Imaging with a Lens
We will explore the image-forming properties of lenses. Thin lenses can be grouped into two categories, positive (converging) lenses, and negative (diverging) lenses. A positive lens is one that causes incident parallel rays to converge at a focal point on the opposite side of the lens. A negative lens is one that causes incident parallel rays to emerge from the lens as though they emanated from a focal point on the incident side of the lens. The two types of lenses are illustrated below.
Use the mouse to determine the position of the lens and of the focal point. The difference in these positions should be the focal length. Select the lens by clicking on it: the focal length f will be shown. Is it about what you expected?
The parallel rays in the figures above represent a point source infinitely far away to the left. (This is a tricky concept! Think about it.) More generally, light rays emanating from any point (the "object") on the central axis will be brought to a focus (the "image", real or virtual) at some other point on the central axis. If the object is not at infinity, but at the finite position p, then a lens of focal length f forms an image at the position q determined by the thin lens equation
Here it is understood that p is measured from the center of the lens, and is considered positive if the object is to the left of the lens, and negative to the right. Similarly q is measured from the center of the lens and is positive to the right of the lens, negative to the left (opposite sign convention from p). What is q in the special case that p goes to infinity? (Compare the figures at the top of this page.)
Experiment with the setup pictured below by dragging the object (source of the light rays) with the mouse. Note that when the object is off-axis the image is off-axis too. If the object is not too far off-axis, then the distances p, q, and f are still given by the thin-lens equation. This determines the horizontal position of the image. The vertical distance of the image from the axis is most easily determined by the special ray through the center of the lens, which is undeviated: as you move the object vertically, the image also moves vertically as if at the opposite end of a lever which pivots at the center of the lens. Try to make sense of these statements in the figure below, and in particular check the thin lens equation.Now replace the "source" in the applet with an "object" (meaning an object of finite extent). The way to do this is to select the source by clicking on it, then push the "Clear Active" button. The active element, that is, the source you selected, should disappear. Now push the "Object" button and click where you want to place the new object (roughly where the source used to be). If necessary, drag the new object a little off-axis so that you can see it better. The object is indicated by a vertical arrow, and the image of the tip of the arrow is constructed by ray-tracing. Only three rays are drawn, but they are enough to see where the image is, and they are the simplest rays, or, as they are called, the principal rays. Notice that the one through the center of the lens is undeviated, and the other two go through the focal points of the lens (select the lens by clicking on it to see this).
Drag the object around and notice how the image behaves. As the object approaches the focal point of the lens, the image retreats to infinity (note the parallel rays). What happens if you bring the object inside the focal length of the lens? Something very interesting! The image reappears as a virtual image on the same side of the lens as the object (i.e., q is now negative: check how the thin lens equation describes this). An eye looking in from the right would receive rays as if from this image. It is called "virtual" because you have to extrapolate the rays backward to find it: the light rays themselves don't actually intersect to form a "real" image. By the way, in this configuration the lens is acting as a magnifying glass: think why this is.
It makes sense to compare the size of the object and the image. We define the magnification M to be the ratio of the size I of the image to the size O of the object, but by similar triangles this is also the ratio of the image distance q to the object distance p: