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A beam of 541-nm light is incident on a diffraction grating that has 400 grooves/mm. (a) Determine the angle of the second-order ray. (b) What If? If the entire apparatus is immersed in water, what is the new second-order angle of diffraction? (c) Show that the two diffracted rays of parts (a) and (b) are related through the law of refraction.

a. $25.6^{\circ}$

b $18.99^{\circ}$

c. $\frac{\sin \theta_{a}}{\sin \theta_{b}}=n$

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Cornell University

Rutgers, The State University of New Jersey

Numerade Educator

McMaster University

So in this problem, the web link of beam being used is 541 nanometer, and the diffraction grating has 400 groups per millimeter, which is gonna be golden four times 10 35 groups. Perimeter. Therefore, the spacing between two groups would be de course water over an and that's gonna give us 2.5 times 10 to the negative six meter. Or you can also see a 2.5 micrometers now, in part, A of the problem in part A. We want to find data for chemicals to the diffraction angle. There are for second order diffraction. So we're gonna use decided cycles and Lambda, but M equals two. So I'm going to write down to Lambda from here. Data for M equals two is going to be called to sign in verse. Sign in verse twice. Lambda over the and this is sign in bursts two times 5 41 times 10 to the negative nine over DE, which is 2.5 times 10 to the negative six. So this is going to give us 25.6 degrees. All right, 25.6 degrees now in part b of the problem. If the entire practice was immersed in water, then the web link off beam in water would chance to different weapons. So if we emerged the entire practice in water and and let's say Lambda W. Is the web went off the beam in water, then it would be called to Lambda over. They refractive index off water. So if you guys know the refractive index, it is 1.33 for the water. So from here, the web linked in water would be lambda W E course Lambda, which is 5 41 times 10 to the negative nine over 1.33 And this is going to give us four points. There are six times 10 to the negative seven meter. Now, in this case, the second order diffraction angle would be Let's call, they had a frame for that one. I can't even call this data to. And the previous one s sorry, the previous one instead or two. And this one s stated to prime in this case that there are two prime would be sine inverse twice. We're blinking water over spacing, and this is going to give us 18.9 degrees. All right, Now, in part C of the problem to show that to diffraction did raise in Parts A and B are related through the law for a fraction. We know that designed ETA two equals two Lambda, right? The sign there are two because to lambda, this for corresponds to the second order to fraction. And similarly, the sign there are two prime it course to Lambda Prime, or you can just write down them That w Now I want to make sure you guys don't get confused. So I'm gonna add here as well, too. Okay, so that that's more clear over here. So from here, if you divide the first relation by the second relation, then we're gonna get scientists to over signed data to prime equals Lambda over Lambda Omega Right now, Lambda over Lambda Water was remember lambda by eat a water so eat eyes. They're effectively next off water. So this was Lambda over. Refractive index off water. Now, these two terms cancel out here and we actually get refractive index of water here. So if you rearrange this, you're going to see that sign there too, because refractive Index off water times sign. There are two prime. Or you can also say one time sign. They had a two. You cause refractive index off water times signed their to prime. And you can see that this is the snail slot. So that proves the last refraction.

University of Wisconsin - Milwaukee