WEBVTT 00:00:00.091 --> 00:00:02.737 This is College Physics Answers with Shaun Dychko. 00:00:03.171 --> 00:00:05.102 With the telescope on Earth here, 00:00:05.314 --> 00:00:08.028 we want to be able to distinguish between two objects 00:00:08.028 --> 00:00:11.331 that are separated by five kilometers over here on the moon 00:00:11.800 --> 00:00:15.000 and the moon is 384,000 kilometers away. 00:00:16.085 --> 00:00:18.577 And so what is this angle here. 00:00:19.085 --> 00:00:22.794 And we're going to use the formula for the really criterion 00:00:23.028 --> 00:00:28.651 and knowing the angle and the wavelength which we're told is about 550 nanometres. 00:00:28.965 --> 00:00:32.480 Figure out what minimum diameter of telescope you need 00:00:32.480 --> 00:00:35.805 in order to be able to resolve those objects 00:00:35.862 --> 00:00:37.485 separated by five kilometers on the moon. 00:00:38.777 --> 00:00:42.657 So theta is approximately going to be this separation divided by x 00:00:42.840 --> 00:00:46.251 and we can take the use this approximation because the angle is really small. 00:00:46.622 --> 00:00:49.857 x is really large compared to delta y in other words. 00:00:50.405 --> 00:00:53.565 So five kilometers divided by 384,000 kilometers 00:00:53.565 --> 00:00:57.605 is 1.302083 times ten to the minus five radians. 00:00:58.331 --> 00:01:02.034 If you don't like this approximation you could instead turn this into 00:01:02.034 --> 00:01:05.834 two right triangles then take the inverse tangent of this 00:01:05.834 --> 00:01:08.771 which would be half of delta y divided by x 00:01:09.371 --> 00:01:12.680 and then multiply that by two 00:01:12.862 --> 00:01:16.731 and then you would get the same answer to this many significant figures. 00:01:19.620 --> 00:01:22.994 So we'll rearrange this really criterion formula 00:01:22.994 --> 00:01:25.489 by multiplying both sides by d over theta 00:01:27.257 --> 00:01:30.091 to solve for the diameter of the telescope that will need. 00:01:30.770 --> 00:01:34.554 So it’s 1.22 times the wavelength over the angle in radians. 00:01:34.925 --> 00:01:38.622 So it's 1.22 times 550 times ten to the minus nine meters wavelength 00:01:38.977 --> 00:01:42.977 divided by 1.30203 times ten to the minus five radians 00:01:43.457 --> 00:01:48.862 giving a diameter of 5.15 centimeters is the minimum needed in our telescope.