Image Formation by Mirrors and Lenses

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Q26.3

Q26.4

Q26.5

Image Formation by
Mirrors and Lenses

CHAPTER OUTLINE

26.1

26.2

26.3

Images Formed by Flat
Mirrors
Images Formed by
Spherical Mirrors
Images Formed by
Refraction
Thin Lenses

ConnectionMedical
Fiberscopes

26.4
26.5 Context

ANSWERS TO QUESTIONS

Q26.1 With a concave spherical mirror, for objects beyond the focal length

the image will be real and inverted. For objects inside the focal
length, the image will be virtual, upright, and magnified. Try a
shaving or makeup mirror as an example.

Q26.2 With a convex spherical mirror, all images of real objects are upright, virtual, and smaller than the

object. As seen in Question 26.1, you only get a change of orientation when you pass the focal
point—but the focal point of a convex mirror is on the non-reflecting side!

The mirror equation and the magnification equation apply to plane mirrors. A curved mirror is
made flat by increasing its radius of curvature without bound, so that its focal length goes to infinity.
q= − . The virtual image is as far behind the mirror

= − ; therefore, p

= we have

From

0

1
p

1
+ =
q

1
f

1
p

1
q

as the object is in front. The magnification is M

= 1 . The image is right side up and actual

q
= − =
p

p
p

Stones at the bottom of a clear stream always appear closer to the surface because light is refracted
away from the normal at the surface. Example 26.5 in the textbook shows that its apparent depth is
three quarters of its actual depth.

For definiteness, we consider real objects ( p > 0 ).

(a)

For M

= − to be negative, q must be positive. This will happen in

=

1
q

1
f

1
− if p
p

f> , if the

object is farther than the focal point.

size.

q
p

q
p

(b)

For M

= − to be positive, q must be negative.

From

=

1
q

1
f

1
p

− we need p

f< . continued on next page 713 714 Image Formation by Mirrors and Lenses (c) (d) (e) (f) For a real image, q must be positive. As in part (a), it is sufficient for p to be larger than f. For q < 0 we need p f< . For M > 1 , we consider separately M < −1 and M > 1 .

If M

= − < −1, we need > 1

q
p

From

1
p

1
+ =
q

1
f

,

+

1
p

1
>
f

q
p

1
p

p
2

or

< f or − >q

p

or

or

or

or

or

q p>

1
q

1
< . p 2 p 1 >
f

f< 2 . p p< − q >
0

1
f

Now if − >q

1

p

we may require q < 0 , since then = 1 p 1 f 1 − q with gives > − as required

1
p

For q < 0 in = − we need 1 f 1 p 1 q 1 q or − >p

q.

f< . p Thus the overall condition for an enlarged image is simply p f< 2 . For M < 1 , we have the reverse of part (e), requiring p f> 2 .

Q26.6

Never.

Always.

Using the same analysis as in Question 26.5 except f < 0 . (a) (b) (c) (d) (e) (f) Always. Always. Never. Never, for light rays passing through the lens will always diverge. Chapter 26 715 Q26.7 We assume the lens has a refractive index higher than its surroundings. For the biconvex lens in Figure 26.20(a), R1 0> and R2

a
0< . Then all terms in n − 1 are positive and f > 0 . For the

F
f
HG

1
R
1

−

1
R

2

I
KJ

>

other two lenses in part (a) of the figure, R1 and R2 are both positive but R1 is less than R2 . Then
1
R
1

1
R
2
For the biconcave lens and the plano-concave lens in Figure 26.20(b), R1

and the focal length is again positive.

