonna be our blue light condition, which is equal to 460 nanometers for our wavelength values. Fantastic. So now at, at this stage, we can equate the two equations for thickness. And when we do that, we will find that it's equal to M one plus one half multiplied by Lambda one F. And I'm using F to denote the film once again, just in case we're cut, in case we're confused, which is all equal to M two plus one half, all multiplied by Lambda two F. Fantastic. So now moving right along here, we now need to convert the wavelengths to a film medium. So as we should recall, the wavelengths of light inside of the soap film is related to the wavelength in the air by a refractive index of N equals 1.40 which is stated to us by the pro itself. So let's start with yellow light first. So let's sulfur yellow light. So we need to consider that lambda where lambda denotes wavelength. So we need to say LAMBDA one F and we need to say that it's equal to 580 nanometers, we divide it by a refractive index. So 1.40. So let me plug that into our calculator. We round to two decimal places. Our final answer for a Lambda one F will be equal to 414.29 nanometers. And once again, that's when we round two decimal places. And then similarly for Lambda two F which noting once again that Lambda two is our blue light condition. So we need to take 460 nanometers and we divide it by 1.40 and also rounding to two decimal places, we will find that our final answer is approximately equal to 328.57 nanometers. Fantastic. So now moving right along. So for this particular problem, we should recall a note that instead of using decimal approximations, we are able to use the ratio of the original wavelengths directly. So with that in mind, we could take the ratio of our two wavelengths in the air which meaning we could write that lambda two divided by lambda one is which is equal to 460 nanometers divided by 580 nanometers, which when we plug that into our calculator and make sure we write our final answer as a fraction or as a ratio, we'll get 23 divided by 29 Fantastic. So that also means that this is all equal to Lambda two F divided by Lambda one F. Fantastic. So we're making a lot of progress here. OK. So now our next step as we need to solve for the values of M one and M two from our thickness equation. So as we should recall, a note, the ratio of the wavelengths, we need to use our value for the ratio of wavelengths and we need to plug it directly into our equation. So using our equation where we equated the two answers together. So that's so let's look right here. So let's call the equation where we equated our thicknesses together which is M one plus one half, all multiplied by Lambda one F which is equal to M two plus one half, all multiplied by Lambda Tua. So we need to take equation one, which is when we equated our thickness values together and we need to substitute in the value for our ratio of wavelengths into that equation. And when we do that, we will find that it's equal to M one plus one half, which is equal to 23 divided by 29 multiplied by and two plus one half. Fantastic. So now in order to make the calculation simpler, let us express both sides in terms of fractions. So when we do that, we will be able to write that M one and we could get rid of our parentheses. We could say M one plus I'll put the fractions in a parenthesis case we're plugging it into a calculator. So you don't get confused. So the M one plus one half is all equal to 23 divided by 29 all multiplied by M two plus one half. Fantastic. So what's happening here is, as you should see, our end goal is we're trying to get M one all by itself. So we're trying to get M one all by itself, isolate and sulfur on one side of our equation using some algebra. So in an effort to save some time, I'm gonna skip several steps and give you the most simplest form for what M one, the expression for M one is, but you'll know you've done your algebra correctly when you get M one all by itself and M one is equal to 23 multiplied by M two minus three, all divided by 29. Fantastic. So moving right along here. So our next step is we need to find the integer values for M one and M two. So in order for M one to be an integer, as we should recall, the numerator 23 multiplied by M two minus three must be, it has to be divisible by 29. So, and at this point, we need to find the smallest value of M two that makes 23 multiplied by M two minus three, a multiple of 29. So in order to do that, let us solve for 23 multiplied by M two minus three, which is E equal to or equivalent to zero. So we need to write that 23 multiplied by M two. Use our three lines here is 23 mud 29. OK. Fantastic. So we need to multiply both sides by the modular inverse of 23 mod 29 and the inverse of 23 modulo 29 is 24. Because, and let's make a little note here because 23 multiplied by 24. Is it a apart equal to one multiplied by mud 29? Fantastic. So that means we're able to easily write that M two or three lines again. Oops, I wrote four is equal to three multiplied by 24 mad 29 Fantastic. Which when we simplify this down to its most simple form, and we'll get that M two is equal to 14 multiplied by mod of 29. So thus the smallest value of M two is 14. Fantastic. So now our last step is we need to calculate the minimum thickness. So now that we know that and let's make a little note of this in blue, since it is very important that M two is equal to 14, we need to substitute this value into the equation for T for either yellow or blue light. So to keep things simple, let's use our blue light value for Lambda one F. So when we do that, we will find that T is equal to when we plug in our known variables 14 plus one half. And we need to multiply it once again by our wavelength of blue light for Lambda one F. So we need to multiply it by 328 0.57. And don't forget that it's units in its units of measurement are nanometers fantastic. And then we need to divide everything by two. Fantastic. So when we plug that into our calculator and we round to one decibel place, we will find that our final answer is approximately equal to two point four, multiplied by 10 to the power of negative six meters. So I've gone ahead and given you the an the final answer in units and meters. But you have to be very careful because when you initially plug in our expression for t into our calculator, you're gonna get an answer in units and nanometers. So you need to use dimensional analysis or quickly look it up, but you need to convert nanometers to meters. So when you convert nanometers to meters, you will get 2.4 multiplied by 10 to the power of negative 6 m. And that is our final answer. Hooray, we did it. So looking back at our multiple choice answers, the correct answer has to be the multiple choice answer letter A which once again is 2.4 multiplied by 10 to the power a negative 6 m. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

A thin oil slick (n_o = 1.50) floats on water (n_w = 1.33). When a beam of white light strikes this film at normal incidence from air, the only enhanced reflected colors are red (653 nm) and violet (392 nm). From this information, deduce the (minimum) thickness t of the oil slick.

Key Concepts

Thin Film Interference

Wavelength and Refraction Index

Constructive and Destructive Interference

Watch next

A thin oil slick (no = 1.50) floats on water (nw = 1.33). When a beam of white light strikes this film at normal incidence from air, the only enhanced reflected colors are red (653 nm) and violet (392 nm). From this information, deduce the (minimum) thickness t of the oil slick.

Key Concepts

Thin Film Interference

Wavelength and Refraction Index

Constructive and Destructive Interference

Watch next

A thin oil slick (n_o = 1.50) floats on water (n_w = 1.33). When a beam of white light strikes this film at normal incidence from air, the only enhanced reflected colors are red (653 nm) and violet (392 nm). From this information, deduce the (minimum) thickness t of the oil slick.