2.4 Thin Lenses

Ray Tracing and Thin Lenses

Ray tracing is the technique of determining or following (tracing) the paths taken by light rays.

Ray tracing for thin lenses is very similar to the technique we used with spherical mirrors. As for mirrors, ray tracing can accurately describe the operation of a lens. The rules for ray tracing for thin lenses are similar to those of spherical mirrors:

1. A ray entering a converging lens parallel to the optical axis passes through the focal point on the other side of the lens (ray 1 in part (a) of Figure 2.21). A ray entering a diverging lens parallel to the optical axis exits along the line that passes through the focal point on the same side of the lens (ray 1 in part (b) of the figure).
2. A ray passing through the center of either a converging or a diverging lens is not deviated (ray 2 in parts (a) and (b)).
3. For a converging lens, a ray that passes through the focal point exits the lens parallel to the optical axis (ray 3 in part (a)). For a diverging lens, a ray that approaches along the line that passes through the focal point on the opposite side exits the lens parallel to the axis (ray 3 in part (b)).

Figure 2.21 Thin lenses have the same focal lengths on either side. (a) Parallel light rays from the object toward a converging lens cross at its focal point on the right. (b) Parallel light rays from the object entering a diverging lens from the left seem to come from the focal point on the left.

Thin lenses work quite well for monochromatic light (i.e., light of a single wavelength). However, for light that contains several wavelengths (e.g., white light), the lenses work less well. The problem is that, as we learned in the previous chapter, the index of refraction of a material depends on the wavelength of light. This phenomenon is responsible for many colorful effects, such as rainbows. Unfortunately, this phenomenon also leads to aberrations in images formed by lenses. In particular, because the focal distance of the lens depends on the index of refraction, it also depends on the wavelength of the incident light. This means that light of different wavelengths will focus at different points, resulting is so-called “chromatic aberrations.” In particular, the edges of an image of a white object will become colored and blurred. Special lenses called doublets are capable of correcting chromatic aberrations. A doublet is formed by gluing together a converging lens and a diverging lens. The combined doublet lens produces significantly reduced chromatic aberrations.

Image Formation by Thin Lenses

We use ray tracing to investigate different types of images that can be created by a lens. In some circumstances, a lens forms a real image, such as when a movie projector casts an image onto a screen. In other cases, the image is a virtual image, which cannot be projected onto a screen. Where, for example, is the image formed by eyeglasses? We use ray tracing for thin lenses to illustrate how they form images, and then we develop equations to analyze quantitatively the properties of thin lenses.

Consider an object some distance away from a converging lens, as shown in Figure 2.22. To find the location and size of the image, we trace the paths of selected light rays originating from one point on the object, in this case, the tip of the arrow. The figure shows three rays from many rays that emanate from the tip of the arrow. These three rays can be traced by using the ray-tracing rules given above.

• Ray 1 enters the lens parallel to the optical axis and passes through the focal point on the opposite side (rule 1).
• Ray 2 passes through the center of the lens and is not deviated (rule 2).
• Ray 3 passes through the focal point on its way to the lens and exits the lens parallel to the optical axis (rule 3).

The three rays cross at a single point on the opposite side of the lens. Thus, the image of the tip of the arrow is located at this point. All rays that come from the tip of the arrow and enter the lens are refracted and cross at the point shown.

After locating the image of the tip of the arrow, we need another point of the image to orient the entire image of the arrow. We chose to locate the image base of the arrow, which is on the optical axis. As explained in the section on spherical mirrors, the base will be on the optical axis just above the image of the tip of the arrow (due to the top-bottom symmetry of the lens). Thus, the image spans the optical axis to the (negative) height shown. Rays from another point on the arrow, such as the middle of the arrow, cross at another common point, thus filling in the rest of the image.

Although three rays are traced in this figure, only two are necessary to locate a point of the image. It is best to trace rays for which there are simple ray-tracing rules.

Figure 2.22 Ray tracing is used to locate the image formed by a lens. Rays originating from the same point on the object are traced—the three chosen rays each follow one of the rules for ray tracing, so that their paths are easy to determine. The image is located at the point where the rays cross. In this case, a real image—one that can be projected on a screen—is formed.

Several important distances appear in the figure. As for a mirror, we define $d_{\text{o}}$ to be the object distance, or the distance of an object from the center of a lens. The image distance $d_{\text{i}}$ is defined to be the distance of the image from the center of a lens. The height of the object and the height of the image are indicated by $h_{\text{o}}$ and $h_{\text{i}}$, respectively. Images that appear upright relative to the object have positive heights, and those that are inverted have negative heights. By using the rules of ray tracing and making a scale drawing with paper and pencil, like that in Figure 2.22, we can accurately describe the location and size of an image. But the real benefit of ray tracing is in visualizing how images are formed in a variety of situations.

