Image Produced by a Magnifying Glass

We are given that $d_{\text{o}}=7\text{.}\text{50 cm}$ and $f=\text{10}\text{.}\text{0 cm}$, so we have a situation where the object is placed closer to the lens than its focal length. We therefore expect to get a case 2 virtual image with a positive magnification that is greater than 1. Ray tracing produces an image like that shown in Figure 25.37, but we will use the thin lens equations to get numerical solutions in this example.

To find the magnification $m$, we try to use magnification equation, $m={\mathrm{–d}}_{\text{i}}/d_{\text{o}}$. We do not have a value for $d_{\text{i}}$, so that we must first find the location of the image using lens equation. (The procedure is the same as followed in the preceding example, where $d_{\text{o}}$ and $f$ were known.) Rearranging the magnification equation to isolate $d_{\text{i}}$ gives

Entering known values, we obtain a value for $\mathrm{1/}{d}_{\text{i}}$:

This must be inverted to find $d_{\text{i}}$:

Now the thin lens equation can be used to find the magnification $m$, since both $d_{\text{i}}$ and $d_{\text{o}}$ are known. Entering their values gives

A number of results in this example are true of all case 2 images, as well as being consistent with Figure 25.37. Magnification is indeed positive (as predicted), meaning the image is upright. The magnification is also greater than 1, meaning that the image is larger than the object—in this case, by a factor of 4. Note that the image distance is negative. This means the image is on the same side of the lens as the object. Thus the image cannot be projected and is virtual. (Negative values of $d_{\text{i}}$ occur for virtual images.) The image is farther from the lens than the object, since the image distance is greater in magnitude than the object distance. The location of the image is not obvious when you look through a magnifier. In fact, since the image is bigger than the object, you may think the image is closer than the object. But the image is farther away, a fact that is useful in correcting farsightedness, as we shall see in a later section.