0< and R2 0> . Then

both terms are negative in

and the focal length is negative. For the middle lens in part (b)

1
R
1

−

1
R

2

of the figure, R1 and R2 are both positive but R1 is greater than R2 . Then

and the focal

1
R
1

< 1 R 2 length is again negative. Both words are inverted. However OXIDE has up-down symmetry whereas LEAD does not. Q26.8 Q26.9 In the diagram, only two of the three principal rays have been used to locate images to reduce the amount of visual clutter. The upright shaded arrows are the objects, and the correspondingly numbered inverted arrows are the images. As you can see, object 2 is closer to the focal point than object 1, and image 2 is farther to the left than image 1. O1 O2 F C I2 I 1 FIG. Q26.9 V C O I V FIG. Q26.10 Q26.10 As in the diagram, let the center of curvature C of the fishbowl and the bottom of the fish define the optical axis, intersecting the fishbowl at vertex V. A ray from the top of the fish that reaches the bowl surface along a radial line through C has angle of incidence zero and angle of refraction zero. This ray exits from the bowl unchanged in direction. A ray from the top of the fish to V is refracted to bend away from the normal. Its extension back inside the fishbowl determines the location of the image and the characteristics of the image. The image is upright, virtual, and enlarged. Q26.11 Because when you look at the inversion clearly displays the name of the AMBULANCE behind you. Do not jam on your brakes when a MIAMI city bus is right behind you. in your rear view mirror, the apparent left-right Q26.12 With the meniscus design, when you direct your gaze near the outer circumference of the lens you receive a ray that has passed through glass with more nearly parallel surfaces of entry and exit. Thus, the lens minimally distorts the direction to the object you are looking at. If you wear glasses, turn them around and look through them the wrong way to maximize this distortion. 716 Image Formation by Mirrors and Lenses Q26.13 An infinite number. In general, an infinite number of rays leave each point of any object and travel in all directions. Note that the three principal rays that we use for imaging are just a subset of the infinite number of rays. All three principal rays can be drawn in a ray diagram, provided that we extend the plane of the lens as shown in Figure Q26.13. O F F I FIG. Q26.13 Q26.14 Absolutely. Only absorbed light, not transmitted light, contributes internal energy to a transparent object. A clear lens can stay ice-cold and solid as megajoules of light energy pass through it. Q26.15 The focal point is defined as the location of the image formed by rays originally parallel to the axis. An object at a large but finite distance will radiate rays nearly but not exactly parallel. Infinite object distance describes the definite limiting case in which these rays become parallel. To measure the focal length of a converging lens, set it up to form an image of the farthest object you can see outside a window. The image distance will be equal to the focal length within one percent or better if the object distance is a hundred times larger or more. Q26.16 Use a converging lens as the projection lens in a slide projector. Place the brightly illuminated slide slightly farther than its focal length away from it, so that the lens will produce a real, inverted, enlarged image on the screen. Q26.17 Make the mirror an efficient reflector (shiny). Make it reflect to the image even rays far from the axis, by giving it a parabolic shape. Most important, make it large in diameter to intercept a lot of solar power. And you get higher temperature if the image is smaller, as you get with shorter focal length; and if the furnace enclosure is an efficient absorber (black). Q26.18 The artist’s statements are accurate, perceptive, and eloquent. The image you see is “almost one’s whole surroundings,” including things behind you and things farther in front of you than the globe is, but nothing eclipsed by the opaque globe or by your head. For example, we cannot see Escher’s index and middle fingers or their reflections in the globe. optical focus. The principal axis will always lie in a line that runs through the center of the sphere and the bridge of your nose. Outside the globe, you are at the center of your observable universe. If you wink at the ball, the center of the looking-glass world hops over to the location of the image of your open eye. The point halfway between your eyes is indeed the focus in a figurative sense, but it is not an Q26.19 The three mirrors, two of which are shown as M and N in the figure to the right, reflect any incident ray back parallel to its original direction. When you look into the corner you see image I3 of yourself. Chapter 26 717 FIG. Q26.19 SOLUTIONS TO PROBLEMS Section 26.1 P26.1 Images Formed by Flat Mirrors m . 