WEBVTT 00:00:00.111 --> 00:00:02.466 This is College Physics Answers with Shaun Dychko. 00:00:03.488 --> 00:00:06.600 We have a diffraction grating with 30,000 lines per centimeter. 00:00:07.111 --> 00:00:09.822 And let's find out the separation between each line. 00:00:10.111 --> 00:00:11.777 So we'll take the reciprocal of that and say 00:00:11.977 --> 00:00:13.955 one centimeter for every 30,000 lines 00:00:14.177 --> 00:00:17.066 and we'll write centimeters as one times ten to the minus two meters. 00:00:17.533 --> 00:00:21.177 This gives us 3.33 times ten to the minus seven meters between each line. 00:00:22.066 --> 00:00:24.533 So with that information we can answer a question a 00:00:24.800 --> 00:00:26.400 which says show that 00:00:26.400 --> 00:00:28.911 none of the wavelengths in the visible spectrum 00:00:29.288 --> 00:00:31.711 will appear in the diffraction pattern. 00:00:33.333 --> 00:00:37.111 So let's hypothetically say that it will appear 00:00:37.110 --> 00:00:40.844 and solve for the angle and we'll divide both sides by d 00:00:41.133 --> 00:00:43.488 and then take the inverse sine of both sides and we get that the angle 00:00:43.660 --> 00:00:45.755 to a maximum is the inverse sine of the order 00:00:45.933 --> 00:00:47.511 times a wavelength divided by d. 00:00:48.466 --> 00:00:54.133 So we want to minimize this number here 00:00:54.130 --> 00:00:55.644 in order to take the inverse sine of it. 00:00:56.066 --> 00:00:58.955 Because what we're trying to avoid is the possibility 00:00:59.311 --> 00:01:00.666 that this number is greater than one 00:01:01.266 --> 00:01:04.066 because the inverse sine is something greater than one is undefined 00:01:04.377 --> 00:01:07.422 because the sine graph is like this. 00:01:08.866 --> 00:01:16.177 And the range of the sine graph is between positive one and negative one, 00:01:16.644 --> 00:01:17.977 then when you take the inverse sine 00:01:18.377 --> 00:01:20.488 of a number you're finding out what is the x value 00:01:20.480 --> 00:01:22.400 corresponding to that particular sine value. 00:01:22.800 --> 00:01:25.133 But if this the sine is greater than one 00:01:25.577 --> 00:01:28.088 there is no x value that would correspond to it. 00:01:32.044 --> 00:01:37.177 So we're gonna take the inverse sine of one times the smallest 00:01:38.400 --> 00:01:42.066 wavelength we can put in here because we want to minimize this value 00:01:42.688 --> 00:01:44.755 and prevent it from going over one. 00:01:45.311 --> 00:01:47.955 So we put in 380 times ten to the minus nine meters. 00:01:47.950 --> 00:01:50.333 That's the wavelength of Violet, 00:01:50.888 --> 00:01:54.400 the minimum wavelength in the visible spectrum. 00:01:55.000 --> 00:01:57.733 And we divide by the separation between lines 00:01:57.730 --> 00:01:59.600 and the diffraction grating of 3.33 00:01:59.688 --> 00:02:00.822 times ten to the minus seven meters. 00:02:01.311 --> 00:02:03.444 But we find that nevertheless this is undefined 00:02:03.688 --> 00:02:06.466 because this is a number greater than one. 00:02:08.400 --> 00:02:10.200 So there is no angle 00:02:11.355 --> 00:02:15.466 in the diffraction pattern for wavelengths that are in the visible spectrum. 00:02:17.177 --> 00:02:20.222 Then part b asks us for 00:02:23.311 --> 00:02:28.200 what wavelength would have a first order maximum 00:02:31.422 --> 00:02:32.466 for this diffraction grating. 00:02:32.933 --> 00:02:37.222 So we're solving this formula now for lambda by dividing both sides bym 00:02:37.755 --> 00:02:39.177 and we get lambda is separation 00:02:39.170 --> 00:02:41.977 between lines in the diffraction grating times sine over m 00:02:43.022 --> 00:02:45.844 and we choose theta to be 90 degrees 00:02:46.222 --> 00:02:52.977 because that's going to give us the maximum possible wavelength 00:02:53.066 --> 00:02:55.422 that will appear in the diffraction pattern 00:02:56.666 --> 00:02:58.577 because when you have, 00:03:02.777 --> 00:03:04.911 let's just consider two slits here for a second 00:03:05.266 --> 00:03:06.355 and you have a screen here 00:03:06.755 --> 00:03:10.088 and you have light going through these mini, mini slits in the diffraction grating 00:03:10.444 --> 00:03:12.888 and you're going to have a maximum appearing at 00:03:14.266 --> 00:03:17.977 the furthest possible angle being 90 degrees way up here. 00:03:19.600 --> 00:03:22.244 And this will maximize lambda 00:03:22.240 --> 00:03:26.888 because sine of 90 is the maximum sine you can possibly have which is one. 00:03:28.400 --> 00:03:30.955 This works out to 333 nanometers. 00:03:32.533 --> 00:03:36.866 So that's the maximum possible wavelength that'll show in the pattern 00:03:37.222 --> 00:03:41.955 and that is nevertheless shorter than the visible spectrum. 00:03:43.933 --> 00:03:49.044 So part c is asking us for the 00:03:49.800 --> 00:03:53.888 maximum number of lines on the diffraction grating. 00:03:54.044 --> 00:04:01.533 That would show a full second order maximum. 00:04:02.022 --> 00:04:05.466 So having a full second order maximum means that 00:04:05.844 --> 00:04:09.088 every wavelength in the visible spectrum will show up. 00:04:09.660 --> 00:04:12.844 And so we're going to choose lambda to be the longest 00:04:14.333 --> 00:04:16.355 wavelength of 700 nanometers. 00:04:17.933 --> 00:04:19.177 So the order is two. 00:04:19.466 --> 00:04:21.866 And so we have two times 700 nanometers written as 00:04:22.000 --> 00:04:23.266 times ten to the minus nine meters. 00:04:23.577 --> 00:04:25.400 And we divide this by sine 90 00:04:25.666 --> 00:04:27.355 And we divide this by sine 90 because we're looking for just this threshold 00:04:27.350 --> 00:04:29.822 when it just barely shows up in the interference pattern. 00:04:31.022 --> 00:04:33.955 So that's 1.4 times ten to the minus six meters per line. 00:04:34.333 --> 00:04:36.488 And it asks us for the lines per centimeter. 00:04:36.911 --> 00:04:39.200 So we take the reciprocal that 00:04:39.200 --> 00:04:42.266 and say there's one line for every 1.4 times ten to the minus six meters 00:04:42.777 --> 00:04:45.022 and change the units into centimeters. 00:04:45.555 --> 00:04:49.733 And then that ends up being 7142 lines per centimeter.