Focal length for a spherical mirror Focal length for a spherical mirror
Mirror equation in terms of focal length
Mirror equation in terms of focal length
24.4 | REFLECTIONS AND IMAGES PRODUCED BY CURVED MIRRORS 813 Mirror
Rays are approximately parallel at mirror.
Object is very far away (so⬇ ).
Figure 24.31 When an object is very far away (so ⬇ ), the rays are approximately parallel to the principal axis. The object is said to be “at infi nity.”
2So far, all our objects have been in front of the mirror, but when two or more mirrors and lenses are combined, it is possible for an object to be behind a mirror. We’ll see such examples in Chapter 26.
(Continued)
concave mirror: f5 R 2
convex mirror: f5 2R
2 (24.16) 5. The object and image heights ho and hi are positive if the object/image is
upright and negative if it is inverted.
We’ll follow these sign conventions (summarized in Fig. 24.32) throughout this book. Other books may follow different sign conventions, however, so you should always check the sign conventions before applying relations involving so, si, and so forth from other books.
CO N C E P T C H E C K 2 4 .7 | Object and Image Distances
Figure 24.27 shows an object and its image as formed by a curved mirror. What are the signs (positive or negative) of (a) the object distance so, (b) the image distance si, and (c) the focal length f?
Applications of the Mirror Equation
To see the mirror equation in action, consider the problem sketched in Figure 24.33 in which an object is placed a distance so 2.5 m in front (to the left) of a concave mirror with R 1.5 m. Let’s fi nd the location of the image and also determine if the image is real or virtual. The distances so and si along with the three rays in our standard ray-tracing diagram are shown in Figure 24.33.
(Compare with Fig. 24.26.) While this diagram gives information about the image, the mirror equation provides quantitative answers.
We fi rst use the given value of R to fi nd the focal length f R/2 0.75 m and show the position of the focal point F in the ray diagram. (We could use the form of the mirror equation in Eq. 24.13 to fi nd the image location si without fi rst fi nding f, but it is usually a good idea to include the focal point F in all ray diagrams.) The mirror equation can then be used to fi nd the image location. Solving Equation 24.15 for 1/si, we get
1 so1 1
si5 1 f 1
si51 f 2 1
so
Substituting values for the focal length and object distance gives 1
si5 1
0.75 m2 1
2.5 m 50.933 m21 si51.07 m
Here we keep three signifi cant fi gures to avoid roundoff problems later in this exam- ple. Since si is positive, we expect to fi nd a real image on the left (in front) of the mirror, which is confi rmed by the ray diagram. The ray diagram also shows that the image is inverted, so according to our sign conventions the image height hi is nega- tive. To fi nd the value of hi, we recall the triangles in Figure 24.30B, which led to Equation 24.11; now we must be careful about the signs of ho and hi. We see from Figure 24.33 that the object is above the principal axis, so according to our sign conventions ho is positive. Since hi is below the principal axis, hi is negative. We thus have
2hi si 5ho
so (24.17) Figure 24.32 Sign conven-
tions for application of the mirror equation.
F Principal axis
CONCAVE MIRROR:
f positive so negative so positive
o and i positive
si negative si positive
o and i negative
F Principal axis
Other conventions as for concave mirror
CONVEX MIRROR:
f negative
Figure 24.33 Using ray tracing together with the mirror equation to fi nd the location of an image.
o
F C
i
Principal axis Object
Image
so 2.5 m si R 1.5 m
Solving for the magnifi cation m hi /ho (Eq. 24.10), we get m5 hi
ho 5 2si
so
(24.18)
m5 21.07 m
2.5 m 5 20.43
The negative value of m means that we have an inverted image (as already noted).
Because @m@ 1, the image is reduced as shown in Figure 24.33.
24.4 | REFLECTIONS AND IMAGES PRODUCED BY CURVED MIRRORS 815
E X A M P L E 2 4 . 5 Applying the Mirror Equation to a Convex Mirror
Consider again the problem sketched in Figure 24.33, but now assume the mirror is convex. (a) Construct the ray diagram. (b) Is the image inverted or upright? (c) Find the image location si. (d) Find the size of the image. Assume the same values of R and so as in Figure 24.33 and assume the object height is ho 1.0 cm.
RECOGNIZE T HE PRINCIPLE
A ray diagram should always be your fi rst step in problems involving image formation.
We fi rst construct a ray diagram, similar to the ray diagram we used in dealing with a concave mirror. We next use our ray diagram to apply the sign conventions for the image and object distances and heights and for the focal length. The mirror equation can then be applied to fi nd the image’s location and properties.
