Geometric Optics, Vision and Optical Instruments

69 Image Formation by Lenses

Learning Objectives

Lenses are among the most important optical devices used in science, medicine, and everyday life. They are found in eyeglasses, contact lenses, cameras, microscopes, telescopes, endoscopes, smartphones, and even in the human eye. By bending light through refraction, lenses allow us to focus images, magnify tiny structures, and correct vision problems. For students pursuing careers in the health sciences, understanding how lenses work is especially important. Medical professionals rely on optical instruments every day—from surgical loupes and operating microscopes to ophthalmoscopes used to examine the retina. Understanding the basic physics of lenses provides the foundation for understanding how these instruments produce clear images. The word lens comes from the Latin word for the lentil bean because the shape of a common convex lens resembles that of a lentil.

A converging (convex) lens is thicker in the middle than at the edges. As shown in Figure 69.1, rays of light entering the lens parallel to its principal axis are refracted toward the axis and meet at a single point on the opposite side of the lens. The principal axis is an imaginary line passing through the center of the lens and perpendicular to its surfaces. The point where parallel rays meet after passing through the lens is called the focal point, labeled F. The distance from the center of the lens to this point is known as the focal length, represented by f.

Parallel rays entering a converging lens are refracted toward the principal axis and meet at the focal point. An expanded view illustrates refraction at both lens surfaces.
Figure 69.1: A converging (convex) lens bends parallel light rays toward the principal axis so they meet at the focal point, F. The distance from the center of the lens to the focal point is called the focal length, f. The inset illustrates how refraction occurs at both surfaces of the lens.

A familiar example of a converging lens is a magnifying glass. When sunlight passes through a magnifying glass, the nearly parallel rays from the Sun are focused into a very small spot, as shown in Figure 69.2. Because the light energy is concentrated into such a small area, the temperature at the focal point can become high enough to ignite paper or other combustible materials.

A magnifying glass focuses nearly parallel rays of sunlight onto a small spot on paper.
Figure 69.2: A converging lens focuses nearly parallel rays from the Sun to its focal point. Concentrating the light into a very small area greatly increases the energy delivered to the paper, allowing it to burn.

The ability of a lens to bend light is described by its optical power. A lens that bends light strongly brings parallel rays to a focus over a short distance and therefore has a short focal length. Conversely, a weaker lens bends light less and has a longer focal length. The optical power P of a lens is defined as the reciprocal of its focal length:

[latex]P=\frac{1}{f}[/latex]

where the focal length f must be expressed in meters. The SI unit of optical power is the diopter (D):

[latex]1\ \mathrm{D}=1\ \mathrm{m}^{-1}[/latex]

Optical power is different from power measured in watts. Instead of describing the rate of energy transfer, diopters measure how strongly a lens converges or diverges light. Optometrists and ophthalmologists prescribe eyeglasses and contact lenses using diopters. For example, a reading glass labeled +2.00 D has a focal length of 0.50 m, while a lens labeled −3.00 D is a diverging lens used to correct nearsightedness.

Example 69.1: Calculating the Power of a Magnifying Glass

A magnifying glass concentrates sunlight to a small spot located [latex]8.00~\text{cm}[/latex] from the center of the lens. Determine:

  1. the focal length of the lens, and
  2. the optical power of the lens.

Strategy

Because the Sun is extremely far from Earth, the sunlight reaching the magnifying glass consists of nearly parallel rays. A converging lens brings parallel rays to its focal point, so the distance from the lens to the focused spot is the focal length. Once the focal length is known, the optical power can be calculated from

[latex]P=\frac{1}{f}[/latex]

The focal length must be expressed in meters before calculating the power in diopters.

