WEBVTT 00:00:00.088 --> 00:00:02.422 This is College Physics Answers with Shaun Dychko. 00:00:03.444 --> 00:00:06.777 The acceptance angle for the objective lens on a microscope 00:00:07.200 --> 00:00:11.022 is the angle at the tip of this cone of light that is entering the lens 00:00:11.733 --> 00:00:14.822 and the bigger this angle is, 00:00:15.111 --> 00:00:18.177 the bigger the cone of light enter in the lens and the better the resolution 00:00:19.533 --> 00:00:21.400 in terms of seeing detail. 00:00:22.000 --> 00:00:24.133 And now, the smaller the angle, 00:00:24.422 --> 00:00:30.933 the better at locating a specific position in the sample. 00:00:31.488 --> 00:00:34.866 So, we're going to find the acceptance angle for these two different lenses 00:00:35.911 --> 00:00:40.022 and we have a 0.10 numerical aperture lens 00:00:40.020 --> 00:00:41.933 and then a 0.65 numerical aperture lens. 00:00:42.244 --> 00:00:45.444 We're assuming that they're both in air, since it doesn't tell us otherwise, 00:00:45.866 --> 00:00:48.066 in which case our index of refraction is one. 00:00:49.644 --> 00:00:51.888 But first, let's find our formula for the acceptance angle. 00:00:52.377 --> 00:00:53.955 So, we're given that the numerical aperture 00:00:53.950 --> 00:00:55.822 is the index of refraction of the medium 00:00:55.820 --> 00:00:57.022 through which the light is traveling, 00:00:57.020 --> 00:00:58.111 which is air in this case 00:00:58.377 --> 00:01:00.110 times sine of alpha, 00:01:00.110 --> 00:01:02.533 which we're told is half of the acceptance angle. 00:01:03.222 --> 00:01:05.800 And so, we'll divide both sides by the index of refraction N 00:01:06.155 --> 00:01:08.422 and we get sine of half the acceptance angle 00:01:08.420 --> 00:01:10.511 is numerical aperture divided by index of refraction 00:01:10.888 --> 00:01:12.355 and then take the inverse sine of both sides 00:01:12.488 --> 00:01:14.822 to get Theta over two is inverse sine of that. 00:01:15.977 --> 00:01:17.377 Then, multiply both sides by two 00:01:18.044 --> 00:01:20.822 and you end up with the acceptance angle Theta is 00:01:21.040 --> 00:01:22.800 two times inverse sine of numerical aperture 00:01:22.800 --> 00:01:24.377 divided by index of refraction. 00:01:26.044 --> 00:01:27.333 So, in the first case, 00:01:27.330 --> 00:01:28.800 we have the acceptance angle is two 00:01:28.800 --> 00:01:31.333 times the inverse sine of 0.1 numerical aperture 00:01:31.644 --> 00:01:33.822 divided by the index of refraction of air which is one, 00:01:34.244 --> 00:01:36.466 giving us an acceptance angle of 11 degrees. 00:01:38.044 --> 00:01:38.933 And then in the second case, 00:01:39.400 --> 00:01:42.533 he acceptance angle is two times inverse sine of 0.65 00:01:42.666 --> 00:01:45.177 numerical aperture divided by one, which is 81 degrees. 00:01:46.511 --> 00:01:51.800 So, we should first use the 0.1 numerical aperture lens 00:01:51.800 --> 00:01:53.755 to locate the object 00:01:53.750 --> 00:01:56.577 because it has a smaller acceptance angle, 00:01:56.570 --> 00:02:00.111 and so it sees only a small piece of the sample at a time 00:02:00.355 --> 00:02:05.888 and so for that reason, it's better to be able to identify the place 00:02:05.880 --> 00:02:08.044 where the target is. 00:02:08.533 --> 00:02:09.666 And then after finding it, 00:02:10.600 --> 00:02:12.066 you would switch to this lens 00:02:12.311 --> 00:02:15.333 to get a higher resolution view of it 00:02:15.330 --> 00:02:16.977 since the acceptance angle is larger, 00:02:18.200 --> 00:02:19.666 it's getting a larger cone of light. 00:02:19.911 --> 00:02:23.488 And with that, it can create a higher resolution image to see more detail.