CHAPTER 25 THE REFLECTION OF LIGHT: MIRRORSlwillia2/21/ch25.pdfof light reflecting from the mirror...

CHAPTER 25 THE REFLECTION OF LIGHT: MIRRORS

ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (e) This is the definition of a wave front (see Section 25.1). 2. (b) Rays are radial lines pointing outward from the source and perpendicular to the wave

fronts. They point in the direction of the velocity of the wave. 3. (c) When diffuse reflection occurs, the surface reflects different light rays in different

directions. 4. (c) The ray of light strikes the mirror four units down from the top of the mirror with a 45°

angle of incidence. The ray reflects from the mirror at an angle of 45° and passes through point C.

5. (a) The image is as far behind the mirror as the object is in front of the mirror. In addition,

the image and the object lie on the horizontal line that is perpendicular to the mirror. 6. (d) The image of your friend is 2 m behind the mirror. The distance between you and the

mirror is 5 m. Thus, the distance between you and your friend’s image is 7 m. 7. (b) Letters and words held up to a mirror are reversed left-to-right and right-to-left. 8. (d) As discussed in Section 25.4, rays that are parallel and near the principal axis of a

concave mirror converge at the focal point after reflecting from the mirror. 9. (a) Parallel rays that are near the principal axis converge at the focal point after reflecting

from a concave mirror. The radius of curvature is twice the focal length (see Equation 25.1), so R = 2f = 36 cm.

10. (d) This is how real and virtual images are defined. See Sections 25.3 and 25.5. 11. (a) Any ray that leaves the object and reflects from the mirror can be used in the method of

ray tracing to locate the image. 12. (c) According to the discussion in Section 25.5, a concave mirror can produce an enlarged

image, provided the object distance is less than the radius of curvature. A convex mirror cannot produce an enlarged image, regardless of where the object is located.

13. (b) A convex mirror always produces a virtual, upright image (see Section 25.5).

1248 THE REFLECTION OF LIGHT: MIRRORS

14. (e) A negative image distance means that the image is behind the mirror and, hence, is a virtual image. See the Reasoning Strategy at the end of Section 25.6.

15. (c) A convex lens always produces an upright image that is smaller than the object. 16. f = 4.0 cm 17. (b) The image distance is di = −mdo = −2(25 cm) = −50 cm (Equation 25.4). 18. f = 90.0 cm

Chapter 25 Problems 1249

CHAPTER 25 THE REFLECTION OF LIGHT:

MIRRORS PROBLEMS 1. REASONING AND SOLUTION Referring to Figure 25.9b and Conceptual Example 2, we

find the following locations for the three images:

Image 1: 2.0 m, 1.0 mImage 2: 2.0 m, 1.0 mImage 3: 2.0 m, 1.0 m

x yx yx y

= − = += + = −= + = +

2. REASONING The drawing shows the situation described. The law of reflection indicates that the angle of incidence θi is equal to the angle of reflection θr.

SOLUTION Referring to the drawing,

we see that for the incident and reflected light

tanθi =d

3.0 cmIncident light

and tanθr =

6.0 cm

Reflected light

But θi = θr, according to the law of reflection, so that tan θi = tan θr, and we have

or =23.0 cm 6.0 cmd d= l l

From the drawing we see that d + = 9.0 cm. Using the fact that = 2.0d, we obtain

d + = d + 2.0d = 9.0 cm or d = 3.0 cm Therefore, the laser should be aimed at the point at x = +3.0 cm . ____________________________________________________________________________________________

Target

+x

+y 6.0 cm

3.0 cm θi θi θr

θr

d

9.0 cm


3. SSM REASONING The drawing shows the image of the bird in the plane mirror, as seen by the camera. Note that the image is as far behind the mirror as the bird is in front of it. We can find the distance d between the camera and the image by noting that this distance is the hypotenuse of a right triangle. The base of the triangle has a length of 3.7 m + 2.1 m and the height of the triangle is 4.3 m.