Solution for (a)

The focal length is the distance from the center of the lens to the focused spot:

[latex]f=8.00~\text{cm}=0.0800~\text{m}[/latex]

Solution for (b)

Using the definition of optical power,

[latex]P=\frac{1}{f}=\frac{1}{0.0800~\text{m}}=12.5~\text{D}[/latex]

Discussion

The magnifying glass has a relatively large positive power of [latex]+12.5~\text{D}[/latex]. The positive sign indicates that it is a converging lens. Optical power measured in diopters should not be confused with power measured in watts. Diopters describe how strongly a lens bends light, whereas watts describe the rate at which energy is transferred.

Diverging Lenses

A diverging lens, also called a concave lens, is thinner in the middle than at the edges. Instead of bringing parallel rays together, it bends them away from the principal axis. As shown in Figure 69.3, rays entering a diverging lens parallel to the principal axis spread apart after passing through the lens. When the outgoing rays are extended backward, they appear to originate from a common point on the same side of the lens as the incoming light. This point is the focal point of the diverging lens. Because the rays do not actually pass through the focal point, the focal point is described as virtual. By convention, both the focal length and optical power of a diverging lens are negative. For example, if a diverging lens has a focal length of

[latex]f=-5.00~\text{cm}=-0.0500~\text{m}[/latex]

then its optical power is

[latex]P=\frac{1}{f}=\frac{1}{-0.0500~\text{m}}=-20.0~\text{D}[/latex]
Parallel rays enter a diverging lens and spread apart after refraction. Dashed backward extensions of the rays meet at a virtual focal point on the same side of the lens as the incoming light. A central ray passes through with negligible deflection.
Figure 69.3: A diverging (concave) lens causes parallel rays to spread apart. The rays appear to originate from the virtual focal point, F, on the same side of the lens as the incoming light. The focal length and optical power of a diverging lens are negative.

Healthcare Connection: Correcting Nearsightedness

Diverging lenses are commonly used to correct myopia, or nearsightedness. In a myopic eye, light from distant objects is focused in front of the retina. A diverging corrective lens spreads the incoming rays slightly before they enter the eye, allowing the cornea and lens of the eye to focus the image correctly on the retina. Because diverging corrective lenses have negative focal lengths, prescriptions for myopia are written with negative powers, such as [latex]-2.00~\text{D}[/latex] or [latex]-4.50~\text{D}[/latex].

Light paths through a lens are reversible. This means that if a ray can travel through a lens in one direction, it can follow the same path in reverse. For example, parallel rays entering a converging lens meet at the focal point, as shown in Figure 69.1. Reversing the path, light emitted from a small source placed at the focal point emerges from the lens as a set of parallel rays.

Light rays spreading from a small source at the focal point of a converging lens emerge from the opposite side as parallel rays.
Figure 69.4: A small light source placed at the focal point of a converging lens produces parallel rays on the opposite side. This is the reverse of the situation shown in Figure 69.1. The same principle is used to produce directed beams in headlights, flashlights, lighthouses, and some medical illumination systems.

Ray Tracing and Thin Lenses

Ray tracing is a graphical method used to determine the paths followed by light rays and locate the images formed by optical systems. For lenses, ray tracing provides a simple way to predict whether an image will be upright or inverted, enlarged or reduced, and real or virtual. Detailed lens systems, such as those found in cameras, microscopes, and medical imaging instruments, may contain many curved surfaces and require computer modeling. However, the behavior of a single lens can often be understood using the thin-lens approximation. A thin lens is an idealized lens whose thickness is small compared with the distances involved in forming an image. Although light is refracted at both surfaces of the lens, we treat the total change in direction as if it occurs at a single plane passing through the center of the lens. This approximation allows us to ignore small effects associated with lens thickness, including some aberrations and dispersion.

A symmetrical thin lens has two focal points, one on each side of the lens. Both focal points are the same distance from the lens, so the focal length has the same magnitude regardless of the direction in which light travels. This symmetry follows from the reversibility of light paths.

Parallel rays enter converging and diverging lenses from the right. The converging lens brings the rays to a focal point on the left, while the diverging lens produces rays that appear to originate from a focal point on the right.
Figure 69.5: A symmetrical thin lens has focal points on both sides at equal distances from its center. (a) Parallel rays entering a converging lens from the right meet at the focal point on the left. (b) Parallel rays entering a diverging lens from the right appear to originate from the focal point on the right.

