Refraction and Lenses Legacy Problem #31 Guided Solution

Here's a difficult problem, both conceptually and mathematically, and like any difficult problem, you'll have to employ the habits of an effective problem solver, reading the problem carefully and identifying known and unknown quantities, and then plotting out a strategy that uses your conceptual understanding and mathematical abilities in order to get from the known information to the unknown information. The question pertains to a bug habitat that has a magnifying glass in it that magnifies the images of ants, crickets, slugs, and other objects, and the focal length of the magnifier is 22 centimeters. If you know much about a magnifying glass, you know that what it does is it magnifies the image of objects and makes virtual images, or upright images. And so F is equal to 22 centimeters, and what we're told is that the image height is 12 times the object height. I could say height divided by hole is equal to 12, and it's a positive 12 because these are upright images, and that's what magnifying glasses do. And so the high hole ratio is known to be positive 12, and the focal length is known to be 22 centimeters, and what I wish to calculate is the distance from the ants to the lens and the ant is the object, so I'm hoping to calculate the object. Now in my mind I realize that I'm probably going to have to use the lens equation, because that's the only equation I know in this unit that has a focal length value in it. So it would be something like 1 over 22 equal 1 over d object plus 1 over d image, but if I wish to solve for d object using that equation, I'm in trouble because I don't know d image. So I'm going to have to make use of this idea that the height divided by hole value is equal to positive 12. Now the height divided by hole value is also equal to the negative di divided by doe value. So I can say positive 12 equal negative di image distance divided by doe object distance, and I can multiply both sides of that equation by doe in order to get an expression for di expressed in terms of doe. That would become di equal negative 12 doe. And now with that expression of di expressed in terms of doe, I can substitute that into the lens equation in place of di. And that turns the lens equation into an equation with just one unknown in it, that unknown being doe. That would go something like this, 1 divided by 22 is equal 1 divided by doe plus 1 divided by, and here's where the substitution comes in, 1 divided by negative 12 doe. Now I can find a common denominator for that equation in order to add the two fractions on the right side together. That common denominator is 12 doe, and I can get that common denominator by taking the first fraction, 1 over doe, and multiplying numerator and denominator by 12, a legal operation in algebra. So now the equation becomes 1 over 22 equal 12 over 12 doe minus 1 over 12 doe. With a common denominator I can now group the two fractions together with one common denominator, and the numerator being simply the sum of the two numerators in the fractions. So that would become 1 over 22 equal 12 minus 1 over 12 doe. 12 minus 1 is of course 11, so I have 1 over 22 equal 11 over 12 doe, and now I can cross multiply. I have 12 doe equal 11 times 22, and I can evaluate for doe by dividing through by 12, and I get 20.1666, and I can round that to the second significant digit, such as 20 decimal place centimeters.