Oblique Parallel Rays and Focal Plane

We have seen that rays parallel to the optical axis are directed to the focal point of a converging lens. In the case of a diverging lens, they come out in a direction such that they appear to be coming from the focal point on the opposite side of the lens (i.e., the side from which parallel rays enter the lens). What happens to parallel rays that are not parallel to the optical axis (Figure 2.23)? In the case of a converging lens, these rays do not converge at the focal point. Instead, they come together on another point in the plane called the focal plane. The focal plane contains the focal point and is perpendicular to the optical axis. As shown in the figure, parallel rays focus where the ray through the center of the lens crosses the focal plane.

Figure 2.23 Parallel oblique rays focus on a point in a focal plane.

Thin-Lens Equation

Ray tracing allows us to get a qualitative picture of image formation. To obtain numeric information, we derive a pair of equations from a geometric analysis of ray tracing for thin lenses. These equations, called the thin-lens equation and the lens maker’s equation, allow us to quantitatively analyze thin lenses.

Consider the thick bi-convex lens shown in Figure 2.24. The index of refraction of the surrounding medium is $n_1$ (if the lens is in air, then $n_1=1.00$) and that of the lens is $n_2$. The radii of curvatures of the two sides are $R_1\text{and}{R}_2$. We wish to find a relation between the object distance $d_{\text{o}}$, the image distance $d_{\text{i}}$, and the parameters of the lens.

Figure 2.24 Figure for deriving the lens maker’s equation. Here, t is the thickness of lens, $n_1$ is the index of refraction of the exterior medium, and $n_2$ is the index of refraction of the lens. We take the limit of $t\to 0$ to obtain the formula for a thin lens.

To derive the thin-lens equation, we consider the image formed by the first refracting surface (i.e., left surface) and then use this image as the object for the second refracting surface. In the figure, the image from the first refracting surface is $Q^{\prime }$, which is formed by extending backwards the rays from inside the lens (these rays result from refraction at the first surface). This is shown by the dashed lines in the figure. Notice that this image is virtual because no rays actually pass through the point $Q^{\prime }$. To find the image distance $d_{\text{i}}^{\prime }$ corresponding to the image $Q^{\prime }$, we use Equation 2.11. In this case, the object distance is $d_{\text{o}}$, the image distance is $d_{\text{i}}^{\prime }$, and the radius of curvature is $R_1$. Inserting these into Equation 2.3 gives

The image is virtual and on the same side as the object, so $d_{\text{i}}^{\prime }<0$ and $d_{\text{o}}>0$. The first surface is convex toward the object, so $R_1>0$.

To find the object distance for the object Q formed by refraction from the second interface, note that the role of the indices of refraction $n_1$ and $n_2$ are interchanged in Equation 2.11. In Figure 2.24, the rays originate in the medium with index $n_2$, whereas in Figure 2.15, the rays originate in the medium with index $n_1$. Thus, we must interchange $n_1$ and $n_2$ in Equation 2.11. In addition, by consulting again Figure 2.24, we see that the object distance is $d_{\text{o}}^{\prime }$ and the image distance is $d_{\text{i}}$. The radius of curvature is $R_2$ Inserting these quantities into Equation 2.11 gives

The image is real and on the opposite side from the object, so $d_{\text{i}}>0$ and $d_{\text{o}}^{\prime }>0$. The second surface is convex away from the object, so $R_2<0$. Equation 2.15 can be simplified by noting that $d_{\text{o}}^{\prime }=\left|d_{\text{i}}^{\prime }\right|+t$, where we have taken the absolute value because $d_{\text{i}}^{\prime }$ is a negative number, whereas both $d_{\text{o}}^{\prime }$ and t are positive. We can dispense with the absolute value if we negate $d_{\text{i}}^{\prime }$, which gives $d_{\text{o}}^{\prime }=\text{−}{d}_{\text{i}}^{\prime }+t$. Inserting this into Equation 2.15 gives

Summing Equation 2.14 and Equation 2.16 gives

In the thin-lens approximation, we assume that the lens is very thin compared to the first image distance, or $t\ll d_{\text{i}}^{\prime }$ (or, equivalently, $t\ll R_1\text{and}{R}_2$). In this case, the third and fourth terms on the left-hand side of Equation 2.17 cancel, leaving us with

Dividing by $n_1$ gives us finally

The left-hand side looks suspiciously like the mirror equation that we derived above for spherical mirrors. As done for spherical mirrors, we can use ray tracing and geometry to show that, for a thin lens,

where f is the focal length of the thin lens (this derivation is left as an exercise). This is the thin-lens equation. The focal length of a thin lens is the same to the left and to the right of the lens. Combining Equation 2.18 and Equation 2.19 gives

which is called the lens maker’s equation. It shows that the focal length of a thin lens depends only of the radii of curvature and the index of refraction of the lens and that of the surrounding medium. For a lens in air, $n_1=1.0$ and $n_2\equiv n$, so the lens maker’s equation reduces to

Using the Thin-Lens Equation

The thin-lens equation and the lens maker’s equation are broadly applicable to situations involving thin lenses. We explore many features of image formation in the following examples.