Another useful property of a thin lens is that a ray passing through its center is deflected by only a very small amount. In ray diagrams, this ray is treated as continuing in a straight line. Together with the behavior of rays parallel to the axis and rays passing through a focal point, this provides a simple set of rules for locating images.

Take-Home Exploration: Examining Eyeglasses

Look through a pair of eyeglasses or reading glasses from both directions. Observe whether objects appear larger or smaller and whether the effect changes when the lenses are reversed. Based on your observations, decide whether each lens is converging or diverging and whether the thin-lens approximation seems reasonable. Do not use someone else's corrective lenses while walking or performing tasks that require accurate vision.

A ray passes through the center of a converging lens and a diverging lens with negligible change in direction.
Figure 69.6: A ray passing through the center of a thin lens is deflected by a negligible amount. In a ray diagram, it is therefore drawn as continuing along a straight line.

Ray tracing can be performed using a pencil, straightedge, and a carefully drawn diagram. Although light leaves every point on an object in many directions, only two rays are needed to locate the corresponding point on the image. A third ray can be added to check the construction. The most useful rays are those whose paths can be predicted using the following rules.

Rules for Ray Tracing Through Thin Lenses

Converging lenses:

  1. A ray entering the lens parallel to the principal axis exits through the focal point on the opposite side of the lens.
  2. A ray that passes through the focal point on the object's side of the lens exits parallel to the principal axis.
  3. A ray passing through the center of the lens continues in a straight line with negligible deflection.

Diverging lenses:

  1. A ray entering the lens parallel to the principal axis exits in a direction that appears to originate from the focal point on the object's side of the lens.
  2. A ray directed toward the focal point on the opposite side of the lens exits parallel to the principal axis.
  3. A ray passing through the center of the lens continues in a straight line with negligible deflection.

When constructing a ray diagram, begin at a specific point on the object, usually the top of an arrow representing the object. Trace at least two of the principal rays through the lens. The image of that point is located where the refracted rays meet or where their backward extensions appear to meet.

  • If the refracted rays actually meet, the lens forms a real image.
  • If the refracted rays remain separated but appear to come from a common point when extended backward, the lens forms a virtual image.

The complete image is formed by repeating this reasoning for every point on the object. In practice, tracing rays from the top of the object is sufficient to determine the image's position, height, and orientation.

Healthcare Connection: Ray Tracing in Vision

Ray tracing provides a simplified model of how the cornea and lens of the eye direct light onto the retina. In a healthy eye, rays from each point on an object are brought together at a corresponding point on the retina, forming a sharp real image. In nearsightedness or farsightedness, the rays meet either in front of or behind the retina. Corrective lenses alter the incoming ray paths so that the eye can focus them accurately on the retinal surface.

Image Formation by Thin Lenses

One of the most important applications of lenses is their ability to form images. Cameras, microscopes, telescopes, projectors, eyeglasses, and the human eye all rely on lenses to produce images that are either projected onto a surface or viewed directly. To determine where an image forms, we use ray tracing. Instead of following every light ray leaving an object, we trace just a few principal rays whose paths are easy to predict. The point where those rays intersect (or appear to intersect) determines the location of the image. Figure 69.7 illustrates image formation by a converging lens. Three principal rays are drawn from the top of the object:

  1. A ray parallel to the principal axis passes through the focal point after leaving the lens.
  2. A ray passing through the center of the lens continues in essentially a straight line.
  3. A ray passing through the focal point before reaching the lens emerges parallel to the principal axis.

All three rays intersect at the same location, showing where the image of the top of the object forms. Repeating this process for every point on the object produces the complete image.

Ray tracing for a converging lens showing three principal rays from the top of an object meeting at the top of a real inverted image.
Figure 69.7: Principal rays used to locate the image formed by a converging lens. In this case the rays meet on the opposite side of the lens, producing a real, inverted image.

