Question
What is the distance between lines on a diffraction grating that produces a second-order maximum for $760-mathrm{nm}$ red light at an angle of $60.0^{circ}$ ?
What is the distance between lines on a diffraction grating that produces a second-order maximum for $760-mathrm{nm}$ red light at an angle of $60.0^{circ}$ ?
Answers
If a diffraction grating produces a third-order bright spot for red light (of wavelength 700 nm) at 65.0$^\circ$ from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength 400 nm)?
Oh, righty. So for this problem, we know the wavelength of our first situation to 700 Nana meters and that the third bright spot happens at 65 degrees and we want to know for that same grading for a way of length of 490 meters. At what angle? This is the second bright spot gonna be So we can use this information from our 1st 1 to figure out what the spacing between the sleds and our diffraction grating is using D signed data equals and lambda. So if we saw for D, you get t equals and Landa over sign data. So for this situation, if we pull in our numbers, this is the third border bright spot or a wavelength of 700 nanometers. Sign of 65 degrees gives us a slip spacing of 23 17.1 mana meters. Now I can do the same thing up here. This guy so decide Data sequel to End Landa. Now we're looking for Seita so sign data equals n lambda over Deep said Seita is the inverse sine m lambda over D. So if I play and my numbers now, we're dealing with the second spot and the new wave length of 400 nanometers and the slit spacing that we just solved for 23 17.1 indiana meters. Make sure your units are the same. Gives me, uh, they two of 20.22 groups.
For this question, we have light with the wavelength. Lambda equals 589 times 10 to the minus nine meters for 589 millimeters and it falls on a diffraction grating and it has a diffraction angle Fatal one of 16.5 degrees. I called this data one because this is the first order where M is equal to one and wants to know how far apart the slits are. And at what angle will the third order be observed? So the first thing you do is figure out the distance between the slits. This is because the equation is d time to sign of the angle Fada is equal to in Lambda. So if we can find D and we know him and Lambda, then we can solve for theta. So the first thing we're gonna do is solve for DC because that chemicals one, um, we know everything in this equation, so D then would be equal to one times lambda, which is just Landa divided by the sine of the angle, uh, of the first maximum, which is data one as written above, complaining those values into this expression, we find the distance between the slits is too 0.7 times 10 to the minus six meters. Which weaken boxing is one part of our solution. Now that we know that we can find the angle when M is equal to three, so we'll start a new page for this. But what we have here is that again using this equation here that I'm pointing to it the arrow we can solve for when M is equal to three. We have beta three is equal to the inverse sine. You have to take the inverse sine of both sides to isolate the angle. So I'd be the inverse sine of, uh, M lambda, where m is equal to three. So this is three times Lambda divided by D. So we know decent. We just saw for it. So plugging those values in and taking the inverse sine of that ratio, we find that this is equal to 58 0.6 degrees. Tweaking boxing is the second part of our solution to this question.
Hello, everyone. This is problem. 63 from Chapter 28 says a diffraction grating has 2500 lines per centimeter. Was the angle between the first quarter maximum for red light on the first order of maximum for blue light. And it gives us the wavelength for both of them. Okay, so our strategy for um for this problem is, of course, to use our diffraction are normal diffraction grating equation decide if theta is equal to em Lambda. But what I actually want to do is find the difference between, uh, the first order maximum angle for red light and the 1st 1 of maximum for blue light. So, um, I'll have two separately. Calculate status of R and theta. So be thing gulf red light in the angle for blue light and then just subtract them. So of course, just rearranging this equation, we could we see that we're just going to have m limb. They are for D, but its arc sine m lambda B over D. Okay, so these are just the wavelengths for their respective colors. And of course, our EMS here are both one and our desire the same. Um, but of course we have 2500 lines per centimeter. So this means that D is if I want to put it into meters 0.1 meter divided by 2500. Okay, so often times these problems involve actually inverting what they give us because it's quite natural for them to give us a number of lines per distance. And what we need is the distance in between the lines. It's often we just have to invert so we can simply carry forth this calculation because now we have d and we have Linda and Linda be. We know the chemicals one, and if we do all that, we get approximately 3.9 degrees.
The grating his 2000 lines person to meet you. So from here we can find a D, which is a working into meters which will become sentimental becomes zero point zero Wonder of the room eaters divided by 2000. This will give us the slip distance five times 10 to the power minus six reaches. Then from here we can find and they therefore the first older maximum lost order next month. Um so theta will be cool too Signed in Worse off Landau divided by D for first or the maximum Substituting the values for Linda we have here is five times dental power minus six meters divided by slipped distance. We have is five times. Sorry, this is minus Linda. So this is a 5.2 old times 10 to the power minus seven meters. This is Linda. So where is the DEA's five times sent about minus six. So simplifying we get a better will be cool too. 5197 degrees