0 8 8 × 3 10 m s − 9 ~ 10 s I stand 40 cm from my bathroom mirror. I scatter light, which travels to the mirror and back to me in time showing me a view of myself as I was at that look-back time. I’m no Dorian Gray! The virtual image is as far behind the mirror as the choir is in front of the mirror. Thus, the image is 5.30 m behind the mirror. The image of the choir is 0 800 from the organist. Using similar triangles: . 5 30 . 6 10 m m m = + . View Looking Down South image of choir mirror or = m ′ h 0 600 . a ′ = h m . 6 10 0.800 m F f HG 6 10 m . 0.800 m I KJ = . 0 600 m . 4 58 m . 0.600 m h' Organist 5.30 m 0.800 m FIG. P26.2 P26.2 P26.4 P26.5 by and (2) (3) 718 Image Formation by Mirrors and Lenses P26.3 The flatness of the mirror is described R = ∞ , f = ∞ = . 0 1 f 1 p 1 + = q 1 f = 0 By our general mirror equation, or p= − . q = M = = 1 − q p ′ h h so ′ = = h 70 0. h inches . FIG. P26.3 Thus, the image is as far behind the mirror as the person is in front. The magnification is then The required height of the mirror is defined by the triangle from the person’s eyes to the top and bottom of his image, as shown. From the geometry of the triangle, we see that the mirror height must be: F HG ′ h KJ = ′F I HG h I KJ = p p 2 p − p q ′ h 2 . Thus, the mirror must be at least 35.0 inches high . A graphical construction produces 5 images, with images I1 and I 2 directly into the mirrors from the object O, , b and O I b and I , 2 ,3 I 4 I 1 , I 5 g g forming the vertices of equilateral triangles. FIG. P26.4 (1) The first image in the left mirror is 5.00 ft behind the mirror, or 10 0. ft from the position of the person. The first image in the right mirror is located 10.0 ft behind the right mirror, but this location is 25.0 ft from the left mirror. Thus, the second image in the left mirror is 25.0 ft behind the mirror, or 30 0. from the person. ft The first image in the left mirror forms an image in the right mirror. This first image is 20.0 ft from the right mirror, and, thus, an image 20.0 ft behind the right mirror is formed. This image in the right mirror also forms an image in the left mirror. The distance from this image in the right mirror to the left mirror is 35.0 ft. The third image in the left mirror is, thus, 35.0 ft behind the mirror, or 40 0. from the person. ft P26.6 The flat mirrors have Chapter 26 719 R → ∞ f → ∞. and The upper mirror M1 produces a virtual, actual sized image I1 according to = = 1 f 1 ∞ = 0 1 p 1 q 1 + 1 q = − 1 p 1 with M 1 = − = + . 1 q 1 p 1 As shown, this image is above the upper mirror. It is the object for mirror M 2 , at object distance p 2 = p 1 + . h The lower mirror produces a virtual, actual- size, right-side-up image according to 1 p 2 q 2 + 1 q = − 2 p = 0 = − b p 1 + g h 2 q p 2 2 It is virtual . Upright With magnification +1 . (a) (b) (c) (d) (e) FIG. P26.6 with M 2 = − = + and M 1 overall = M M 1 2 = 1. Thus the final image is at distance p behind the lower mirror. + h 1 It does not appear to be reversed left and right. In a top view of the periscope, parallel rays from the right and left sides of the object stay parallel and on the right and left. 720 Image Formation by Mirrors and Lenses Images Formed by Spherical Mirrors Section 26.2 P26.7 For a concave mirror, both R and f are positive. We also know that f R= 2 = 10 0. cm . (a) 1 q = − = 1 f 1 p 1 cm . 10 0 − 1 cm . 40 0 = 3 cm . 40 0 and q = 13 3. cm = M = − q p cm . 13 3 40.0 cm = − . 0 333 . The image is 13.3 cm in front of the mirror, real, and inverted . (b) 1 q = − = 1 p 1 10 0 cm . − 1 20 0 cm . = 1 20 0 cm . 1 f and q = 20 0. cm = M = − q p 20 0 cm . 20.0 cm = − . 1 00 . The image is 20.0 cm in front of the mirror, real, and inverted . (c) 1 q = − = 1 p 1 10 0 cm . − 1 10 0 cm . = 0 1 f Thus, q = infinity. No image is formed . The rays are reflected parallel to each other. *P26.8 1 q = − = − 1 f 1 p 1 . 0 275 m − 1 . 10 0 m gives q = −0 267 . m . Thus, the image is virtual . = M = − − q p − 0 267 . 10 0 m . = 0 026 7 . Thus, the image is upright +Ma f and h . diminished M < 1 c