WEBVTT 00:00:00.085 --> 00:00:02.845 This is College Physics Answers with Shaun Dychko. 00:00:03.217 --> 00:00:07.274 We're told that the letters on a page are both three and a half millimeters high. 00:00:07.485 --> 00:00:09.097 That is the object height. 00:00:09.800 --> 00:00:12.040 We also know what the object distance is. 00:00:12.040 --> 00:00:14.777 The book is held 30 centimeters in front of the eye. 00:00:15.114 --> 00:00:19.148 And so, here's the lens of the eye and then there's the book here. 00:00:20.011 --> 00:00:22.925 That's the object distance with 30 centimeters. 00:00:23.560 --> 00:00:28.125 And, we're told also that the lens retina distance 00:00:28.314 --> 00:00:32.200 should be considered two centimeters for all the problems in this chapter. 00:00:32.520 --> 00:00:35.068 And so, we have these three pieces of information to work with. 00:00:35.862 --> 00:00:37.857 Now, we have to find out what the image height is. 00:00:38.017 --> 00:00:43.234 And so, what height will this three and a half millimeter letter on the page have 00:00:43.234 --> 00:00:44.794 when it's on the retina. 00:00:46.165 --> 00:00:49.754 So, we know that magnification is image height divided by object height, 00:00:49.874 --> 00:00:53.891 and that equals also negative of image distance divided by object distance. 00:00:53.891 --> 00:00:57.137 And, we have three of these things and we need to solve for Hi. 00:00:57.731 --> 00:01:00.097 So, we multiply both sides by Ho. 00:01:01.514 --> 00:01:03.777 So, the image height then is negative of image distance 00:01:03.937 --> 00:01:06.108 times object height divided by object distance. 00:01:07.011 --> 00:01:08.925 So, that's negative of two centimeters 00:01:08.925 --> 00:01:10.925 written as times ten to the minus two meters 00:01:11.222 --> 00:01:13.285 times three and a half times ten to the minus three meters 00:01:13.537 --> 00:01:15.982 divided by 30 times ten to the minus two meters. 00:01:16.228 --> 00:01:18.228 And, this gives this many meters, 00:01:18.440 --> 00:01:23.434 which we'll convert in to millimeters and it's negative 233 millimeters. 00:01:24.720 --> 00:01:29.125 So, this negative sign means that the letter is flipped on the retina, 00:01:29.274 --> 00:01:33.451 but then your brain is smart enough to flip it back right side up again. 00:01:35.988 --> 00:01:40.634 Now, part B is asking compare the size of the print 00:01:40.685 --> 00:01:43.405 to the size of the rods and cones in the fovea 00:01:43.405 --> 00:01:46.154 and the fovea is the center part of your eye. 00:01:46.691 --> 00:01:51.388 And, we'll see how much resolution could be possible. 00:01:52.571 --> 00:01:57.622 So, now the possible resolution will depend on 00:01:57.622 --> 00:02:03.680 how the area of a cone compares with the size of the image that's on the retina. 00:02:04.217 --> 00:02:06.342 So, we need to figure out the area of one cone. 00:02:07.057 --> 00:02:09.937 Now, the information we are given in section 26.3 00:02:10.142 --> 00:02:13.228 is that there are six million cones in the fovea 00:02:13.228 --> 00:02:15.782 and the fovea has the size of about five millimeters. 00:02:16.428 --> 00:02:22.085 So, we'll find the area of the cone is going to be the area of the fovea 00:02:22.160 --> 00:02:24.554 divided by the number of cones that are there. 00:02:25.737 --> 00:02:32.062 And, we can also say that the area of a cone is pi times its diameter squared over four. 00:02:33.022 --> 00:02:35.685 And then, we'l figure out the diameter of the cone down here 00:02:35.685 --> 00:02:37.131 and find that it's two micrometers. 00:02:37.914 --> 00:02:44.165 So, the area of the fovea then is pi times the fovea's radius squared 00:02:44.297 --> 00:02:45.634 divided by the number of cones. 00:02:46.057 --> 00:02:48.308 And, we don't know what the radius is but we know the diameter, 00:02:48.308 --> 00:02:51.691 so we'll substitute that in, diameter over two, in place of radius. 00:02:52.045 --> 00:02:54.925 This works out to pi times diameter of the fovea squared 00:02:55.125 --> 00:02:56.971 over four times the number of cones. 00:02:57.880 --> 00:03:02.960 And, the area of a cone also is pi times diameter of a cone squared over four. 00:03:03.382 --> 00:03:07.954 So, we can equate this with this because they're both equal to area of the cone. 00:03:08.131 --> 00:03:09.217 And, we did that here. 00:03:10.685 --> 00:03:16.628 And then, we can solve for Dc squared by multiplying both sides by four over pi. 00:03:18.754 --> 00:03:20.331 And then, take the square root of both sides. 00:03:21.880 --> 00:03:24.874 So, the diameter of a cone is the diameter of the fovea, 00:03:24.874 --> 00:03:26.182 which we know is five millimeters, 00:03:26.394 --> 00:03:28.982 and divided by the square root of the number of cones. 00:03:29.114 --> 00:03:30.508 So, that's the square root of six million, 00:03:30.731 --> 00:03:34.480 which is 0.002 millimeters, which is two micrometers. 00:03:35.440 --> 00:03:40.988 So, two micrometers is much smaller than the 233 micrometer image size. 00:03:41.765 --> 00:03:45.982 This 0.223 millimeters is the same as 223 micrometers. 00:03:46.777 --> 00:03:50.194 And so, because the cones are smaller than the image, 00:03:50.422 --> 00:03:53.302 that means you will have a good resolution.