The image shown in Figure 69.7 is a real image. A real image forms when light rays actually converge at the image location. Because the rays physically meet, the image can be projected onto a screen, camera sensor, photographic film, or the retina of the eye.

Figure 69.8 shows two familiar examples. A camera lens projects a real image onto the image sensor (or film), while the lens of the eye projects a real image onto the retina, where specialized photoreceptor cells convert light into electrical signals that are sent to the brain.

Healthcare Connection: Vision and the Retina

The image formed on the retina is real, inverted, and usually smaller than the object being viewed. Although the retinal image is upside down, the brain automatically interprets the visual information so that we perceive the world as upright. Many vision disorders occur because the eye cannot focus this real image precisely on the retina. Corrective lenses change the paths of incoming rays so that the image forms at the proper location.

Comparison of image formation by a camera lens and the human eye, both producing real inverted images on the image sensor or retina.
Figure 69.8: (a) A camera forms a real image on its image sensor or photographic film. (b) The eye forms a real image on the retina, where photoreceptors detect the light.

Several distances and dimensions are used to describe image formation:

  • Object distance ([latex]d_o[/latex]): the distance from the object to the center of the lens.
  • Image distance ([latex]d_i[/latex]): the distance from the center of the lens to the image.
  • Object height ([latex]h_o[/latex]): the height of the object.
  • Image height ([latex]h_i[/latex]): the height of the image.

By convention, upright images have positive heights, while inverted images have negative heights. Although carefully drawn ray diagrams provide an intuitive understanding of image formation, quantitative calculations are performed using the thin lens equation and the magnification equation.

Thin Lens Equation

[latex]\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}[/latex]

Magnification Equation

[latex]m=\frac{h_i}{h_o}=-\frac{d_i}{d_o}[/latex]

The magnification describes how much larger or smaller the image is compared with the object. A positive magnification corresponds to an upright image, while a negative magnification corresponds to an inverted image.

These two equations apply to all ideal thin lenses and allow us to predict the position, orientation, and size of images formed by cameras, microscopes, telescopes, magnifying glasses, and the human eye.

Example 69.2: Finding the Image of a Light Bulb Using Ray Tracing and the Thin Lens Equation

A clear light bulb is placed [latex]0.750~\text{m}[/latex] in front of a converging lens with a focal length of [latex]0.500~\text{m}[/latex]. Use ray tracing to estimate the image location. Then calculate:

  1. the image distance
  2. the magnification using the thin lens equations.
A light bulb placed 0.750 meters in front of a converging lens forms a real inverted image on a screen 1.50 meters from the lens.
Figure 69.9: Ray tracing predicts that the converging lens forms a real, inverted image on a screen.

Strategy

Because the object is farther from the lens than its focal length ([latex]d_o>f[/latex]), a real image will form on the opposite side of the lens. Ray tracing provides an estimate, while the thin lens equation gives the exact location. Given:

[latex]d_o=0.750~\text{m} \qquad f=0.500~\text{m}[/latex]

Solution

The principal rays intersect approximately [latex]1.50~\text{m}[/latex] from the lens. The image is inverted and about twice as tall as the object. Using the thin lens equation:

[latex]\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}[/latex]

Solving for [latex]d_i[/latex],

[latex]\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o}[/latex]
[latex]\frac{1}{d_i}=\frac{1}{0.500}-\frac{1}{0.750}=2.00-1.33=0.667~\text{m}^{-1}[/latex]
[latex]d_i=\frac{1}{0.667}=1.50~\text{m}[/latex]

The magnification is

[latex]m=-\frac{d_i}{d_o}=-\frac{1.50}{0.750}=-2.00[/latex]

Discussion

The negative magnification indicates that the image is inverted. The magnitude of the magnification shows that the image is twice as tall as the object. The agreement between ray tracing and the thin lens equation illustrates how the graphical and mathematical approaches complement one another.

