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A student wishes to predict the magnification of an image given the distance from the object to a converging lens with an unknown index of refraction. What data must the student collect in order to make such a prediction for any object distance?
  1. A specific object distance and the image distance associated with that object distance.
  2. A specific image distance and a determination of whether the image formed is upright or inverted.
  3. The diameter and index of refraction of the lens.
  4. The radius of curvature of each side of the lens.
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OpenStax College Physics for AP® Courses Solution, Chapter 25, Problem 15 (Test Prep for AP® Courses) (2:04)

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This is College Physics Answers with Shaun Dychko. A student wants to be able to take any object distance from the lens and figure out what the magnification of the image will be. Well, magnification is the negative of the image distance divided by the object distance. But, we don't know what the image distance will be in each case, but we can substitute for image distance using the focal length of the lens. And, we can figure out the focal length given a one particular object distance and image distance combination. So, let's substitute for <i>Di</i> using this thin lens equation which has object distance and focal length in it. And, we'll subtract one over object distance from both sides and that gives us one over image distance is one over <i>F</i> minus one over <i>Do</i>, which we can write as a single fraction by multiplying this by <i>Do</i> over <i>Do</i> and this by <i>F</i> over <i>F</i>. And, we end up with a common denominator <i>Do</i> times <i>F</i> and in the numerator we have <i>Do</i> minus <i>F</i>. And then, the image distance is going to be the reciprocal of that fraction then. And so, image distance is <i>Do</i> times <i>F</i> over <i>Do</i> minus <i>F</i>, which we can then substitute in to our magnification formula and we place <i>Di</i> with <i>Do</i> <i>F</i> over <i>Do</i> minus <i>F</i>. And so, we have magnification then after the <i>Do</i>s cancel here. This one over <i>Do</i> being from here. The magnification, then, is negative of the focal length divided by object distance minus focal length, which maybe looks a little cleaner by writing <i>F</i> over <i>F</i> minus <i>Do</i>. Either way, if you can figure out what the focal length of the lens is given a specific object distance and image distance from that point on, you can always predict what the magnification will be for any object distance by substituting the object distance into this formula for magnification. And so, the answer is A.