Only the simplest magnifiers are built with a single lens of the sort we analyzed in Chapter 34. Optical instruments, such as microscopes and cameras, are invariably built with multiple lenses. The reason, as we’ll see, is to improve the image quality.

The analysis of multi-lens systems requires only one new rule: The image of the first lens acts as the object for the second lens. To see why this is so, FIGURE 35.1 shows a simple telescope consisting of a large-diameter converging lens, called the objective, and a smaller converging lens used as the eyepiece. (We’ll analyze telescopes more thoroughly later in the chapter.) Highlighted are the three special rays you learned to use in Chapter 34:

• A ray parallel to the optical axis refracts through the focal point.

• A ray through the focal point refracts parallel to the optical axis.

• A ray through the center of the lens is undeviated.

FIGURE 35.1 Ray-tracing diagram of a simple astronomical telescope.

The rays passing through the objective converge to a real image at $\text{P}\prime$, but they don’t stop there. Instead, light rays diverge from $\text{P}\prime$ as they approach the second lens. As far as the eyepiece is concerned, the rays are coming from $\text{P}\prime$, and thus $\text{P}\prime$ acts as the object for the second lens. The three special rays passing through the objective lens are sufficient to locate the image $\text{P}\prime$, but these rays are generally not the special rays for the second lens. However, other rays converging at $\text{P}\prime$ leave at the correct angles to be the special rays for the eyepiece. That is, a new set of special rays is drawn from $\text{P}\prime$ to the second lens and used to find the final image point $\text{P}\text{″}$.

EXAMPLE 35.1 A magnifying lens

A magnifying “lens” (used in the singular) is usually a combination of two or more lenses. The combination has much better optical properties than a single lens because, as you’ll learn later in the chapter, the combination is able to reduce or eliminate aberrations that impair a lens’s ability to form sharp images. Consider a magnifying lens in which light passes through first a diverging lens, with $f_1=-10.0\text{mm,}$ and then a converging lens, with $f_2=20.0\text{mm,}$ with the two lenses spaced 12.0 mm apart. Suppose you use this combination lens to view a small bug 20.0 mm from the diverging lens while you look through the converging lens. Where is the bug’s image, and what is its lateral magnification?

MODEL Each lens is a thin lens. The image of the first lens is the object for the second lens.

VISUALIZE FIGURE 35.2 shows the geometry and a few of the rays. The image is an upright, virtual image far to the left of the diverging lens.

FIGURE 35.2 Pictorial representation of a combination lens.

FIGURE 35.2 Full Alternative Text

SOLVE For the first lens, the bug is at object distance $s_1=20.0\text{mm}$. Its image is found from the thin-lens equation:

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{s_1^{\prime }}}& =\hfill & {\displaystyle \frac{1}{f_1}}-{\displaystyle \frac{1}{s_1}}={\displaystyle \frac{1}{-10.0\text{ mm}}}-{\displaystyle \frac{1}{20.0\text{ mm}}}=-0.150{\text{ mm}}^{-1}\hfill \\ \hfill s_1^{\prime }& =\hfill & -6.67\text{ mm}\hfill \end{array}$$

This is a virtual image 6.67 mm to the left of the first lens. The magnification of the first lens is

$$m_1=-\frac{s_1^{\prime }}{s_1}=-\frac{-6.67\text{ mm}}{20.0\text{ mm}}=+0.333$$

The image of the diverging lens now acts as the object for the converging lens. It is distance $s_2=6.67\text{mm}+12.0\text{mm}=18.67\text{mm}$ from the second lens. Another application of the thin-lens equation yields

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{s_2^{\prime }}}& =\hfill & {\displaystyle \frac{1}{f_2}}-{\displaystyle \frac{1}{s_2}}={\displaystyle \frac{1}{20.0\text{ mm}}}-{\displaystyle \frac{1}{18.67\text{ mm}}}=-0.00357{\text{ mm}}^{-1}\hfill \\ \hfill s_2^{\prime }& =\hfill & -280\text{ mm}\hfill \end{array}$$

This is a virtual image 280 mm to the left of the converging lens with magnification

$$m_2=-\frac{s_2^{\prime }}{s_2}=-\frac{-280\text{ mm}}{18.67\text{ mm}}=+15.0$$

The second lens magnifies the image of the first lens, so the total magnification is the product of the individual magnifications:

$$m_{\text{total}}=m_1{m}_2=+5.00$$

Thus an upright image 280 mm in front of the converging lens has a total lateral magnification of $+5.00$. You see this image when you look at the bug through the converging lens.

REVIEW A hand lens like the one in the photo usually consists of two lenses (a doublet) or three (a triplet). The magnification marking $\times 10$ is the angular magnification, which we’ll introduce in Section 35.4, not the lateral magnification that we calculated in this example.

A combination lens, such as a magnifier or a camera lens or the eyepiece lens for a telescope, can be characterized by an effective focal length. For example, a camera lens might be specified as a “35 mm lens,” which means an effective focal length of 35 mm. The calculation of effective focal lengths for real lenses is the subject of an advanced optics course, but the result for a combination of two thin lenses is simple. For two thin lenses with focal lengths $f_1$ and $f_2$ separated by distance $L$, their effective focal length $f_{\text{eff}}$ is given by

For example, we can calculate that the magnifying lens of Example 35.1 has effective focal length $f_{\text{eff}}=100\text{mm}$. The idea of effective focal length will be especially important when we look at camera lenses in the next section.

Two thin lenses in contact have separation $L=0$, and thus the inverse of the effective focal length is the sum of the inverses of the individual focal lengths. We can state this more clearly by defining the power of a lens as the inverse of its focal length:

A lens with more power (a shorter focal length) causes light rays to refract through a larger angle. The SI unit of lens power is the diopter, abbreviated D, defined as $1\text{D}=1\text{m}{}^{-1}$. Thus a lens with $f=50\text{cm}=0.50\text{m}$ has power $P=2.0\text{D}$.

You can see that $P_{\text{eff}}=P_1+P_2$ for thin lenses in contact. That is, the total refractive power is simply the sum of the individual refractive powers. But be careful to apply this only to lenses in contact; you still need to use Equation 35.1 for the more general situation.