Today we looked at the results of the diverging lens data. Here is a quick pic of some results:

As you can see, the -1 slope still comes up, but now the line is in quadrant IV of the Cartesian coordinate system. The inverse of the focal length is still the focal length, but now it is negative, which is the convention used for diverging lenses. So the thin lens equation still applies.

Once we discussed this, I proposed the following challenge: Find the object location for your diverging lens that will result in the image being 1/5th the object size. I used 1/5 instead of 5x because the image for a diverging lens is always smaller than the object. Something to do with the pesky geometry of the lens.

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