Refraction and Lenses Legacy Problem #29 Guided Solution

A good problem solver is going to read a problem carefully and pick up on little cues or hints within the problem statement that make all the difference in the world between a right and wrong answer. Here we read about an upright image and we read that it is reduced to one-fourth of the object's height. In effect, what they are saying in the first clause of the first sentence is that the height of the image is one-fourth the height of the object. Upright image is reduced in height to one-fourth of the object's height. So we can stay from that little statement that height, image height, is equal to one-fourth, one over four, times the object height. Another way to write that is that height divided by height is equal to one-fourth. Now we are told this occurs whenever the object is placed 26.9 centimeters from the lens. So d-object or d-o is equal to 26.9 centimeters. What we are asked is to calculate the focal length and I know that the only equation I have that has f or focal length in it is the lens equation that says one over f is equal to one over d-object plus one over d-image. What I need to know to calculate f is object distance, easy enough, it is stated, and image distance which I do not know. But what I can say is I can say that the height divided by hole is equal to one-fourth but it is also equal to the negative die over doe ratio. That is the case because height divided by hole is equal to the negative of the die divided by doe. And so what I am going to do is I am going to take one-fourth and say it is equal to negative die divided by doe. And then I am going to put 26.9 centimeters into the denominator of that ratio and I am going to multiply both sides by 26.9 so that the die is equal to negative one-fourth multiplied by 26.9. The die comes out to be negative 6.725 centimeters and that is expected that it would be negative because this is a virtual image and virtual image distances are negative. Now I know the two things I need to know in order to calculate the focal length. I say one over f is equal to one over 26.9 plus one over the negative 6.725 and I evaluate that on my calculator and I find out what f is. It comes out to be negative 8.9667. I can round that to the third significant digit, negative 8.97. The negative of this focal length value simply indicates that this is a virtual, that this virtual image is being created by a diverging lens. We would expect that because any image which is virtual or upright and also reduced in size must have been created by a diverging lens because a converging lens would create a magnified virtual image.