Today the kids worked on completing their challenge. I was asked if they should use the graphically determined focal length or if they should use what they thought it should be. I let them decide. I personally would have used the graphically determined value. Yes, it might make the math a little messier, but using the data collected for the setup, when you will be using that setup, makes more sense to me.
Results were mixed. 2 groups got 50% error, once group got 0% error, and one had such a slightly divergent lens, that do came out to be about +500 cm. Oops.
Now you might ask, how did we confirm the size of the virtual image? Well, with two lenses of course! I had a lens of known focal length (f = 20 cm.) Once the diverging lens and object were placed, I asked the students for their di.
With di in hand, I set my converging lens 40 cm away from their di. Why 40 cm? well, a lens of focal length = 20 cm has 2F at 40 cm. As we learned before, an object at 2f will form an image that is equal in size. So by placing my converging lens 40 cm away from where their virtual image formed, I could project the image onto a screen, and measure the size of my real image. this would be the same size as the virtual image.
At the end of the hour, I assigned them the task of redoing their double lens problems from last week.