A real image such as this is produced whenever a converging lens has a positive focal length and the object is located farther from the lens than the focal length ([latex]d_o>f[/latex]). Cameras, projectors, and the human eye all operate under these conditions. A different situation occurs when the object is moved closer to the lens than its focal length. This is how a magnifying glass works. Instead of producing a real image on the opposite side of the lens, the refracted rays continue to diverge. However, when extended backward, they appear to originate from a point on the same side of the lens as the object.

A converging lens forms either a real inverted image when the object is beyond the focal point or a virtual magnified image when the object is inside the focal length.
Figure 69.10: (a) When the object is beyond the focal length, the converging lens forms a real inverted image. (b) When the object is inside the focal length, the lens acts as a magnifier, producing a virtual enlarged image.

This second case is illustrated in Figure 69.11. The refracted rays never actually meet, but they appear to come from a common point on the object's side of the lens.

Ray tracing for an object inside the focal length of a converging lens producing a virtual upright magnified image.
Figure 69.11: When the object is inside the focal length of a converging lens, the refracted rays diverge but appear to originate from a common point, producing an upright, magnified virtual image.

Example 69.3: Image Produced by a Magnifying Glass

A book page is held [latex]7.50~\text{cm}[/latex] from a converging lens with a focal length of [latex]10.0~\text{cm}[/latex]. Determine:

  1. the image distance, and
  2. the magnification produced by the lens.

Strategy

The object is closer to the lens than the focal length:

[latex]d_o=7.50~\text{cm}[/latex]. Therefore, the converging lens acts as a magnifying glass and produces a virtual, upright, enlarged image. We first use the thin lens equation to find the image distance. We then use the magnification equation to determine the image size relative to the object.

Solution for (a)

Begin with the thin lens equation:

[latex]\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}[/latex]

Rearranging to solve for the image distance gives

[latex]\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o}[/latex]

Substitute the known values:

[latex]\frac{1}{d_i} = \frac{1}{10.0~\text{cm}} - \frac{1}{7.50~\text{cm}} = -0.0333~\text{cm}^{-1}[/latex]

Taking the reciprocal gives

[latex]d_i=-30.0~\text{cm}[/latex]

Solution for (b)

The magnification is

[latex]m=-\frac{d_i}{d_o}[/latex]

Substituting the image and object distances gives

[latex]m = -\frac{-30.0~\text{cm}}{7.50~\text{cm}} = 4.00[/latex]

Discussion

The image distance is negative, so the image is virtual and located on the same side of the lens as the object. The positive magnification indicates that the image is upright, while the value [latex]m=4.00[/latex] means that the image appears four times as tall as the object. This is the behavior expected from a magnifying glass: the object is placed inside the focal length, causing the lens to produce an enlarged virtual image that can be viewed by the eye but cannot be projected onto a screen.

Images Formed by Diverging Lenses

A diverging lens forms a virtual, upright, and reduced image for any position of a real object. Looking through eyeglasses designed to correct nearsightedness provides a familiar example. Objects viewed through these lenses appear smaller but remain upright, as shown in Figure 69.12.

A car viewed through a diverging lens appears upright and smaller than the actual car.
Figure 69.12: A diverging lens produces an upright, reduced virtual image. This type of lens is commonly used to correct nearsightedness. (Credit: Daniel Oines, Flickr.)

The ray tracing for a diverging lens is illustrated in Figure 69.13. A ray entering parallel to the principal axis is refracted so that it appears to originate from the focal point on the object's side of the lens. A second ray passes through the center of the lens with negligible deflection. The refracted rays do not actually meet. However, their backward extensions intersect on the same side of the lens as the object, locating the virtual image. Because the image is upright and smaller than the object, its magnification is positive but less than 1.

