It has been a while since I posted last. Life has been getting in the way. Namely the upcoming gun deer season and planning school around my future trip to Trees for Tomorrow.
As such, as of now, I am going to focus on Physics for this blog. I only teach this curriculum every other year, and as it digs deeper into the Light models than probably most classes, I am going to try and share what I can about this class.
We wrapped up looking at lens action, and with the conclusion of the Converging lens lab, we did a problem set dealing with quantitative lens problems. The problem set has some nice features:
– a set of problems that gives the students do, di, hi, ho, f, or any combination thereof. They have to pose a verbal solution to the question, do the algebra to solve for stuff, then a scale drawing for comparison.
– some questions that force the students to deal with the fact that M=hi/ho=-di/do. That negative sign (oh you gotta love the negative signs if you love physics) shows up in a few problems that only work IF you don’t ignore it. By the way, that sign is an indication that the image formed is inverted compared to the object.
-there is also a few questions that make the students deal with the diverging lenses, but just enough to kind of get them thinking.
The last few questions involve adding a second lens, specifically a converging lens. Most students, due to lack of another option, use the image formed by the first lens, as the object for the second lens. One question in particular though, makes them deal with an image that forms beyond the second lens, so that their is no light coming from a real image and passing through the screen. To add to the challenge, the image forms directly on the focal point.
Most kids say that there is no image as a result. They know from previous experience that if an object sits on the focal point, then no image forms. The problem is that there is no object or image forming on that focal point. Because the second lens is dropped in before the light that forms the image from the first lens can converge, it creates a problem. Still, that second lens acts like there is an object 20 cm from it even though it is not there.
Doesn’t that sound kind of I like a virtual image? Only this time, it’s acting like an object. So letters treat it like a “virtual object.” In this case do is negative. Using that along with f, you can use the thin lens equation to find the image placement.
We develop this solution because it leads to yet another possibility: can we use two lenses to figure out where a virtual image forms, such as with a diverging lens. So we set it up, and take a look. Sure enough we can set up a light source, diverging lens, and converging lens and get an image to for on a screen. How do you use this to figure out where the virtual image formed with the diverging lens?
Take the diverging lens out. Ok, now your image on the screen is all blurry. But if you slide the light source forward, all of a sudden, the image reforms. Why? Because the light source is where the virtual image was! From this, you can get the image distance for the virtual image.
What if we took a bunch of data for do and di for the diverging lens. Does the thin lens equation still work? We find out tomorrow.