SOLUTION The distance d from the camera to the image of the bird can be obtained by using the Pythagorean theorem:

( ) ( )2 23.7 m + 2.1 m 4.3 m 7.2 md = + = 4. REASONING According to the discussion about relative velocity in Section 3.4, the

velocity vIY of your image relative to you is the vector sum of the velocity vIM of your image relative to the mirror and the velocity vMY of the mirror relative to you: vIY = vIM + vMY. As you walk perpendicularly toward the stationary mirror, you perceive the mirror moving toward you in the opposite direction, so that vMY = –vYM. The velocities vYM and vIM have the same magnitude. This is because the image in a plane mirror is always just as far behind the mirror as the object is in front of it. For instance, if you move 1 meter perpendicularly toward the mirror in 1 second, the magnitude of your velocity relative to the mirror is 1 m/s. But your image also moves 1 meter toward the mirror in the same time interval, so that the magnitude of its velocity relative to the mirror is also 1 m/s. The two velocities, however, have opposite directions.

SOLUTION According to the discussion in the REASONING,

vIY = vIM + vMY = vIM – vYM (1)

Remembering that the magnitudes of both velocities vYM and vIM are the same and that the

direction in which you walk is positive, we have

vYM = +0.90 m/s and vIM = –0.90 m/s The velocity vIM is negative, because its direction is opposite to the direction in which you

walk. Substituting these values into Equation (1), we obtain

vIY = (–0.90 m/s) – (+0.90 m/s) = –1.80 m/s

3.7 m

4.3 m

2.1 m Bird

Hedge

2.1 m

4.3 m

Image


5. SSM REASONING The geometry is shown below. According to the law of reflection, the incident ray, the reflected ray, and the normal to the surface all lie in the same plane, and the angle of reflection θr equals the angle of incidence θ i . We can use the law of reflection and the properties of triangles to determine the angle θ at which the ray leaves M2.

γ

φ

65° 120°α

β

θ

M2

M1 SOLUTION From the law of reflection, we know that φ = 65°. We see from the figure that

φ +α = 90°, or α = 90°œφ = 90°œ65° = 25°. From the figure and the fact that the sum of the interior angles in any triangle is 180°, we have α +β +120° =180°. Solving for β , we find that β =180° œ(120° + 25°) = 35°. Therefore, since β + γ = 90°, we find that the angle γ is given by γ = 90°œβ = 90°œ35° = 55°. Since γ is the angle of incidence of the ray on mirror M2, we know from the law of reflection that θ = 55° .

6. REASONING The drawing at the right

shows the laser beam after reflecting from the plane mirror. The angle of reflection is α, and it is equal to the angle of incidence, which is 33.0°. Note that the angles labeled β in the drawing are also 33.0°, since they are angles formed by a line that intersects two parallel lines. Knowing these angles, we can use trigonometry to determine the distance dDC, which locates the spot where the beam strikes the floor.

SOLUTION Applying trigonometry to triangle DBC, we see that

BC BCDC

DCtan or

tand d

dd

ββ

= = (1)

Floor

1.80 m

1.10 m

33.0°

A

B

C D

α

β

β

L


The distance dBC can be determined from BC AB1.80 md d= − , which can be substituted into Equation (1) to show that

BC ABDC

1.80 mtan tand d

dβ β

−= = (2)

The distance dAB can be found by applying trigonometry to triangle LAB, which shows that

( )ABABtan or 1.10 m tan

1.10 md

dβ β= = Substituting this result into Equation (2), gives

( ) ( )ABDC

1.80 m 1.80 m 1.10 m tan 1.80 m 1.10 m tan33.01.67 m

tan tan tan33.0d

dβ

β β− − − °

= = = =°

7. REASONING AND SOLUTION The two arrows, A and B are located in front of a plane

mirror, and a person at point P is viewing the image of each arrow. As discussed in Conceptual Example 1, light emanating from the arrow is reflected from the mirror and is reflected toward the observer at P. In order for the observer to see the arrow in its entirety, both rays, the one from the top of the arrow and the one from the bottom of the arrow, must pass through the point P.

P Plane mirror

AB

P Plane mirror

According to the law of reflection, all rays will be reflected so that the angle of reflection is equal to the angle of incidence. The ray from the top of arrow A strikes the mirror and reflects so that it passes through point P. Likewise, the ray from the bottom of the arrow is reflected such that it too passes through point P. Therefore, the observer at P

sees the arrow at A in its entirety . Similar reasoning shows that the ray from the top of arrow B passes through point P.

