Physics 103

Imaging with a Lens

Introduction

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.

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: