Illustrate image formation in a flat mirror.Explain with ray diagrams the formation of an image using spherical mirrors.Determine focal length and magnification given radius of curvature, distance of object and image.

You are watching: If you are working with a convex mirror (f<0), which of the following describes the image?

We only have to look as far as the nearest bathroom to find an example of an image formed by a mirror. Images in flat mirrors are the same size as the object and are located behind the mirror. Like lenses, mirrors can form a variety of images. For example, dental mirrors may produce a magnified image, just as makeup mirrors do. Security mirrors in shops, on the other hand, form images that are smaller than the object. We will use the law of reflection to understand how mirrors form images, and we will find that mirror images are analogous to those formed by lenses.

Figure 1 helps illustrate how a flat mirror forms an image. Two rays are shown emerging from the same point, striking the mirror, and being reflected into the observer’s eye. The rays can diverge slightly, and both still get into the eye. If the rays are extrapolated backward, they seem to originate from a common point behind the mirror, locating the image. (The paths of the reflected rays into the eye are the same as if they had come directly from that point behind the mirror.) Using the law of reflection—the angle of reflection equals the angle of incidence—we can see that the image and object are the same distance from the mirror. This is a virtual image, since it cannot be projected—the rays only appear to originate from a common point behind the mirror. Obviously, if you walk behind the mirror, you cannot see the image, since the rays do not go there. But in front of the mirror, the rays behave exactly as if they had come from behind the mirror, so that is where the image is situated.


Figure 2. (a) Parallel rays reflected from a large spherical mirror do not all cross at a common point. (b) If a spherical mirror is small compared with its radius of curvature, parallel rays are focused to a common point. The distance of the focal point from the center of the mirror is its focal length f. Since this mirror is converging, it has a positive focal length.

Just as for lenses, the shorter the focal length, the more powerful the mirror; thus, P=frac1f\ for a mirror, too. A more strongly curved mirror has a shorter focal length and a greater power. Using the law of reflection and some simple trigonometry, it can be shown that the focal length is half the radius of curvature, or f=fracR2\, where R is the radius of curvature of a spherical mirror. The smaller the radius of curvature, the smaller the focal length and, thus, the more powerful the mirror

The convex mirror shown in Figure 3 also has a focal point. Parallel rays of light reflected from the mirror seem to originate from the point F at the focal distance f behind the mirror. The focal length and power of a convex mirror are negative, since it is a diverging mirror.

Ray tracing is as useful for mirrors as for lenses. The rules for ray tracing for mirrors are based on the illustrations just discussed:


Figure 4. A case 1 image for a mirror. An object is farther from the converging mirror than its focal length. Rays from a common point on the object are traced using the rules in the text. Ray 1 approaches parallel to the axis, ray 2 strikes the center of the mirror, and ray 3 goes through the focal point on the way toward the mirror. All three rays cross at the same point after being reflected, locating the inverted real image. Although three rays are shown, only two of the three are needed to locate the image and determine its height.

Example 1. A Concave Reflector

Electric room heaters use a concave mirror to reflect infrared (IR) radiation from hot coils. Note that IR follows the same law of reflection as visible light. Given that the mirror has a radius of curvature of 50.0 cm and produces an image of the coils 3.00 m away from the mirror, where are the coils?

Strategy and Concept

We are given that the concave mirror projects a real image of the coils at an image distance di=3.00 m. The coils are the object, and we are asked to find their location—that is, to find the object distance do. We are also given the radius of curvature of the mirror, so that its focal length is f=fracR2=25.0 ext cm\ (positive since the mirror is concave or converging). Assuming the mirror is small compared with its radius of curvature, we can use the thin lens equations, to solve this problem.


Since di and f are known, thin lens equation can be used to find do: frac1d_ exto+frac1d_ exti=frac1f\.

Rearranging to isolate do gives frac1d_ exto=frac1f-frac1d_ exti\.

Entering known quantities gives a value for frac1d_ exto\frac1d_ exto=frac10.250 ext m-frac13.00 ext m=frac3.667 extm\.

This must be inverted to find do: d_ exto=frac1 ext m3.667=27.3 ext cm\.


Note that the object (the filament) is farther from the mirror than the mirror’s focal length. This is a case 1 image (do > f and f positive), consistent with the fact that a real image is formed. You will get the most concentrated thermal energy directly in front of the mirror and 3.00 m away from it. Generally, this is not desirable, since it could cause burns. Usually, you want the rays to emerge parallel, and this is accomplished by having the filament at the focal point of the mirror.

Note that the filament here is not much farther from the mirror than its focal length and that the image produced is considerably farther away. This is exactly analogous to a slide projector. Placing a slide only slightly farther away from the projector lens than its focal length produces an image significantly farther away. As the object gets closer to the focal distance, the image gets farther away. In fact, as the object distance approaches the focal length, the image distance approaches infinity and the rays are sent out parallel to one another.

Example 2. Solar Electric Generating System

One of the solar technologies used today for generating electricity is a device (called a parabolic trough or concentrating collector) that concentrates the sunlight onto a blackened pipe that contains a fluid. This heated fluid is pumped to a heat exchanger, where its heat energy is transferred to another system that is used to generate steam—and so generate electricity through a conventional steam cycle. Figure 5 shows such a working system in southern California. Concave mirrors are used to concentrate the sunlight onto the pipe. The mirror has the approximate shape of a section of a cylinder. For the problem, assume that the mirror is exactly one-quarter of a full cylinder.