However, as the drawing shows, the ray from the bottom of the arrow does not pass through P. This conclusion is true no matter where the bottom ray strikes the mirror. The observer

does not see the arrow at B in its entirety .


y

x

θ

θ

θ

8. REASONING The drawing shows the ray of light reflecting from the mirror and striking the floor. The angle of reflection θ is the same as the angle of incidence. The angle θ is also the angle that the light ray makes with the floor (see the drawing). Therefore, we can use the inverse tangent function to find θ as a function of y and x;

1tan yx

θ − ⎛ ⎞= ⎜ ⎟⎝ ⎠. Since the mirror is facing

east and the sun is rising, the angle of incidence θ becomes larger (see the drawing). As the angle θ becomes larger, the distance x becomes smaller.

SOLUTION When the horizontal distance is x1 = 3.86 m, the angle of incidence θ1 is

1 1

11

1.80 mtan tan 25.03.86 m

yx

θ − −⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

When the horizontal distance is x2 = 1.26 m, the angle of incidence θ2 is

1 1

22

1.80 mtan tan 55.01.26 m

yx

θ − −⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

As the sun rises, the change in the angle of incidence is 55.0° − 25.0° = 30.0°. Since the earth

rotates at a rate of 15.0° per hour, the elapsed time between the two observations is

( ) 1 hElapsed time = 30.0 2.00 h15.0⎛ ⎞° =⎜ ⎟°⎝ ⎠

9. REASONING a. The incident ray from the laser travels a distance of L = 50.0 km due north to the mirror (see the drawing, which is not to scale). The angle θ we seek, that between the normal to the surface of the mirror and due south, is the angle of incidence. The angle of incidence must equal the angle of reflection, so the angle α between the incident and reflected rays is bisected by the normal to the surface of the mirror. Therefore, we have that

α θ

θ

Laser Detector

Mirror North

East

South

d

L


12θ α= (1)

Because the laser, the mirror, and the detector sit at the corners of a right triangle, we will use ( )1tan d Lα −=

(Equation 1.6) to determine the angle α, where d = 117 m is

the distance between the laser and the detector.

b. Turning the surface normal too far from due south increases both the angle of incidence and the angle of reflection, in this case by 0.004° each. Thus, the angle α between the incident and reflected rays increases by twice that amount. We will use the new angle α, with tan D Lα = (Equation 1.3), to determine the distance D between the laser and the reflected beam (see the drawing). The distance x by which the reflected beam misses the detector is the distance D minus the distance d between the laser and the detector:

x D d= − (2)

We note that the distance d is given in meters, while the distance L is given in kilometers, where 1 km = 103 m.

SOLUTION a. Substituting ( )1tan d Lα −=

(Equation 1.6) into Equation (1) yields

1 11 1 1

2 2 2 3117 mtan tan 0.0670

50.0 10 mdL

θ α − −⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠o

b. When the mirror is misaligned, the angle of incidence is larger than the result found in part (a) by 0.004°. From Equation (1), then, the angle α between the incident and reflected rays becomes

( )2 2 0.0670 0.004 0.142α θ= = + =o o o Solving tan D Lα = (Equation 1.3) for the distance D between the laser and the reflected beam, we obtain tanD L α= (3) Substituting Equation (3) into Equation (2), and using 0.142 ,α = o we find that ( )3tan 50.0 10 m tan 0.142 117 m 7 mx D d L dα= − = − = × − =o

L

Laser Detector

Mirror

x

D

d

α θ θ


10. REASONING The drawing shows two plane mirrors that intersect at an angle of 50°. An incident light ray reflects from one mirror and then the other. The various angles are labeled θ, α, β, γ , ε and φ. We wish to find the numerical value of the angle θ . We will do this using the law of reflection, which states that when light is reflected from a surface, the angle of incidence is equal to the angle of reflection. We will also use the fact that the sum of the interior angles of a triangle is 180°.

