FIGURE 34.30 shows a lens and an object whose object distance s from the lens is larger than the focal length. Rays from point P on the object are refracted by the lens so as to converge at point $\text{P}\prime$ on the opposite side of the lens at image distance $s^{\prime }$. If rays diverge from an object point P and interact with a lens such that the refracted rays converge at point $\text{P}\prime$, actually meeting at $\text{P}\prime$, then we call $\text{P}\prime$ a real image of point P. Contrast this with our prior definition of a virtual image as a point from which rays—which never meet—appear to diverge. For a real image, the image distance $s^{\prime }$ is positive.

FIGURE 34.30 Rays from an object point P are refracted by the lens and converge to a real image at point ${\text{P}}^{\prime }$.

All points on the object that are in the same plane, the object plane, converge to image points in the image plane. Points Q and R in the object plane of Figure 34.30 have image points $\text{Q}\prime$ and $\text{R}\prime$ in the same plane as point $\text{P}\prime$. Once we locate one point in the image plane, such as $\text{P}\prime$, we know that the full image lies in the same plane.

There are two important observations to make about Figure 34.30. First, the image is upside down with respect to the object. This is called an inverted image, and it is a standard characteristic of real-image formation with a converging lens. Second, rays from point P fill the entire lens surface, and all portions of the lens contribute to the image. A larger lens will “collect” more rays and thus make a brighter image.

FIGURE 34.31 is a close-up view of the rays very near the image plane. The rays don’t stop at $\text{P}\prime$ unless we place a screen in the image plane. When we do so, we see a sharp, well-focused image on the screen. To focus an image, you must either move the screen to coincide with the image plane or move the lens or object to make the image plane coincide with the screen. For example, the focus knob on a projector moves the lens forward or backward until the image plane matches the screen position.

FIGURE 34.31 A close-up look at the rays near the image plane.

Figure 34.30 highlights three “special rays” based on the three situations of Figure 34.29. These three rays alone are sufficient to locate the image point $\text{P}\prime$. That is, we don’t need to draw all the rays shown in Figure 34.30. The procedure known as ray tracing consists of locating the image by the use of just these three rays.

EXAMPLE 34.7 Finding the image of a flower

A 4.0-cm-diameter flower is 200 cm from the 50-cm-focal-length lens of a camera. How far should the light detector be placed behind the lens to record a well-focused image? What is the diameter of the image on the detector?

MODEL The flower is in the object plane. Use ray tracing to locate the image.

VISUALIZE FIGURE 34.32 shows the ray-tracing diagram and the steps of Tactics Box 34.2. The image has been drawn in the plane where the three special rays converge. You can see from the drawing that the image distance is $s\prime \approx 67\text{cm}$. This is where the detector needs to be placed to record a focused image.

FIGURE 34.32 Ray-tracing diagram for Example 34.7.

FIGURE 34.32 Full Alternative Text

The heights of the object and image are labeled $h$ and $h\prime$. The ray through the center of the lens is a straight line, thus the object and image both subtend the same angle $\theta$. Using similar triangles,

$${\displaystyle \frac{h\prime }{s\prime }}={\displaystyle \frac{h}{s}}$$

Solving for $h\prime$ gives

$$h\prime =h{\displaystyle \frac{s\prime }{s}}=(4.0\text{cm}){\displaystyle \frac{67\text{cm}}{200\text{cm}}}=1.3\text{cm}$$

The flower’s image has a diameter of 1.3 cm.

REVIEW We’ve been able to learn a great deal about the image from a simple geometric procedure.