Consider a thin converging lens. Where does the image form and what type of image is formed as the object approaches the lens from infinity? This may be seen by using the thin-lens equation for a given focal length to plot the image distance as a function of object distance. In other words, we plot

for a given value of f. For $f=1\text{cm}$, the result is shown in part (a) of Figure 2.25.

Figure 2.25 (a) Image distance for a thin converging lens with $f=1.0\text{cm}$ as a function of object distance. (b) Same thing but for a diverging lens with $f=\mathrm{-1.0}\text{cm}$.

An object much farther than the focal length f from the lens should produce an image near the focal plane, because the second term on the right-hand side of the equation above becomes negligible compared to the first term, so we have $d_{\text{i}}\approx f.$ This can be seen in the plot of part (a) of the figure, which shows that the image distance approaches asymptotically the focal length of 1 cm for larger object distances. As the object approaches the focal plane, the image distance diverges to positive infinity. This is expected because an object at the focal plane produces parallel rays that form an image at infinity (i.e., very far from the lens). When the object is farther than the focal length from the lens, the image distance is positive, so the image is real, on the opposite side of the lens from the object, and inverted (because $m=\text{−}{d}_{\text{i}}\text{/}{d}_{\text{o}}$). When the object is closer than the focal length from the lens, the image distance becomes negative, which means that the image is virtual, on the same side of the lens as the object, and upright.

For a thin diverging lens of focal length $f=\mathrm{-1.0}\text{cm}$, a similar plot of image distance vs. object distance is shown in part (b). In this case, the image distance is negative for all positive object distances, which means that the image is virtual, on the same side of the lens as the object, and upright. These characteristics may also be seen by ray-tracing diagrams (see Figure 2.26).

Figure 2.26 The red dots show the focal points of the lenses. (a) A real, inverted image formed from an object that is farther than the focal length from a converging lens. (b) A virtual, upright image formed from an object that is closer than a focal length from the lens. (c) A virtual, upright image formed from an object that is farther than a focal length from a diverging lens.

To see a concrete example of upright and inverted images, look at Figure 2.27, which shows images formed by converging lenses when the object (the person’s face in this case) is place at different distances from the lens. In part (a) of the figure, the person’s face is farther than one focal length from the lens, so the image is inverted. In part (b), the person’s face is closer than one focal length from the lens, so the image is upright.

Figure 2.27 (a) When a converging lens is held farther than one focal length from the man’s face, an inverted image is formed. Note that the image is in focus but the face is not, because the image is much closer to the camera taking this photograph than the face. (b) An upright image of the man’s face is produced when a converging lens is held at less than one focal length from his face. (credit a: modification of work by “DaMongMan”/Flickr; credit b: modification of work by Casey Fleser)

Work through the following examples to better understand how thin lenses work.

When solving problems in geometric optics, we often need to combine ray tracing and the lens equations. The following example demonstrates this approach.

Example 2.5

Choosing the Focal Length and Type of Lens

To project an image of a light bulb on a screen placed a distance of 1.50 m from the lens, you need to choose what type of lens to use (converging or diverging) and its focal length (Figure 2.28). The distance between the lens and the lightbulb is fixed at 0.75 m. Also, what is the magnification and orientation of the image?

Strategy

The image must be real, so you choose to use a converging lens. The focal length can be found by using the thin-lens equation and solving for the focal length. The object distance is $d_{\text{o}}=0.75\text{m}$ and the image distance is $d_{\text{i}}=1.5\text{m}$.

Solution

Solve the thin lens for the focal length and insert the desired object and image distances:

$$\begin{array}{cc}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}\hfill & =\frac{1}{f}\hfill \\ \hfill f& ={\left(\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}\right)}^{\mathrm{-1}}\hfill \\ & ={\left(\frac{1}{0.75\text{m}}+\frac{1}{1.5\text{m}}\right)}^{\mathrm{-1}}\hfill \\ & =0.50\text{m}\hfill \end{array}$$

The magnification is

$$m=\text{−}\frac{d_i}{d_{\text{o}}}=\text{−}\frac{1.5\text{m}}{0.75\text{m}}=\mathrm{-2.0}.$$

Significance

The minus sign for the magnification means that the image is inverted. The focal length is positive, as expected for a converging lens. Ray tracing can be used to check the calculation (see Figure 2.28). As expected, the image is inverted, is real, and is larger than the object.

Figure 2.28 A light bulb placed 0.75 m from a lens having a 0.50-m focal length produces a real image on a screen, as discussed in the example. Ray tracing predicts the image location and size.