I was starting to type up the next Laser Smackdown entry, when it occurred to me that this was a good point to talk about a neat little thing from optics. It further occurred to me that this would be a good poll/quiz topic, to see what people think before I give you the real answer.

So, here’s the scenario: You have a brightly illuminated object, the canonical example being a vertical arrow (it could also be a shadow cast in a beam of light). You take a lens, and put it in front of the arrow, projecting an image of the arrow on a screen, like so:

i-876298025040b017b1b70255acf73b27-LensQuiz.jpg

As you can see from the picture, the image that is formed on the screen is inverted– the arrow points down, rather than up.

Now, imagine taking a card (the black rectangle below the lens) and positioning it right in front of the lens, so it blocks the lower half of the lens. What happens to the image?

This is a classical problem, so you have to pick one and only one answer. You can easily check your answer, if you have access to a flashlight and a magnifying glass, but please don’t do that until after you enter your answer, just so we can see what people arrive at from pure reason.

Comments

  1. #1 Excited State
    April 15, 2010

    I was tempted to choose the last option, in honor of my students. At least I’m not teaching pre-meds this semester.

  2. #2 Matt Springer
    April 15, 2010

    Using the card on the other side of the lens is one of the canonical ways to very precisely locate the focus of a laser. On the near side one half is blocked, on the far side the other is blocked, and exactly at the focus the whole thing disappears in a blob.

  3. #3 Evan Berkowitz
    April 15, 2010

    Will there be a solution set, so that we can practice for future quizzes?

  4. #4 AnyEdge
    April 15, 2010

    I’m not getting out pencil and paper. My official guess is:

    You’re still going to get an image of the entire arrow, because the uncovered half of a lense can still ‘see’ the whole thing (sloppy, but you know what I mean). The lens will still bend the light in the same way as before, i.e., all the light will be inverted and shifted down on the incident surface.

    So you will get an inverted arrow, half the size as before, and entirely below the imaginary line dividing the incident surface in half.

  5. #5 Dave X
    April 15, 2010

    Could it also be a shadow cast in a beam of collimated light?

  6. #6 hibob
    April 15, 2010

    for that teachable moment: ask what happens if you put the card next to the arrow (between the arrow and the lens), then ask what happens to the image as the card is moved until it is next to the lens.

  7. #7 Tercel
    April 15, 2010

    You will get the same image of the arrow. The only change will be that the image quality is slightly reduced by blurring.

  8. #8 Tercel
    April 15, 2010

    In fact, the blurring I mentioned is only due to diffraction effects, so if we are talking about purely geometrical optics then the image would be completely unchanged.

    Also it would be half as bright in either case (blur or no blur).

  9. #9 ADD
    April 15, 2010

    Tercel: If diffraction is not important, the image will be less blurry. When you make the lens smaller, you reduce some sources of aberration.

  10. #10 Gray Gaffer
    April 15, 2010

    Blurring: the question asserted the classical case, so I guess diffraction effects are not considered, therefore you get the same image at half the illumination. However, the blurring can be a significant factor when considering, say, obstacles such as secondary mirrors in folded optics telescopes. Not just blurring of objects, the secondary also degrades the overall image contrast.

    I have two somewhat more complicated suggestions that I think might make for interesting optics discourses:

    1 (simpler one) where is the Nodal Point of a lens? Consider different types, especially single refractor and folded optics geometries. Of practical importance is application to the growing practice of making panoramic pictures with multiple exposures, and how the results are affected by lens choice.

    2 (more complicated, but also more fun – I built one of these in my bedroom as a high school student) a Schlieren system. Today’s question is almost half of a Schlieren system, which is what made me think of it. The fun part is one has to consider multiple points of view simultaneously to understand how and why it works.

The site is currently under maintenance and will be back shortly. New comments have been disabled during this time, please check back soon.