Refraction and Lenses Legacy Problem #28 Guided Solution

Here's a very difficult problem, both conceptually and mathematically, and one in which you're going to have to employ the habits of an effective problem solver. You're going to have to read the problem very carefully, identifying the known and unknown quantities, and then using all your conceptual and mathematical knowledge in order to plot out a strategy as to how to get from the known to the unknown information. We read about an image being produced that is a real image magnified by a factor of five. It's created by a lens that has a focal length of 12.0 centimeters. And so what we know is that f is equal to 12.0 centimeters. What we're looking for is the object. Now, in order to solve this problem, you need to really dissect that first sentence, which says that it's a real image magnified by a factor of five. That means that the high to whole ratio is either plus five or minus five. And the big part of the problem is that you have to decide, is it plus or minus? So high per whole magnification is going to be negative five here, because the image is real. And real images are inverted images. That is, they have negative heights for a positive object type. And so we say high per whole equal negative five, and the high per whole ratio is equal to the negative die per doe ratio. So if I know high per whole, I also know the die per doe ratio. High per whole is equal to negative five, and die divided by doe, the image distance divided by object distance, is equal to the negative of negative five or is equal to five. So I write down two things, that die divided by doe equal positive five, and that f equal 12.0 centimeters. And I'm looking for the object distance. Now the only equation that I have that has the focal length in it is the Lenz equation, one over f equal one over doe plus one over die. Now I don't know die. I'm looking for doe. It would help if I knew die, I could calculate it quite easily. But what I do know about die is I know that the die divided by doe is equal to five. Now if I multiply both sides of that statement by d o, I would have d i equal five times d o. And that's an expression for d i expressed in terms of d o. And I can substitute that into my Lenz equation in order to solve for the object distance. That's my strategy. Here's how it goes. I'm going to write the Lenz equation, one over twelve equal one divided by doe plus one divided by die. But instead of putting die in there, I'm going to put five doe in the denominator. The right side of the equation can be simplified to five over five doe plus one over five doe. I took the first term and just multiplied numerator and denominator by five. That's a legal operation in algebra. Now what I have is a common denominator of five doe on the right side of the equation. So I'm going to group my two numerators over the same common denominator. That's five plus one over five doe. Or six over five doe. And all of that is equal to one over twelve. Now I cross multiply. Five doe is equal to six times twelve or seventy-two. And I divide through by five and I get my answer, fourteen point four centimeters.