SOLUTION We can see in the

drawing that 180θ φ+ = ° , and solving for θ we find that

180θ φ= °− (1)

Since the sum of the interior angles of

a triangle is 180º, we can see in the drawing that 2 2 180α β φ+ + = ° , and solving for ϕ, we find that

( )180 2φ α β= °− + (2)

Incidentray

Outgoingrayθα

γα

ε

β

50°

φ

Substituting Equation (2) into Equation (1) gives

( ) ( )180 180 180 2 2θ φ α β α β= °− = °− °− + = +⎡ ⎤⎣ ⎦ (3) To determine the quantity α β+ in Equation (3), we refer to the triangle formed by the

intersection of the mirrors and the first reflected ray. For this triangle we know that

50 180γ ε° + + = ° (4) In the drawing, the normal to the mirror that the incident ray strikes is the dashed line and

makes a 90° angle with that mirror. Therefore, we know that 90γ α= ° − . Using the same reasoning for the normal to the other mirror, we can also conclude that 90ε β= ° − . Substituting these expressions into Equation (4) gives

50° + 90°−α( )γ

+ 90°− β( )

ε

= 180° or α + β = 50°

Substituting this value for α β+ into Equation (3) gives

( ) ( )2 2 50 100θ α β= + = ° = ° ______________________________________________________________________________


11. REASONING

a. The smallest angle of incidence such that the laser beam hits only one of the mirrors is shown in the left drawing. Here the laser beam strikes the top mirror at an angle of incidence θ1 and just passes the right edge of the lower mirror without striking it. The angle θ1 can be obtained by using trigonometry.

b. The smallest angle such that the laser beam hits each mirror only once is shown in the right drawing. The laser beam strikes the top mirror at an angle of incidence θ2, reflects from the top and bottom mirrors, and just passes the right edge of the upper mirror without striking it. The angle θ2 can also be obtained by using trigonometry.

SOLUTION a. Note from the left drawing that the length of the side opposite the angle θ1 is ( )1

2 17.0 cm , because the angle of reflection equals the angle of incidence. The length of the side adjacent to θ1 is 3.00 cm. Using the inverse tangent function, we find that

( )1

1 21

17.0 cmtan 70.6

3.00 cmθ − ⎡ ⎤

= = °⎢ ⎥⎣ ⎦

b. From the right drawing we see that the length of the side opposite the angle θ2 is

( )13 17.0 cm , because the angle of reflection equals the angle of incidence. The length of the side adjacent to θ2 is 3.00 cm. Again, using the inverse tangent function, we obtain

( )1

1 32

17.0 cmtan 62.1

3.00 cmθ − ⎡ ⎤

= = °⎢ ⎥⎣ ⎦

θ1

17.0 cm

3.00 cm θ2

17.0 cm

3.00 cm


12. REASONING AND SOLUTION We can see from the upper triangle in the drawing that

tan θ = (L − x)/L We also see from the lower triangle that

tan θ = x/(L/2) Equating these two expressions gives

131

2or =L x x x L

L L− =

Therefore, 1

1 1 31 12 2

tan tan 33.7Lx

L Lθ − −⎛ ⎞ ⎛ ⎞

= = = °⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

13. REASONING When an object is located very far away from a spherical mirror (concave or

convex), the image is located at the mirror’s focal point. Here, the image of the distant object is located 18 cm behind the convex mirror, so that the focal length f of the mirror is f = −18 cm and is negative since the mirror is convex.

SOLUTION In constructing a ray diagram, we will need to know the radius R of the mirror.

The focal length of a convex mirror is related to the radius by 12

f R= − (Equation 25.2).

We can use this expression to determine the radius: ( )1

2 or 2 2 18 cm 36 cmf R R f= − = − = − − =

In the ray diagram that follows, we denote the focal point by F and the center of curvature by

C. Note that the horizontal and vertical distances in this drawing are to scale. This means that the mirror is represented by a circular arc that is also drawn to scale. Note that we have used only rays 1 and 3 in constructing this diagram. Only two of the three rays discussed in the text are needed.

From the drawing, we see that the image is located 6.0 cm behind the mirror .

F C 18 cm 6.0 cm

1

3

L − x

θ θ

L

x

L/2


14. REASONING Since the car is “very distant,” we can assume that it is infinitely far from the mirror. Therefore, when paraxial rays leave the car and travel parallel to the principal axis, they appear to come from the focal point of the mirror after reflection. In effect, then, the given location of the image tells us the focal length of the convex mirror. The focal length f is related to the radius R of the mirror according to 1

2f R= − (Equation 25.2).

