The previous section considered a converging lens with the object at distance $s>f$. That is, the object was outside the focal point. What if the object is inside the focal point, at distance $s<f?$ FIGURE 34.33 shows just this situation, and we can use ray tracing to analyze it.

FIGURE 34.33 Rays from an object at distance $s<f$ are refracted by the lens and diverge to form a virtual image.

The special rays initially parallel to the axis and through the center of the lens present no difficulties. However, a ray through the near focal point would travel toward the left and would never reach the lens! Referring back to Figure 34.29b, you can see that the rays emerging parallel to the axis entered the lens along a line passing through the near focal point. It’s the angle of incidence on the lens that is important, not whether the light ray actually passes through the focal point. This was the basis for the wording of step 4b in Tactics Box 34.2 and is the third special ray shown in Figure 34.33.

You can see that the three refracted rays don’t converge. Instead, all three rays appear to diverge from point $\text{P}\prime$. This is the situation we found for rays reflecting from a mirror and for the rays refracting out of an aquarium. Point $\text{P}\prime$ is a virtual image of the object point P. Furthermore, it is an upright image, having the same orientation as the object.

The refracted rays, which are all to the right of the lens, appear to come from $\text{P}\prime$, but none of the rays were ever at that point. No image would appear on a screen placed in the image plane at $\text{P}\prime$. So what good is a virtual image?

Your eye collects and focuses bundles of diverging rays; thus, as FIGURE 34.34a shows, you can see a virtual image by looking through the lens. This is exactly what you do with a magnifying glass, producing a scene like the one in FIGURE 34.34b. In fact, you view a virtual image anytime you look through the eyepiece of an optical instrument such as a microscope or binoculars.

FIGURE 34.34 A converging lens is a magnifying glass when the object distance is less than $f$.

FIGURE 34.34 Full Alternative Text

As before, the image distance $\mathit{s}\text{′}$ for a virtual image is defined to be a negative number $\mathbf{(}\mathit{s}\text{′}\mathbf{<}\mathbf{0}\mathbf{)}$, indicating that the image is on the opposite side of the lens from a real image. With this choice of sign, the definition of magnification, $m=-s\prime /s$, is still valid. A virtual image with negative $s\prime$ has $m>0$, thus the image is upright. This agrees with the rays in Figure 34.33 and the photograph of Figure 34.34b.

EXAMPLE 34.8 Magnifying a flower

To see a flower better, a naturalist holds a 6.0-cm-focal-length magnifying glass 4.0 cm from the flower. What is the magnification?

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

VISUALIZE FIGURE 34.35 shows the ray-tracing diagram. The three special rays diverge from the lens, but we can use a straightedge to extend the rays backward to the point from which they diverge. This point, the image point, is seen to be 12 cm to the left of the lens. Because this is a virtual image, the image distance is a negative $s\prime =-12\text{cm}$. Thus the magnification is

$$m=-{\displaystyle \frac{s\prime }{s}}=-{\displaystyle \frac{-12\text{cm}}{4\text{.0 cm}}}=3.0$$

The image is three times as large as the object and, because $m$ is positive, upright.

FIGURE 34.35 Ray-tracing diagram for Example 34.8.

FIGURE 34.35 Full Alternative Text