It took some time to prepare white boards yesterday, so the white board presentation spilled over into today.
These problems were pretty challenging. I will talk here about 3 of them.
the first had the students figure out where to put an object and two lenses (both f = 30 cm) in order to get a final image that is 12x and inverted compared to the original object.
They figured out pretty fast that the first image had to be virtual to get the proper orientation of the final image. What came next though was a lot of trial and error. In order to get the 12x they knew that one lens would have to magnify it part way, and the other finish it. so they tested 3/4, 4/3, 1/12, 12/1, 6/2 and finally 2/6. Of course only the last set up where the virtual image was 2x the object and the final image was 6x the virtual image would work!
The next question asks the students to figure out two places an object can be placed to get an image that is 8x bigger. That is all it says.
With the focal length being 30 cm, it means the data on the right is for a real image, and the data on the right is for a virtual image.
Finally we have a situation where the image formed by the first lens is actually beyond the second lens. How do you deal with that? Well, here there is no actual light coming from the first image/percieved object and into the second lens. As such, we can look at it as being like a virtual “object.” and use -do.
Kids didn’t like this idea at first, and even though you can punch it all in the calculator, why did it work? Well the light that should be forming that perceived object, is not yet fully converged. When it passes through the second lens it bends even faster. This forms the image between the 2nd lens and the perceived object.
“But that perceived object is right on the focal point for that second lens. so there should be no image.”
“That would be true IF the perceived object was actually there! Remember once that second lens shows up, it really isn’t there. It is virtually there.”