If we wish to place the fluid-carrying pipe 40.0 cm from the concave mirror at the mirror’s focal point, what will be the radius of curvature of the mirror?Per meter of pipe, what will be the amount of sunlight concentrated onto the pipe, assuming the insolation (incident solar radiation) is 0.900 k W/m2?If the fluid-carrying pipe has a 2.00-cm diameter, what will be the temperature increase of the fluid per meter of pipe over a period of one minute? Assume all the solar radiation incident on the reflector is absorbed by the pipe, and that the fluid is mineral oil.Strategy

To solve an Integrated Concept Problem we must first identify the physical principles involved. Part 1 is related to the current topic. Part 2 involves a little math, primarily geometry. Part 3 requires an understanding of heat and density.

Solution to Part 1

To a good approximation for a concave or semi-spherical surface, the point where the parallel rays from the sun converge will be at the focal point, so = 2= 80.0 cm.

Solution to Part 2

The insolation is 900 W /m2. We must find the cross-sectional area A of the concave mirror, since the power delivered is 900 W /m2 × A. The mirror in this case is a quarter-section of a cylinder, so the area for a length L of the mirror is extA=frac14left(2pi extR ight) extL\. The area for a length of 1.00 m is then

displaystyle extA=fracpi2Rleft(1.00 ext m ight)=fracleft(3.14 ight)2left(0.800 ext m ight)left(1.00 ext m ight)=1.26^2\

The insolation on the 1.00-m length of pipe is then

displaystyleleft(9.00 imes10^2frac extW extm^2 ight)left(1.26 ext m^2 ight)=1130 ext W\

Solution to Part 3

The increase in temperature is given by mcΔT. The mass m of the mineral oil in the one-meter section of pipe is

eginarraylllm&=& ho extV= hopileft(fracd2 ight)^2left(1.00 ext m ight)\ ext &=&left(8.00 imes10^2 ext kg/m^3 ight)left(3.14 ight)left(0.0100 ext m ight)^2left(1.00 ext m ight)\ ext &=&0.251 ext kgendarray\

Therefore, the increase in temperature in one minute is

eginarraylllDeltaT&=&fracQmc\ ext &=&fracleft(1130 ext W ight)left(60.0 ext s ight)left(0.251 ext kg ight)left(1670 ext Jcdot ext kg/^circ ext C ight)\ ext &=&162^circ extCendarray\

Discussion for Part 3

Figure 7. Case 3 images for mirrors are formed by any convex mirror. Ray 1 approaches parallel to the axis, ray 2 strikes the center of the mirror, and ray 3 approaches toward the focal point. All three rays appear to originate from the same point after being reflected, locating the upright virtual image behind the mirror and showing it to be smaller than the object. (b) Security mirrors are convex, producing a smaller, upright image. Because the image is smaller, a larger area is imaged compared to what would be observed for a flat mirror (and hence security is improved). (credit: Laura D’Alessandro, Flickr)

Section Summary

The characteristics of an image formed by a flat mirror are: (a) The image and object are the same distance from the mirror, (b) The image is a virtual image, and (c) The image is situated behind the mirror.Image length is half the radius of curvature: f=fracR2\A convex mirror is a diverging mirror and forms only one type of image, namely a virtual image.

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Conceptual Questions

What are the differences between real and virtual images? How can you tell (by looking) whether an image formed by a single lens or mirror is real or virtual?Can you see a virtual image? Can you photograph one? Can one be projected onto a screen with additional lenses or mirrors? Explain your responses.Is it necessary to project a real image onto a screen for it to exist?At what distance is an image always located—at do, di, or f?Under what circumstances will an image be located at the focal point of a lens or mirror?What is meant by a negative magnification? What is meant by a magnification that is less than 1 in magnitude?Can a case 1 image be larger than the object even though its magnification is always negative? Explain.Figure 8 shows a light bulb between two mirrors. One mirror produces a beam of light with parallel rays; the other keeps light from escaping without being put into the beam. Where is the filament of the light in relation to the focal point or radius of curvature of each mirror?

Figure 8. The two mirrors trap most of the bulb’s light and form a directional beam as in a headlight.

The two mirrors trap most of the bulb’s light and form a directional beam as in a headlight.Two concave mirrors of different sizes are placed facing one another. A filament bulb is placed at the focus of the larger mirror. The rays after reflection from the larger mirror travel parallel to one another. The rays falling on the smaller mirror retrace their paths.Devise an arrangement of mirrors allowing you to see the back of your head. What is the minimum number of mirrors needed for this task?If you wish to see your entire body in a flat mirror (from head to toe), how tall should the mirror be? Does its size depend upon your distance away from the mirror? Provide a sketch.It can be argued that a flat mirror has an infinite focal length. If so, where does it form an image? That is, how are di and do related?Why are diverging mirrors often used for rear-view mirrors in vehicles? What is the main disadvantage of using such a mirror compared with a flat one?


converging mirror: a concave mirror in which light rays that strike it parallel to its axis converge at one or more points along the axis

diverging mirror: a convex mirror in which light rays that strike it parallel to its axis bend away (diverge) from its axis

law of reflection: angle of reflection equals the angle of incidence

Selected Solutions to Problems & Exercises

1. +0.667 m

3. (a) −1.5 × 10−2 m; (b) −66.7 D

5. +0.360 m (concave)

7. (a) +0.111; (b) −0.334 cm (behind “mirror”); (c) 0.752cm

9. m=frach_ extih_ exto=-fracd_ extid_ exto=-frac-d_ extod_ exto=fracd_ extod_ exto=1Rightarrow h_ exti=h_ exto\