Refraction and Lenses Legacy Problem #32 Guided Solution

An effective problem solver reads a problem carefully, identifying the known and unknown information, and then plots out a strategy as to how to get from the known to the unknown quantities. Here in this question, we read about an inverted image which is produced, and its size is one-fifth the size of the object. I can take that first sentence and restate it in equation form as the h-i image height is one-fifth times h-o. But I must give attention to the fact that the h-i, or image height, is an inverted height. And being an inverted height, it has a negative image height. And so I have to take that little statement, h-i equal one-fifth h-o, and insert a negative sign in front of it, because the image height is negative for an inverted image. I'm also told that the focal length equal 22.8 centimeters, f equal 22.8 centimeters. And what I'm looking for is the object distance, d object. In my mind, I tell myself that I'm going to have to use the Lenz equation to calculate object distance. That Lenz equation stated as 1 over focal length, or 1 over f, which is known, is equal to 1 over d object, what I'm looking for, plus 1 over d image, which I do not know. So what I have is a situation where I'm going to have to make use of the h-i equal negative one-fifth h-o statement in order to get some information about image distance. That information gets plugged into the Lenz equation, and I solve for object distance ultimately. So here is how I proceed. I say h-i over h-o equal negative one-fifth, but I also know the equation that the ratio of high to low is equal to the ratio of negative di per do. So the negative one-fifth is equal to negative di per do. Then I can take that statement, and I can say that the do is equal to positive five di. I did that by cross-multiplying on the statement. So do equal positive five di is an expression for do expressed in terms of di, and I can substitute that into my Lenz equation. The Lenz equation then becomes 1 over 22.8 is equal to 1 over do, but I don't write do. Instead I write 5 dot, 1 over 5 dot, plus 1 over di. So now I have two fractions on the right side that I'm going to have to add together, and to do that I have to get a lowest common denominator. That lowest common denominator would be the denominator of 5 dot. I can get it if I take the second fraction right side of the equation, multiply numerator and denominator by 5. Now the equation becomes 1 over 22.8 is equal to 1 over 5 dot plus 5 over 5 dot. With the common denominator I can now group the two fractions together as one, with the same common denominator in the numerator being just simply the sum of the two individual numerators. So now the equation becomes 1 over 22.8 is equal to 1 plus 5 over 5 di. Of course 1 plus 5 is equal to 6. I have 1 over 22.8 equals 6 over 5 dot. Now what I can do is cross-multiply, and that gives me 5 times di is equal to 6 times 22.8, also known as 136.8. I can divide through by 5, and that gives me d image is equal to 27.36 centimeters. Problem is I'm not looking for d image, I'm looking for d object, but previously I've stated that d object is equal to 5 times the d image, and so I can take the 27.36 centimeters and multiply by 5, I get 136.8 centimeters, and I can round that to 3 significant digits as 137 centimeters.