SOLUTION a. Since the image of the very distant car is located 12 cm behind the mirror, we know that

the focal length is 12 cmf = − , the minus sign being required because the mirror is a convex mirror. Using Equation 25.2, we find that

( )12

or 2 2 12 cm 24 cmf R R f= − = − = − − =

b. The ray diagram is similar to that shown in Figure 25.16. ______________________________________________________________________________ 15. SSM REASONING The object distance (do = 11 cm) is shorter than the focal length

(f = 18 cm) of the mirror, so we expect the image to be virtual, appearing behind the mirror. Taking Figure 25.18a as our model, we will trace out: three rays from the tip of the object to the surface of the mirror, then three reflected rays, and finally three virtual rays extending behind the mirror and meeting at the tip of the image. The scale of the ray tracing will determine the location and height of the image. The three sets of rays are:

1. An incident ray from the object to the mirror, parallel to the principal axis and then reflected through the focal point F. 2. An incident ray from the object to the mirror, directly away from the focal point F and then reflected parallel to the principal axis. (The incident ray cannot pass from the object through the focal point, as this would take it away from the mirror, and it would not be reflected.) 3. An incident ray from the object to the mirror, directly away from the center of curvature C, then reflected back through C.


SOLUTION

a. The ray diagram indicates that the image is 28 cm behind the mirror . b. We see from the ray diagram that the image is 7.6 cm tall .

16. REASONING AND SOLUTION The ray diagram

is shown in the figure (Note: f = 5.0 cm and do = 15.0 cm).

a. The ray diagram indicates that the

image distance is 7.5 cm in front of the mirror. b. The image height is 1.0 cm , and the image is

inverted relative to the object.

Object

CF

f

Real Image

do

di

17. REASONING AND SOLUTION a. A ray diagram, which will look similar, but not identical, to that in Figure 25.21a, reveals

that

ithe image distance is 20.0 cm behind the mirror, or 20.0 cmd = − . b. The ray diagram also shows that the image height is 6.0 cm , and the image is upright

relative to the object.

F

C

1

1

2 2

2 3

3

28 cm 11 cm

Image (virtual)

Object 3.0 cm tall

Scale: 6.0 cm

1

3 7.6 cm tall


18. REASONING AND SOLUTION A plane mirror faces a concave mirror ( f = 8.0 cm). The following is a ray diagram of an object placed 10.0 cm in front of the plane mirror.

Scale:

5.0 cm

F

Concave mirror

Plane mirror

Image fromconcave mirror

Object

Reflected rayIncident ray

Image fromplane mirror

The ray diagram shows the light that is first reflected from the plane mirror and then the

concave mirror. The scale is shown in the figure. For the reflection from the plane mirror, as discussed in the text, the image is upright, the same size as the object, and is located as far behind the mirror as the object is in front of it. When the reflected ray reaches the concave mirror, the ray that is initially parallel to the principal axis passes through the focal point F after reflection from the concave mirror. The ray that passes directly through the focal point emerges parallel to the principal axis after reflection from the concave surface. The point of intersection of these two rays locates the position of the image. By inspection, we see that the image is located at 10 9. cm from the concave mirror.

19. SSM REASONING This problem can be solved using the mirror equation, Equation 25.3. SOLUTION Using the mirror equation with d i cm= +26 and f = 12 cm , we find

1 1 1 22d f d

do i

o1

12 cm– 1

26 cm or cm= = = +–

20. REASONING We have seen that a convex mirror always forms a virtual image as shown in

Figure 25.21a of the text, where the image is upright and smaller than the object. These characteristics should bear out in the results of our calculations.

SOLUTION The radius of curvature of the convex mirror is 68 cm. Therefore, the focal

length is, from Equation 25.2, f R= =–( / ) –34 cm1 2 . Since the image is virtual, we know that d i cm= –22 .

a. With d i cm= –22 and f = –34 cm , the mirror equation gives


1 1 1d f d

do i

o1

–34 cm– 1

–22 cm or 62 cm= = = +–

b. According to the magnification equation, the magnification is

mdd

= = =– – –22i

o

cm62 cm

+0.35

c. Since the magnification m is positive, the image is upright .

d. Since the magnification m is less than one, the image is smaller than the object. 21. REASONING a. We are dealing with a concave mirror whose radius of curvature is 56.0 cm. Thus, the

focal length of the mirror is 12 28.0 cmf R= = (Equation 25.1) The object distance is

do = 31.0 cm. With known values for f and do, we can use the mirror equation to (Equation 25.3) find the image distance.

b. To determine the image height hi, recall that it is related to the object height ho by the

magnification m; hi = mho. The magnification is related to the image and object distances by the magnification equation, i o/m d d= − (Equation 25.4). Since we know values for ho, di, and do, we can find the image height hi.

