﻿ RAY TRACING IN THIN LENSES

# Ray Tracing in Thin Lenses

## THIN LENSes

A convergent lens can be depicted like this: While a divergent lens can be depicted as such: However, because the lenses that we will be using are so thin, we will depict them as a line in the images below. They are the vertical axes in the applets below. In addition, we are ignoring what happens within the lens (explained in another section).

## Convergent Lens

Click and drag on points to change the size of the object, image, and focal length.

For this section, we will look through a lens toward the image (from the right of the picture). There are two rays which allow us to find the image from a convergent ray. The first ray travels from the object parallel to the axis toward the lens. Once it hits the lens, it goes toward the focus on the opposite side of the image.

For the second ray, the direction it travels depends on the location of the object. If the object is within the focus, then the second ray starts from the focus, and goes toward the top of the object. Once this ray hits the lens, it travels parallel to the axis. Extend the lines back if you need to to find where the two intersect.

If the object is outside of the focus, then the second ray starts at the top of the object, and goes toward the focus, until it meets the lens. Again, once this occurs, the image travels parallel to the axis. When the object is on the focus, the image is infinitely large.

The intersection of the two rays is where the image is. In the applet, CD is the object (in green), AB is the image, and O and J are the focal points.

We can use similar triangles to approximate the height and position of the image given the focal length, and the height and position of the object.

Ex. The object is within the focus, and the object height is 2.5, object position 3, and the focus 5.

Because , we can use a proportion to determine the height of the image ( ). , where , , f, and are the object position, object height, focal length, and image height, respectively.

Thus, , based on the diagram. As we know , , and f, we can now calculate , which is equivalent to . So, if = 3, = 2.5, and f = 5, we would have the proportion , and so = 4.167.

Next, we need to find the image position using another pair of similar triangles: , so we can find the position ( ). , where is the image position.

Therefore, , and can now find , which is . If we plug in the numbers again, we would have , so = 8.333.

## Divergent Lens

Click and drag on point B to change the size and position of the object and points E and C to change the focal length.

Using similar triangles, we can see that , so . is the same length as because is parallel to the horizontal axis, so it has height (subscript I for incident).
Therefore, . Plugging values in, we get and . , so . Consequently, and , relative to the focal point.