SK E TCH T HE PROBLEM
(a) We follow our recommended procedure for ray tracing applied to spherical mir- rors. Step 1: Figure 24.34A shows the mirror with its principal axis, center of cur- vature C, and focal point F. According to our sign conventions, the focal length for a convex mirror is negative, so F is behind the mirror. The focal length is (from Eq. 24.16) f R/2 0.75 m. Step 2: We add the object to our picture. Step 3: The focal ray, parallel ray, and central ray are added (Fig. 24.34B). The focal ray is initially directed toward the focal point; after refl ecting from the mirror, it travels parallel to the axis. The parallel ray initially travels parallel to the principal axis and then refl ects from the mirror so that it appears to emanate from the focal point. The central ray is initially directed toward the center of curvature of the mirror and refl ects back toward the object. Step 4: Extrapolations of the rays in Figure 24.34B intersect at the image point, which is behind the mirror.
IDENT IF Y T HE REL AT IONSHIPS A ND SOLV E
(b) The ray diagram in Figure 24.34B shows that the image is behind the mirror (a virtual image), is upright, and is smaller than the object.
(c) We next use our sign conventions for image location. The object is in front of the mirror, so so is positive with so2.5 m. Using this value in the mirror equation (Eq. 24.15) along with the focal length determined above, we get
1 si5 1
f 2 1
so 5 1
20.75 m2 1 2.5 m si5 0.58 m
The negative value of si indicates that the image is to the right of the mirror (behind), so it is a virtual image as we noted in connection with the ray diagram in Figure 24.34B.
(d) We can fi nd the image height hi using Equation 24.17; we get 2 hi
si 5 ho so
Figure 24.34 Example 24.5.
Ray diagram for a convex mir- ror. A A sketch of just the object, mirror, focal point, and center of curvature, in preparation for ray tracing. B Same as part A, but now with the parallel, focal, and central rays included.
F C
Principal axis
Image Object
Object
F C
A
B
f R
so
si R/2
Note that in this example, hi is positive while si is negative. Solving for hi and using our result for si gives
hi5 2hoasi
sob 5 2hoa20.58 m
2.5 m b 5 10.23ho 0.23 cm (1) where we have used the given value of ho 1.0 cm.
According to our sign conventions, a positive value of ho corresponds to an upright object as confi rmed in Figure 24.34B. A positive value of ho in Equation (1) gives a positive value for hi, so we again fi nd that the image is upright .
What does it mean?
When an object is placed in front of a convex mirror, the solution of the mirror equation for si is always negative, so the image will always be virtual. In addition, if the object is upright, the image will always be upright. This property is handy in many applications, one of which is discussed in Example 24.6.
E X A M P L E 2 4 . 6 Designing a Car Rearview Side Mirror
The external rearview side mirrors on most cars carry the phrase “objects in the mir- ror are closer than they appear.” (a) What does this phrase tell us about the magni- fi cation of the image? (b) Is this mirror convex or concave? (c) Estimate the radius of curvature of a typical rearview mirror.
RECOGNIZE T HE PRINCIPLE
(a) The “apparent size” of an object gets smaller as the object is moved farther and far- ther away. When viewed by its refl ection from a mirror, this apparent size is also just the image size. When judging the distance from your eye to an object, your brain compares the known size of the object (as stored in your memory) with the size of the image (the apparent size). So, when you judge an object to be farther away that it actually is, the image must be smaller than expected for that particular object distance. Hence, the image formed by the car’s rearview side mirror must have a magnifi cation @m@ 1 . SK E TCH T HE PROBLEM
(b) The next step is to draw a ray diagram, but should this diagram be for a concave mirror or a convex mirror? A car’s external rearview mirror always produces an upright (noninverted) image, for any object distance. (You should check for yourself.) We have seen that the image produced by a concave mirror can be upright or inverted, depending on the object distance (see Figs. 24.26 and 24.28), so a rearview mirror cannot be concave. We have also seen that a convex mirror always produces a virtual image, as can be seen from the ray diagrams in Figure 24.29. Or, if we insert a nega- tive value of f (as appropriate for a convex mirror) into the mirror equation together with a positive so, we fi nd that si is always negative, which means that the image is behind the mirror and is thus a virtual image. (See also Example 24.5.) So, a rearview mirror must be convex , and the ray diagram will look like Figure 24.34B.
IDENT IF Y T HE REL AT IONSHIPS
(c) Because f R/2 (Eq. 24.16 for a convex mirror), we can fi nd the radius of curva- ture of a rearview mirror if we can estimate the focal length. If we know something about the object and image distances, the value of f can be found from the mirror equation. We have already argued that the magnifi cation is less than 1, but what is its value? A value of m 0.1 would be a very small image, whereas (based on the author’s experience) the image is only a little smaller than the true object size. We therefore estimate a magnifi cation of about m ⬇13 for a car that is so 30 m from the mirror. These estimates are only rough, but they should not be off by more than a fac- tor of three. (You should think about how you could check these values for yourself.) The magnifi cation is related to the object and image distances by Equation 24.18,
giving m si/so or simso. Inserting this information into the mirror equation, we fi nd
1 so 1 1
si5 1 f 5 1
so 2 1 mso and, solving for f, we get
f5soa m m21b SOLV E
Using our estimates for m and so gives f5soa m
m21b 5 130 m2 a1 13
321b 5 215 m The radius of curvature of the mirror is thus
R 2f 30 m What does it mean?
