Refraction and Lenses Legacy Problem #27 Guided Solution

You�ve probably done this before. You�ve got your friends together and you want to snap a picture and you�re trying to get them all close together so that you can fit their image up onto this picture. There�s a good deal of physics in that moment and it�s the physics associated with this problem. We�re told the camera lens is a 51.4mm camera lens. That�s describing the focal length of this converging lens on the camera. So F is equal to 51.4mm. And we�re told that we�re trying to fit the image of all these friends onto a film negative that�s 35.8mm wide. So that�s the dimension of the image. We�ll just call that HI even though it�s really not a height but a width. So HI is equal to 35.8mm. And we�re told that Jack and the friends are located 4.62m from the camera lens. That�s an object distance. DO equals 4.62. So I know three bits of information here. What I�ve asked to calculate is what�s the maximum distance that they must squeeze together in order to completely fit all of them in the developed picture. They�re asking me to find an object dimension. We�ll call it HO even though again it�s not a height but a width. So HO is my unknown. Now in order to find HO I�ll have to use the magnification equation. And I�ll have to know HI and DI and DO. And once I do know those three things I can say HI over O equals negative DI over DO and rearrange it to solve for hole. But I don�t know the value of DI. So the first step is to use the lens equation in order to calculate the value of DI. The focal length is known, 51.4mm. And the object distance is known and it�s 4.62m. Now you�ll notice right away that these are two different units. So part of my task involves making sure that I have consistent units between the DO value and the F value. So I�m just going to take the 4.62m for DO and I�m going to convert it to mm, 4,620mm. And then I�m going to substitute into the lens equation and rearrange to solve for DI. 1 over DI becomes equal to 1 over the 51.4mm minus 1 over the 4.620mm. Evaluating for DI and solving for DI I get 51.9783mm. Now that I know DI along with DO and along with HOE I can calculate HOE. I take the magnification equation, HI over HOE is equal to negative DI over DO and I rearrange it. It becomes HOE is equal to negative HI times DO divided by DI. My HI value is 35.8mm. My DO is 4,620mm. And my DI I just calculated is 51.9783mm. When I calculate HOE it comes out to be 31.82.02mm and I can round that to three significant digits such that it�s 3.18m or 3,180mm.