Ray tracing for a diverging lens. A parallel ray refracts as though it came from the near focal point, while a central ray continues straight. Their backward extensions meet to form an upright, reduced virtual image between the object and the lens.
Figure 69.13: A diverging lens produces a virtual, upright, reduced image on the same side of the lens as the object. The image is closer to the lens than the object.

Images Formed by Diverging Lenses

For a real object, a diverging lens always produces an image that is:

  • virtual, because the refracted rays do not actually meet;
  • upright, so the magnification is positive;
  • reduced, so the magnitude of the magnification is less than 1; and
  • located between the object and the lens on the object's side.

Example 69.4: Image Produced by a Diverging Lens

A book page is held [latex]7.50~\text{cm}[/latex] from a diverging lens with a focal length of [latex]-10.0~\text{cm}[/latex], similar to a lens used to correct nearsightedness. Determine:

  1. the image distance, and
  2. the magnification.

Strategy

The procedure is identical to that used for a converging lens, except that the focal length of a diverging lens is negative. We first use the thin lens equation to determine the image distance and then calculate the magnification.

Solution for (a)

Begin with the thin lens equation:

[latex]\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}[/latex]

Solving for the image distance gives

[latex]\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o}[/latex]

Substituting the known values,

[latex]\frac{1}{d_i} = \frac{1}{-10.0~\text{cm}} - \frac{1}{7.50~\text{cm}} = -0.2333~\text{cm}^{-1}[/latex]

Taking the reciprocal,

[latex]d_i=-4.29~\text{cm}[/latex]

Alternatively, the image distance may be found directly using

[latex]d_i=\frac{fd_o}{d_o-f}[/latex]

Substituting the same values gives

[latex]d_i= \frac{(-10.0)(7.50)}{7.50-(-10.0)} = \frac{-75.0}{17.5} = -4.29~\text{cm}[/latex]

Solution for (b)

The magnification is

[latex]m=-\frac{d_i}{d_o}[/latex]

Substituting the values gives

[latex]m = -\frac{-4.29~\text{cm}}{7.50~\text{cm}} = 0.571[/latex]

Discussion

The image distance is negative, indicating that the image is virtual and located on the same side of the lens as the object. The positive magnification shows that the image is upright, while the value [latex]m=0.571[/latex] indicates that the image is approximately 57% as large as the object. These characteristics are true for all images formed by a diverging lens. Regardless of where the object is placed, a diverging lens always produces a virtual, upright, reduced image located between the lens and the object. This behavior is the reason diverging lenses are used to correct nearsightedness, allowing the eye to focus distant objects properly.

Images Formed by Thin Lenses

Table 69.1 summarizes the three types of images produced by thin lenses.

Only real images can be projected onto a screen because the light rays physically converge at the image location. Virtual images cannot be projected directly; they can be seen only by looking through the lens or by using an additional optical system that forms a real image.

Table 69.1: Three Types of Images Formed by Thin Lenses
Case Condition Image Type Image Distance (di) Magnification (m)
Case 1 Converging lens ([latex]f>0[/latex]) with [latex]d_o>f[/latex] Real Positive Negative (inverted)
Case 2 Converging lens ([latex]f>0[/latex]) with [latex]f>d_o[/latex] Virtual Negative Positive, [latex]m>1[/latex] (upright and enlarged)
Case 3 Diverging lens ([latex]f0[/latex]) Virtual Negative Positive, [latex]m1[/latex] (upright and reduced)

In the next section, we will see that mirrors produce the same three categories of images as thin lenses. The same sign conventions and many of the same mathematical relationships apply to spherical mirrors.

Take-Home Experiment: Measuring the Focal Length of a Lens

Collect several inexpensive lenses, such as magnifying glasses or old camera lenses, and determine whether they are converging or diverging. In general, lenses that are thicker in the center than at the edges are converging, while those that are thinner in the center are diverging. On a sunny day, use a converging lens to focus sunlight onto a sheet of paper. Measure the distance between the lens and the smallest, brightest spot of light. This distance is the focal length of the lens.

