Like

Report

A spacer is cut from a playing card of thickness $2.90 \times 10^{-4} \mathrm{m}$ and used to separate one end of two rectangular, optically flat, $3.00-\mathrm{cm}$ long glass plates with $n=1.55$ , as in Figure P 24.24 Laser light at 594 nm shines straight down on the top plate. The plates have a length of 3.00 cm. (a) Count the number of phase reversals for the interfering waves. (b) Calculate the separation between dark interference bands observed on the top plate.

a) 976

b) $3.07 \times 10^{-5} \mathrm{~m}$

You must be signed in to discuss.

University of Michigan - Ann Arbor

Numerade Educator

University of Washington

McMaster University

For part of our question, we are asked to count the number of phase reversals for the interfering waves when the index of refraction and is equal to 1.55 The spacer is cut from playing cards of thickness T equal to 2.9 10% of minus four. Um and they're separated by distant L equal to three centimeters or 30.3 meters and then the wavelength that were considering here from the laser light is lambda equal to 594 nano meters or 594 times 10 to the minus nine meters. So just be sure that whenever you're using, um, these different values that you use the same unit, so you don't want to use something that has meters and something else that has nano meters. So before you, you know, add, multiply, divide anything like that, you need to convert 594 nano meters into 594 times 10 to the minus nine meters. So the condition for destructive interference for part A is, uh, two. Timed in the index of refraction times, the thickness is equal to him times lambda or in other words we could just solve for in and M is equal to two times in times T divided by lambda and remember, convert lambda and two meters from nano meters, and you find that this is equal to approximately 976. And it has to be a whole number here because you're asking for the number of phase reversals. So that's 5 976 is the number of days reversals, and that could be boxing is our solution for a for part B, it says, count the number of phase reversals for the excuse me for part B s is calculate the separation distance between dark interference bands observed on the top plate. So I drew out a little triangle here to represent this for Ella's. The total length T here is that Thickness X is the distance to that, uh, point where it's that big Misty and were asked to find D. And so I made different Triangle Syria Triangle, ABC, right? And you also have triangle a Q P. Okay, so from the destructive interference, the thickness of the film ah is t, of course, and it's equal to in times Lambda over, too. It's one thing to keep in mind, and the X coordinate from the empath inthe band of the dark fringe is X, and that's equal to mm times a think misty. So when N. Is equal to one, X is just equal that they just teeth. Okay? And now we're gonna use the properties of triangles, um, specifically the property of the similar Triangle ABC, which is like a p Q. So therefore, from using property of triangles, we have be time see, divided by a time, see is equal to p. Q. Divided by a P. Those ratios are equivalent, but BC is just equal to think misty right? That's obvious from there, and a C is equal to the distance. X p Q is equal to D, and then a p is equal to l. So we can solve that for D. And we find D is equal to l Times T divided by X. So this is equal to L. And then we said T is equal to in land over to So this is M times lambda, divided by two and then x is m times that they misty. So the EMS cancel here left with El Lambda divided by two t. So plugging in the values for L Lambda too, and the think misty, Remember that land that needs to be converted two meters. Everything needs to be in meters. We find that this is equal to three 0.7 times 10 to the minus five meters, and this can be boxed in as our solution for part B.

University of Kansas