Introduction to Basic Ray Tracing

Introduction to Basic Ray Tracing

In this section, we are going to discuss the basics of ray tracing through lens systems. The first important point to understand is described visually in the following figure

It can be shown that the ray depicted in the figure is ONLY laterally translated and its angle remains unchanged. All rays passing through a lens whose ray is not changed will also pass through what is called the optical center. It is important to note, that in the case of a thin lens, the ray is also not laterally shifted because we make the approximation that the thickness of a thin lens is zero.

The reason we examine the case of an off axis ray passing through the optical center is that we will most often times choose this to be one of the rays we trace in an optical system.

For the most part, we will always dray at least two rays. Sometimes it is necessary to draw three rays, but in the following examples, two will suffice.

Ray 1: The undeviated ray through the center of the lens

Ray 2/3: Will be drawn using the fact that a ray passing through the focal point will emerge from the lens parallel to the central axis and vice versa.

That last part ("and vice versa") is actually quite important. It means we can draw a ray that passes from the object, through the back focal point of the lens, which then exits the lens parallel to the central axis. Or we can draw a ray incident on the lens that is parallel to the central axis, which will then pass through the front focal point of the lens.

Positive Lenses

Take these examples for instance:

This figures shows an object (a blue ringed octopus). The first ray we trace is the ray exiting the top (or bottom) of the object and passing through the optical center.

The second ray we trace is a ray exiting the same point on the object but parallel to the central axis. The ray emerges on the other side of the lens and passes through the second focal point. It is important to point out that we could have drawn the second ray passing from the the same point on the object but passing through the first focal point, in which case it would have emerged from the lens parallel to the central axis. Either ray will work.

Putting both of these rays on the same diagram, we see that they cross somewhere beyond the second focal point of the lens. This is where the object point (the bottom part of the octopus) is imaged. The distances in this figure are labeled so and s_i and obey the equation we derived earlier i.e.,

$\frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f}$

Also take note that the image is flipped up down, this is the case for an image produced by a single, positive lens.

Negative Lenses

Another example is helpful although it can be a little confusing at first. Examine the following figure.

In the case of a negative lens, we obey the same rules although they may appear visibly different. We still trace a ray that passes through the optical center of the lens. However, the ray exiting the object parallel to the central axis intersects the focal point of a negative lens on the same side as it originates. The place where these two rays cross is again the image point, but in the case of a single, negative lens, the image is on the same side of the lens as the object and is virtual i.e., cannot be seen using only this lens. We would have to place another, positive lens after the negative lens to image the virtual image into a real image.

Magnification

The last concept we review here is the idea of magnification of an optical system. I'll just state the magnification of an optical system as given by

$M=-s_i/s_o$

that is, the magnification of an optical system is the ratio of the image distance to the object distance. This is straightforward to see in the following figure

The magnification of a system is always given by the ratio of the image size, y_i, to the object size y_o. In this case, using the ray passing through the optical center (and after locating the image by tracing ray 1 and 2) we see that

$\tan (\theta )=\frac{y_o}{s_o}=\frac{y_i}{s_i}$

so that

$M=\frac{y_i}{y_o}=\frac{s_i}{s_o}$

however, we see here that the y-distances must have opposite sign, because one of them is below the x-axis and one of them is above the x-axis so we write

$M=-\frac{s_i}{s_o}$

Clearly the negative sign indicates if the image is flipped or not. It is important to understand that the Magnification can be less than one or greater than one. If it is greater than one, the image has been enlarged. If the magnification is less than one, the image is smaller than the original object.

It is also interesting to note the inverse relations ship between the magnification and the object distance. Apparently, moving the object closer to a positive lens (but still farther than the first focal length) increases the size of the image.