SOLUTION a. The image distance is given by the mirror equation as follows:

ii o

1 1 1 1 1 so 290 cm28.0 cm 31.0 cm

dd f d

= − = − = (25.3)

b. Using the magnification equation, we find that the image height is

( )ii o o

o

290 cm 0.95 cm 8.9 cm31.0 cm

dh mh h

d⎛ ⎞⎛ ⎞= = − = − = −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

(25.4)

c. Since hi is negative, the image is inverted relative to the object. Thus, to make the picture

on the wall appear normal, the slide must be oriented upside down in the projector. 22. REASONING For an image that is in front of a mirror, the image distance is positive. Since

the image is inverted, the image height is negative. Given the image distance, the mirror equation can be used to determine the focal length, but to do so a value for the object distance is also needed. The object and image heights, together with the knowledge that the image is inverted, allows us to calculate the magnification m. The magnification m is given


by m = –di/do (Equation 25.4), where di and do are the image and object distances, respectively.

SOLUTION According to Equation 25.4, the magnification is

i oi io

o o i or

d hh dm d

h d h= = − = −

Substituting this result into the mirror equation, we obtain

1f= 1

do

+ 1di

= −hi

diho

+ 1di

= 1di

−hiho

+1⎛

⎝⎜

⎞

⎠⎟

= 113 cm

⎛⎝⎜

⎞⎠⎟

−−1.5 cm( )3.5 cm( ) +1

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 0.11 cm−1 or f = 9.1 cm

23. SSM REASONING Since the image is behind the mirror, the image is virtual, and the

image distance is negative, so that i 34.0 cmd = − . The object distance is given as

o 7.50 cmd = . The mirror equation relates these distances to the focal length f of the mirror. If the focal length is positive, the mirror is concave. If the focal length is negative, the mirror is convex.

SOLUTION According to the mirror equation (Equation 25.3), we have

( )o i

o i

1 1 1 1 1 or 9.62 cm1 1 1 17.50 cm 34.0 cm

ff d d

d d

= + = = =+ +

−

Since the focal length is positive, the mirror is concave . 24. REASONING a. We are given that the focal length of the mirror is f = 45 cm and that the image distance is

one-third the object distance, or 1i o3d d= . These two pieces of information, along with the

mirror equation, will allow us to find the object distance. b. Once the object distance has been determined, the image distance is one-third that value,

since we are given that 1i o3d d= .

SOLUTION a. The mirror equation indicates that

o i

1 1 1d d f

+ = (25.3)


Substituting 1i o3d d= and solving for do gives

( )o1

o o o3

1 1 1 4 1 or or 4 4 45 cm 180 cmd fd d f d f

+ = = = = =

b. The image distance is

( ) 11 1i o3 3 180 cm 6.0 10 cmd d= = = ×

25. REASONING Since the focal length f and the object distance do are known, we will use the

mirror equation to determine the image distance di. Then, knowing the image distance as well as the object distance, we will use the magnification equation to find the magnification m.

SOLUTION a. According to the mirror equation (Equation 25.3), we have

( )i

i o

o

1 1 1 1 1 or 4.3 m1 1 1 17.0 m 11 m

dd f d

f d

= − = = = −− −

−

The image distance is negative because the image is a virtual image behind the mirror. b. According to the magnification equation (Equation 25.4), the magnification is

( )i

o

4.3 m0.39

11 md

md

−= − = − =

26. REASONING The magnification m is given by m = –di/do (Equation 25.4), where di and do

are the image and object distances, respectively. The object distance is known, and we can obtain the image distance from the mirror equation:

o i

1 1 1d d f

+ = (25.3)

SOLUTION Solving the mirror equation (Equation 25.3) for the image distance di gives

1 1 1 1 1 1d d f d f d

d ffd

dfdd fo i i o

o

oi

o

o

or = or + = = −−

=−

Substituting this result into the magnification equation (Equation 25.4) gives

m = −

dido

= −fdo / do − f( )

do

= ff − do


Using this result with the given values for the focal length and object distances, we find

