WEBVTT 00:00:00.111 --> 00:00:02.511 This is College Physics Answers with Shaun Dychko. 00:00:03.511 --> 00:00:06.400 We have a thin lens with a focal length of 100 millimeters 00:00:06.933 --> 00:00:10.444 and our object is placed at 103 millimeters from the lens. 00:00:11.244 --> 00:00:12.333 And the question is 00:00:12.622 --> 00:00:13.911 where will the image be? 00:00:14.244 --> 00:00:17.422 Because that's where one would place the projection screen 00:00:18.933 --> 00:00:21.333 and it's going to be an inverted 00:00:21.711 --> 00:00:23.688 and magnified image. 00:00:24.600 --> 00:00:26.955 So in other words the question is what is di? 00:00:27.288 --> 00:00:30.311 So we use the thin lens equation and we're going to solve it for di 00:00:31.244 --> 00:00:34.600 so we'll subtract one over d naught from both sides. 00:00:35.644 --> 00:00:38.444 And we have one over image distance is one over 00:00:38.577 --> 00:00:41.066 focal length minus one over object distance 00:00:41.800 --> 00:00:44.511 and then we'll take the reciprocal of both sides 00:00:47.000 --> 00:00:49.377 so the image distance then is one over the focal length 00:00:49.400 --> 00:00:50.888 minus one over the object distance 00:00:51.177 --> 00:00:53.377 and then take that difference to the power negative one. 00:00:54.088 --> 00:00:57.177 So that's one over 100 millimeters minus one over 103 millimeters 00:00:57.444 --> 00:00:58.422 all to the negative one 00:00:58.888 --> 00:01:01.066 which gives 3433 00:01:01.060 --> 00:01:04.533 and make sure you notice that since I plugged in millimeters 00:01:04.530 --> 00:01:08.000 into the equation here our answer is also in millimeters 00:01:08.577 --> 00:01:11.444 and we'll multiply that by one meter for every thousand millimeters 00:01:11.800 --> 00:01:13.866 and get 3.43 meters. 00:01:15.044 --> 00:01:17.222 So that's the distance that the screen should be from the lens 00:01:17.220 --> 00:01:18.288 on the other side 00:01:18.466 --> 00:01:19.377 compared to the object. 00:01:21.370 --> 00:01:25.888 Now we want to figure out what the dimensions of the image will be 00:01:26.888 --> 00:01:28.755 given the known dimensions of the object. 00:01:29.533 --> 00:01:31.311 So magnification which is the 00:01:31.488 --> 00:01:33.200 image height divided by the object height 00:01:33.488 --> 00:01:35.688 we know is the negative of the image distance 00:01:35.750 --> 00:01:36.955 divided by the object distance 00:01:37.333 --> 00:01:38.844 and so we'll find the image height 00:01:38.840 --> 00:01:40.555 by multiplying both sides 00:01:40.550 --> 00:01:42.000 by the object height that we know 00:01:43.555 --> 00:01:45.600 and then we solve for hi 00:01:46.066 --> 00:01:48.688 and it’s object height times negative image distance 00:01:48.680 --> 00:01:49.555 divided by object distance. 00:01:50.000 --> 00:01:52.555 And we know this image distance and object distance. 00:01:54.733 --> 00:01:57.444 So that's negative 3.433 meters 00:01:57.733 --> 00:02:00.266 divided by 103 times ten to the minus three meters. 00:02:00.422 --> 00:02:01.600 This is the object distance. 00:02:02.222 --> 00:02:06.577 And this gives a height in the image of negative 33.33 00:02:06.822 --> 00:02:09.022 multiplied by whatever the object dimension is. 00:02:10.266 --> 00:02:12.044 So we’ve found the magnification in other words 00:02:12.733 --> 00:02:15.733 and then we use that to figure out the length and width of the image. 00:02:17.288 --> 00:02:20.866 Now we don't really care about the negative sign here, 00:02:20.860 --> 00:02:21.866 the fact that it's inverted, 00:02:23.066 --> 00:02:25.533 so it does take the absolute value of that magnification. 00:02:26.155 --> 00:02:28.933 So the length of the image is going to be 33.33 times 00:02:28.930 --> 00:02:30.066 24 millimeters. 00:02:30.777 --> 00:02:32.666 And that's going to work out to this many millimeters 00:02:32.660 --> 00:02:35.444 which we multiply by one meter for every thousand millimeters 00:02:36.488 --> 00:02:38.222 giving length that 0.8 meters. 00:02:39.822 --> 00:02:43.466 And then the width will be 33.33 times 36 millimeters 00:02:43.866 --> 00:02:47.177 giving an image width of 1.20 meters.