Refraction and Lenses Legacy Problem #25 Guided Solution

Oftentimes, the difficulties which a student has in solving a physics problem is not related to the mathematical manipulation of equations, or even the task of finding the appropriate equation. Rather, it is related to a misconception, a faulty idea. And that could very well be the case in this problem. The fact is, in many physics problems, that there are more than just numbers and equations. And that is the case here. The focal point of a lens is located 17.8 centimeters away from its surface. Most students quickly jump to the idea that f equals 17.8 centimeters. But don't go so fast, because a lens could be a diverging lens, and the focal point could be 17.8 centimeters from it. And therefore, the f would be negative 17.8 centimeters. And if that were the case, that small negative sign would make the difference between a right and a wrong answer. So we have to read on. Because a lens produces a virtual image, which is 38.9 centimeters from the lens. So there is some sort of image a distance from the lens, 38.9 centimeters. Now, most students would quickly jump to the fact that d i is equal to positive 38.9 centimeters. And again, don't be so quick, because it could be positive 38.9 centimeters, or negative 38.9 centimeters. In both of these instances, the difference between the plus and the minus is a conceptual difference. We must understand the concept in order to get whether it's positive or negative. So in the case of this virtual image, the d i value, which describes the distance from the lens to the image, is negative 38.9 centimeters. Because when a virtual image is formed, we refer to the image distance as being negative. The q is the word virtual. Now, for the f value, it happens to be positive 17.8 centimeters. Positive focal lengths indicate converging lenses, while negative focal lengths produce our characteristics of diverging lenses. And the reason I picked positive 17.8 centimeters for f is because when a diverging lens creates an image, the image is located between the lens itself and the focal point. That means that for a 17.8 centimeter focal point diverging lens, the image distance should be less than 17.8 centimeters, and it isn't. It's greater than 17.8 centimeters. So what I know from the onset is that f equals positive 17.8 centimeters, and that d i equals negative 38.9 centimeters. Now the question becomes a math problem, but not until you first do all the necessary physics on this question like the physics I just did. Now I can say 1 over 17.8 is equal to 1 over object distance, plus 1 over negative 38.9 centimeters. And from there, I can evaluate this equation by getting 1 over the object by itself on one side of the equation, and then evaluating what 1 over the object distance is. And that would be equal to 0.081887, and then I can take the reciprocal of that number, and I get 12.2 as the object distance.