Refraction and Lenses Legacy Problem #23 Guided Solution

A good problem solver uses effective habits in order to approach every problem. In this question, what I need to do is read the problem carefully and try to develop a mental picture of what's going on, identifying the known and the unknown quantities and expressing them in terms of the variables of commonly used physics equations. So I read of a screen, the largest screen in the world, at least as of this writing, which is 30.6 meters wide. And images from a 35 millimeter wide film are projected onto the screen. So those first two numbers that I have have to do with an object dimension and an image dimension. Now the images are projected onto the screen, and so this 30.6 meters is an image dimension. We could call it the image height, hi. So I write down hi equal 30.6 meters, and the 35 millimeters is what I would call the object dimension. I could call it ho. Ho equals 35 millimeters. Now right away I notice a problem. The problem is that I have meters for image distance or image height, and I have millimeters for object height. So I'm going to do a quick rearrangement before I go further on this problem. I'm just going to say that the ho equals 35 millimeters can also be stated as ho equal 0.035 meters. Now suppose that the screen of the theater is located at a distance of 46 meters from the projector. That sentence tells me that the 46 meters is the distance that the image is from the projector lens. After all, that's where the images show up on the screen, and it's 46 meters from the projector. So that is what we call di. Di equal 46 meters. Now what I'm to do here is to determine the magnification of the image and the focal length of the lens. To do so, I really have to use my thinking cap. Now when I approach this idea of the focal length, the second part of the question, I know right away that what I need to know is d object and d image. I don't have the d object distance, but I can calculate that because after all, the magnification, the first part of the question, is the ratio of hi to ho, and that's also equal to the ratio of negative di to do. So knowing hi, ho, and di, I should be able to calculate do, and then in the second part of the question, I should be able to find the focal length using the lens equation. So that's my strategy. Now there's one more little issue that's going to become important here, and it doesn't come out quite obviously in this problem, and that's the issue that when you have a projected image onto a screen, the image actually gets inverted. I know when you look on the screen, it looks right side up, and that's simply because the object itself is upside down, the film strip itself is upside down, so there's this inversion that takes place. So what I'm going to have to do is call one of my ho's or hi's negative in this case. Usually I just call ho's positive and hi's negative for an inverted image. So when I go to calculate the magnification, I'm going to go the 30.6 meters, and I'm just going to call it the negative 30.6, and I'm going to divide it by .035 meters for the object dimension, and that gives me the magnification, and it comes out to be negative 874.2857, and I can simply round that to negative 870. Now I'm going to use this magnification value to calculate the value of do using di equal to 46 meters, and so I say this magnification of negative 874.2857 is equal to the negative di over do, which is the negative 46 over do. I can rearrange that to solve for do when I get do equal m over negative di, and when I plug in my negative 874.2857 for m, and 46 meters for my di, and I include the negative sign, I get a do equal .052614, which is a very small number, about 52.6 millimeters. And now what I can do is take this .052614 and this 46 meters, and I can substitute that into the Lenz equation. I say 1 over f is equal to 1 over the do of .052614 divided, or plus 1 over the 46, and I find that 1 over f evaluates to 19.0280. Now I take the reciprocal of that number and I get f, and f comes out to be .052550 meters, and I can do a couple of things with that. First of all, I can round that to the second significant digit, so instead it becomes .053 meters, and if I wish I can express it in units of millimeters, it's 53 millimeters. And that's how you approach this problem.