Suppose a pair of reading glasses found on the rack in a pharmacy has a power of 1.95 d.

10. A soap film (n = 1.33) is formed in loop of wire that is mounted vertically is not exactly uniform in thickuess because of its weight. but wedge shaped in cross section. When the film is illuminated with red light of wavelength 665 nm; One sees that . distance of 3.80 cm separates two bright fringes that are orders apart. By how much does the film differ in thickness Over this distance? Llill

Soap Film Light of wavelength $624 \mathrm{~nm}$ is incident perpendicularly on a soap film (with $n=1.33$ ) suspended in air. What are the least two thicknesses of the film for which the reflections from the film undergo fully constructive interference?

Are they locate the first red interference band? So we have bright bands that will be observed at two, and T. Is equal to M plus one half lambda. So the first bright band when M. Is equal to zero, means N. T. Is equal to lambda over four. And since we have the ratio of X one over X two is the ratio of T one over T. Two, then X two is equal to X one Landed two over λ one. We have a ramp to is equal to 680 nm. Number one is 420 nm. X one is equal to three centimeters. This gives us a value of X two of 4.86 centimetres, puppy determined the film thickness at the positions of the violet and red bands. So if it's violet T. One is equal to lambda 1/4. N. And we have n. is equal to 1.33. So this gives us a thickness of 78.9 nm. Similarly, Thickness two is going to be 128 nm for red bands and then see what is the wedge angle. So wedge angle fada is approximately equal to tangent of theta, Which is equal to T. one over X one, So this is 2.63, 10 to the negative six radiant.

Question, 67 is asking us about film of soap in a rectangular wire frame being held vertically and illuminated by reflected white light. Let me draw that out here. Now, as it's being held vertically, the soap film doesn't have a uniformed thickness. As you go down on the wire frame, the film actually gets thicker. As you get to the bottom, I'll draw a view of the soap film in Profile Toe. Better illustrate what I'm talking about. We'll call the thickness of the film T, and the distance from top to bottom on the film will be called X. Now we have on this film a violet interference band whose wavelength is 420 nanometers and its position is at three centimeters from the top of the soap film were asked a few questions about this soap film in the rectangular frame, the first of which is were asked to find out where the first Red Interference Band is. This red band has a wavelength of 680 nanometers now, because the problem is indicating bright interference bands, we're gonna have constructive interference here, so the formula for that in thin films is to anti is equal Tow am plus one half Lambda. Now n is the refractive index for our soap film. It's given to us as 1.33 T is the film's thickness. M is the order number of the fringes. And Lambda, of course, is the wavelength. Now there are going to be two expressions for constructive interference here because of the two wavelengths, and we're looking for the first bright interference band, so our value for em will be zero. In both cases, we Congar oh ahead and forget about the M in both cases and just say that are constructive interference. Expression will be to NT is equal toe one half Landau or simplifying it down further. We could just say N t is equal toe lambda over four. Now, as is made clear by that triangle I've drawn, they're the thickness of the soap. Film is proportional to the distance from the top of the frame where a fringe occurs, so the relationship between thickness and distance on the rectangular frame can be related by this expression. X one over X two is equal toe t one over T to the thickness of the film is also directly related to the wavelength of the interference bands found there. So we can also say that X one over X two is equal Toe Landau one over Lambda, too. Now we actually have values for the two wavelengths and one of the two distances, so we can just go ahead and plug those in here. And so for the remaining X, we'll call the violet wavelength and the distance of its interference banned from the top of the film, respectively, Lambda one and X one and will call the Red Interference bands Wavelength Lambda two and its position from the top of the Film X two. So let's go ahead and plug those in there. We'll have three centimeters over X two is equal Thio 420 nanometers, over 680 nanometers. If we cross, multiply and then divide, we find that three centimeters times 680 nanometers, divided by 420 nanometers is what X two is equal to. When we solve that, we find that X two has a value of 4.86 centimeters. Part B of our question wants us to find the film's thickness at each of our to interference bands. We'll start by first finding the thickness at the Violet Interference Band. Remember that our expression for constructive interference here Waas n t is equal toe lambda over four. That means that the thickness of our film at the Violet Interference Band I'm gonna call T's of Violet will be equal toe the wavelength of the Violet Interference Band divided by four n What we get when we plug in our values here will be 420 nanometers, divided by four times 1.33 the index of refraction in the soap film that equals 78.95 nanometers. Which is how thick the film is that the Violet Interference Band. Now let's see what happens when we're at the Red Interference Band. This time we're gonna be finding the thickness at the Red Band. We'll call that T's of red, and it will be equal to the wavelength of the red Band divided by four n. Once again, we'll plug in our values and see what we get. This will be 680 nanometers, divided by four times 1.33 which works out to a value of 127.82 nanometers, the thickness of the film at the Red Interference Band. Now for Part C, we're asked to determine what the wedge angle is for the soap film. Remember how we were treating the soap wedged like a triangle when we viewed it in profile? The wedge angle is gonna be this angle here, defined by the thickness. What we can do to find that angle theta is used any of our values for X and T to determine that angle so theta will be equal to the inverse tangent of X over tea. Let's use the T and X values for the Red Interference Band, so we will have the inverse tangent of 127.82 times 10 to the minus 9 m, 127.82 nanometers, divided by 4.86 times 10 to the minus 2 m, the 4.86 centimeter value of X. We'll get our answer in radiance and theta turns out to be equal toe 2.63 times 10 to the minus six radiance

So we have got a soap film over here with different even this end. Given these light undergoes our face sleep but not the light reflected in this direction. So the optical paths indifference to MTU eighties The thickness for destructive indifference is given by M Zambia the effective indexes given by 1.35 so two times and becomes 2.70 It was m dando. So from here we have t equals I m for 506 100 nanometers. So they squatted m one for 500 nanometer divided by dew 0.7 on I am too for 600 divided by 2.7. So from here we can see that the closest solution is M one equal six and M Bleak was five. So he comes out to be six. That's 500 divided by 2.7 equals 3000 divided by 2.7. It was 1111 Nadal meter or 1.1 micron. Are there any other visible way event that are missing? Ah, so far that we have to use t equals I am than, uh on two. Antique was in lambda, so to empty comes out to be. He's either 111 animator times 2.7 That's 3000. So them the equals juicing Emmick was four. We have so were then missing. Choosing M equals four. We get them that equals 7 50 which is also visible light that is missing for seven forced five and six. We already have 600. 500 on choosing M equals seven will give us using Amicable seven will give us the under ik was 3000 divided by seven. It was 428 nanometers. Now what was inserted? Sector strongest. For that, we have to use two nd equals 303,000 nanometer right? 3000 nanometer will come out to the automatic shut off half event So lambda will be 6000 nanometer, divided by odd number to M plus one. So so for M equals nine, we have 667. Nah, no mater. Uh, m equals nine if we choose. And it was 11. If we choose M equals 11. We have 5 40 five nano meter three 6000 divided by 13 with give us 461 animator for 13 And these are all the weapons in that

In this problem, the wavelength of the light in this whole film is given by linda. N is equal to linda by N. Which is equal to 550 nm by 1.32 On solving it, I get the value of claimed A N. Is equal to 4 20 nm. Now the thickness of the soap film is given by the formula to T is equal to a man named on further simplification. Either I developed is equal to m named by two. Now solving it further, I can write the express an edge. Just look at it carefully. I can write the expression 40 which is equal to two, multiplication 4 20 nm by two, which I'm simplification. I get the value of the thickness of the diet friends at 4 20 nm. So I hope you understand how I calculate the thickness of the diet fringe in this problem.

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