This large value of R means that the mirror has just a very slight curvature, which is why it is easily mistaken for a plane mirror. But what is the purpose for using a curved mirror in the fi rst place? Why not just use a fl at mirror? The reason is that a curved mirror gives a larger “fi eld of view,” allowing the driver to see objects over a wider angle and thus spot cars and other things that might be missed with a plane mirror (Fig. 24.35).
2 4 . 5 | L E N S E S
We have seen how a mirror uses refl ection to form an image. A lens uses refraction to accomplish the same thing. Typical lenses are composed of glass or plastic, and the refraction of light rays as they pass from the air into the lens and then back into the air causes the rays to be redirected as shown Figure 24.36. Refraction occurs at the two separate surfaces as shown in Figure 24.36A, but for simplicity our ray diagrams will usually be drawn as in parts B and C of the fi gure, showing a single defl ection of each ray as it passes through the lens.
In the simplest case, the lens surfaces are spherical (Fig. 24.37). A careful treat- ment of the refraction of rays that are close to the principal axis shows that if the incident rays are parallel to the principal axis of the lens, then after passing through the lens the rays all pass through a point on the axis called the focal point, F. This is true for parallel rays incident from either side of the lens (parts B and C of Fig.
24.36). Hence, there are two focal points, and they are at identical distances from the lens. This symmetry is related to the reversibility principle of light propagation (Section 24.1).
Lenses can be classifi ed according to the curvatures of their two surfaces (Fig.
24.38). Given the radius of curvature of each surface (Fig. 24.37), it is possible to use Snell’s law to calculate the focal length. (We’ll see how that works in Section 24.6.) For most applications, we only need to know two things about a lens: the value of the focal length f and whether the lens is converging or diverging. The lenses in Figure 24.36 are converging lenses: all the incoming rays intersect at the focal point on the opposite side. A ray diagram for a diverging lens is shown in Figure 24.39. In that case, parallel rays from the left diverge after passing through the lens, but they all extrapolate back to the focal point on the left (the same side as the incident rays).
24.5 | LENSES 817 A
B
Refraction occurs at both of these surfaces.
Simplified ray diagram showing a single deflection of each ray.
Focal point
Axis
Focal point
F F
C
Axis
F F
Axis
F
Figure 24.36 A Light rays are refracted at both surfaces of a lens.
When parallel light rays strike a converging lens, the refracted rays converge at the focal point F of the lens. B For simplicity, most ray diagrams with lenses show only a single “defl ection” of a ray as it passes through the lens. C If light strikes the lens from the opposite side, the rays converge at a second focal point F on the left side of the lens.
Figure 24.35 Example 24.6.
View through a car’s rearview side mirror.
© Dynamic Graphics/Jupiterimages
Forming an Image with a Lens: Ray Tracing
Our next job is to analyze the images produced by a lens. We’ll start with a ray-tracing analysis and then consider how to calculate quantitatively the image properties. Let’s begin with the case of a converging lens. Figure 24.40A shows the object along with the focal points on the left (FL) and right (FR) sides of the lens. In a typical case, we know the distance so from the object to the lens and the object’s height ho, and we want to fi nd the location of the image and its size. Figure 24.40B shows many rays emanating from the object; these rays are refracted by the lens and then intersect to form the image. For simplicity, this diagram shows only rays that emanate from the tip of the object, but rays also emanate from other points on the object and inter- sect to form other points on the image. In Figure 24.40C, we show only three rays that emanate from the tip of the object; these three rays are particularly useful in a ray tracing analysis. The parallel ray is initially parallel to the principal axis of the lens. The lens refracts this ray so that it passes through the focal point FR on the right. The focal ray passes through the focal point on the left. When the focal ray strikes the lens, the ray is refracted to be par- allel to the axis on the right. The central ray passes straight through the cen- ter of the lens (denoted as C). If the lens is very thin (a good approximation for many lenses), this ray is not defl ected by the lens. These three rays come together at the tip of the image on the right of the lens. In this case, the image is inverted, and because the rays pass through the image, the image is real.
This approach to ray tracing with lenses is described by the following rules.
Ray tracing applied to lenses
1. Construct a fi gure showing the lens and the principal axis of the lens.