Safety note: Concentrated sunlight can become hot enough to burn paper or skin. Perform this activity carefully and never look directly at the Sun through a lens.

Problem-Solving Strategy: Thin Lenses

Many lens problems can be solved systematically by following these steps:

  1. Identify the optical system. Determine that the problem involves a thin lens and decide whether it is converging or diverging.
  2. Sketch the situation. Draw the lens, principal axis, focal points, object, and image if known. Label all given distances and heights.
  3. Identify the unknowns. Decide whether you are solving for image distance, image height, magnification, focal length, or lens power.
  4. List the known quantities. Record all given values and determine which image case (Case 1, 2, or 3) applies. Table 69.1 provides useful guidance for predicting the signs of the variables.
  5. Use ray tracing when appropriate. A carefully drawn ray diagram often predicts whether the image is real or virtual, upright or inverted, and larger or smaller than the object.
  6. Apply the thin lens equations. Use
    [latex]\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}[/latex]

    and

    [latex]m=\frac{h_i}{h_o}=-\frac{d_i}{d_o}[/latex]

    Solve algebraically before substituting numerical values whenever possible.

  7. Check your answer. Verify that the signs and magnitudes of your results agree with the expected image type. For example, a virtual image should have a negative image distance, while a real image should have a positive image distance.

Common Misconceptions

  • Every point on an object emits light in many directions. A lens forms an image by collecting rays from every point on the object—not just from a single ray.
  • A complete image can still be formed even if part of a lens is covered. Covering half the lens makes the image dimmer because less light reaches the image, but the entire image remains visible.

Conceptual Questions

  1. A flat sheet of glass, such as a window, can be thought of as a lens with an extremely long focal length. If this is the case, where would it form an image? How are the object distance [latex]d_o[/latex] and image distance [latex]d_i[/latex] related?
  2. When looking through a window, especially if the room on the opposite side is dark, you may notice a faint double reflection. Explain why two reflected images can sometimes be seen.
  3. When focusing a camera, the lens moves closer to or farther from the image sensor. If the camera lens behaves like a thin lens, why can't the lens remain at a fixed distance from the sensor for both nearby and distant objects?
  4. An ideal thin lens has two focal points, one on each side, at equal distances from its center. Such a lens should behave the same regardless of which side light enters. Look through a pair of eyeglasses from both directions and discuss whether they behave like ideal thin lenses.
  5. How would the focal length of a lens change if it were completely submerged in water instead of being surrounded by air? Explain your reasoning in terms of refraction.

Problems & Exercises

  1. What is the optical power, in diopters, of a camera lens with a focal length of [latex]50.0~\text{mm}[/latex]?
  2. A camera's zoom lens has an adjustable focal length ranging from [latex]80.0~\text{mm}[/latex] to [latex]200~\text{mm}[/latex]. Determine the corresponding range of lens powers.
  3. Reading glasses are labeled [latex]+1.75~\text{D}[/latex]. What is the focal length of these lenses?
  4. Your eyeglass prescription is [latex]-4.50~\text{D}[/latex]. Determine the focal length of the corrective lenses.
  5. A camera uses a lens with a focal length of [latex]35.0~\text{mm}[/latex] to photograph a flower located [latex]75.0~\text{cm}[/latex] away. Determine how far behind the lens the image sensor (or film) must be placed. Clearly identify the knowns, unknowns, and equations used.
  6. A slide projector has a lens with a focal length of [latex]100~\text{mm}[/latex].
    1. If the slide is positioned [latex]103~\text{mm}[/latex] from the lens, how far away should the screen be to produce a sharp image?
    2. If the slide measures [latex]24.0~\text{mm}\times36.0~\text{mm}[/latex], determine the dimensions of the projected image.
  7. A dermatologist examines a mole using a magnifying glass with a focal length of [latex]15.0~\text{cm}[/latex], holding the lens [latex]13.5~\text{cm}[/latex] from the skin.
    1. Where is the image located?
    2. What is the magnification?
    3. If the mole has a diameter of [latex]5.00~\text{mm}[/latex], what is the diameter of its image?
  8. Your parent's reading glasses have a power of [latex]+2.25~\text{D}[/latex]. How far from a sheet of paper should you hold the glasses to focus sunlight onto the paper?
  9. A camera with a [latex]50.0~\text{mm}[/latex] focal length lens photographs a person standing [latex]3.00~\text{m}[/latex] away.
    1. How far behind the lens must the image sensor be placed?
    2. If the image sensor is [latex]36.0~\text{mm}[/latex] tall, what fraction of a [latex]1.75~\text{m}[/latex]-tall person will fit in the image?
    3. Comment on whether your answer is consistent with everyday photography.
  10. A macro camera lens has a focal length of [latex]22.0~\text{mm}[/latex]. The greatest distance it can be positioned from the image sensor is [latex]33.0~\text{mm}[/latex].
    1. What is the closest object that can be brought into focus?
    2. What is the corresponding magnification?
  11. A camera uses a lens with focal length [latex]50.0~\text{mm}[/latex] positioned [latex]51.0~\text{mm}[/latex] from the image sensor.
    1. How far away is the object that is in focus?
    2. If the image height is [latex]2.00~\text{cm}[/latex], determine the object's height.
  12. A magnifying glass is held [latex]5.00~\text{cm}[/latex] from a rare coin and produces a magnification of [latex]3.00[/latex].
    1. Determine the focal length of the magnifying glass.
    2. Calculate its optical power in diopters.
    3. Compare this value with the powers of typical reading glasses (about [latex]1.0\text{–}4.0~\text{D}[/latex]). Should a magnifier have a larger or smaller power?
  13. A diverging lens has a power of [latex]-4.00~\text{D}[/latex]. What magnification is produced when an object is placed [latex]25.0~\text{cm}[/latex] from the lens?
  14. In Example 69.3, a book placed [latex]7.50~\text{cm}[/latex] from a magnifying glass with focal length [latex]10.0~\text{cm}[/latex] produced a magnification of 4.00.
    1. Determine the magnification when the book is held [latex]8.50~\text{cm}[/latex] from the lens.
    2. Repeat for an object distance of [latex]9.50~\text{cm}[/latex].
    3. Describe the trend in magnification as the object approaches the focal point.
  15. A telephoto lens with focal length [latex]200~\text{mm}[/latex] is used to photograph mountains located [latex]10.0~\text{km}[/latex] away.
    1. Where is the image formed?
    2. What is the height of the image of a [latex]1000~\text{m}[/latex]-tall cliff?
  16. A camera with a [latex]100~\text{mm}[/latex] focal length lens photographs the Sun. The Sun has a diameter of [latex]1.40\times10^6~\text{km}[/latex] and is approximately [latex]1.50\times10^8~\text{km}[/latex] from Earth. Determine the diameter of the Sun's image formed on the image sensor.
  17. Starting with the thin lens equation and the definition of magnification, show that the magnification can also be written as
    [latex]m=\frac{f}{f-d_o}[/latex]

Glossary

converging lens
A convex lens that causes light rays traveling parallel to its principal axis to converge at a focal point on the opposite side of the lens.
diverging lens
A concave lens that causes parallel light rays to spread apart, appearing to originate from a focal point on the same side of the lens as the incoming light.
focal point
The point where parallel rays converge after passing through a converging lens, or the point from which rays appear to diverge after passing through a diverging lens.
focal length
The distance from the center of a lens to its focal point.
magnification
The ratio of image height to object height, indicating how much larger or smaller the image is compared with the object.
power
The reciprocal of the focal length (measured in meters), expressed in diopters (D).
real image
An image formed where light rays actually converge and that can be projected onto a screen.
virtual image
An image formed where light rays only appear to originate. A virtual image cannot be projected onto a screen.
definition

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Introductory Physics for the Health and Life Sciences II Copyright © 2012 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.