Smaller object distance m = ff − do

= −27.0 cm−27.0 cm( )− 9.0 cm( ) = 0.750

Greater object distance m = ff − do

= −27.0 cm−27.0 cm( )− 18.0 cm( ) = 0.600

27. SSM REASONING When paraxial light rays that are parallel to the principal axis strike a

convex mirror, the reflected rays diverge after being reflected, and appear to originate from the focal point F behind the mirror (see Figure 25.16). We can treat the sun as being infinitely far from the mirror, so it is reasonable to treat the incident rays as paraxial rays that are parallel to the principal axis.

SOLUTION a. Since the sun is infinitely far from the mirror and its image is a virtual image that lies

behind the mirror, we can conclude that the mirror is a convex mirror . b. With d i = –12.0 cm and do =∞, the mirror equation (Equation 25.3) gives

1 1 1 1 1 1f d d d d

= + =∞+ =

o i i i

Therefore, the focal length f lies 12.0 cm behind the mirror (this is consistent with the

reasoning above that states that, after being reflected, the rays appear to originate from the focal point behind the mirror). In other words, f = –12.0 cm. Then, according to Equation 25.2, f R= – 12 , and the radius of curvature is

R f= =–2 –2(–12.0 cm) = 24.0 cm

28. REASONING The magnification of the mirror is related to the image and object distances

via the magnification equation. The image distance is give, and the object distance is unknown. However, we can obtain the object distance by using the mirror equation, which relates the image and object distances to the focal length, which is also given.

SOLUTION According to the magnification equation, the magnification m is related to the

image distance di and the object distance do according to

i

o

dm

d= − (25.4)

According to the mirror equation, the image distance di and the object distance do are related

to the focal length f as follows:


o i o i

1 1 1 1 1 1 or d d f d f d

+ = = − (25.3)

Substituting this expression for 1/do into Equation 25.4 gives

i ii

o i

1 1 1

36 cm 1 2.012 cm

d dm d

d f d f⎛ ⎞

= − = − − = − +⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞= − + = −⎜ ⎟⎝ ⎠

29. REASONING We need to know the focal length of the mirror and can obtain it from the

mirror equation, Equation 25.3, as applied to the first object:

1 1 1d d f

fo1 i1

114.0 cm

1–7.00 cm

or –14.0 cm+ = + = =

According to the magnification equation, Equation 25.4, the image height hi is related to the

object height ho as follows: hi = mho = –di / do( )ho .

SOLUTION Applying this result to each object, we find that h hi2 i1= , or

–di2do2

⎛

⎝⎜

⎞

⎠⎟ ho2 =

–di1do1

⎛

⎝⎜

⎞

⎠⎟ ho1

Therefore,

di2 = do2

di1do1

⎛

⎝⎜

⎞

⎠⎟

ho1ho2

⎛

⎝⎜

⎞

⎠⎟

Using the fact that h ho o12 2= , we have

di2 = do2

di1do1

⎛

⎝⎜

⎞

⎠⎟

ho1ho2

⎛

⎝⎜

⎞

⎠⎟ = do2

–7.00 cm14.0 cm

⎛⎝⎜

⎞⎠⎟

ho12ho1

⎛

⎝⎜

⎞

⎠⎟ = –0.250 do2

Using this result in the mirror equation, as applied to the second object, we find that

1 1 1

o2 i2d d f+ =

or

1 1–0.250

1–14.0 cmo2 o2d d

+ =

Therefore, do2 42 0 cm= + .


30. REASONING We will determine the type of mirror from the fact that the image is enlarged.

To determine the focal length f, we will use the mirror equation, which is o i

1 1 1d d f

+ =

(Equation 25.3), where do is the object distance and di is the image distance. To determine

the magnification m, we will use the magnification equation, which is i

o–d

md

= (Equation

25.4). The algebraic sign of the value for m will tell us whether the image is upright (+) or inverted (−) with respect to the object.

SOLUTION a. Since the image of the tooth is enlarged, it cannot be a plane mirror, for which the object

and the image would have the same size. Convex mirrors produce smaller images in all cases. Therefore, the enlarged image means that the mirror must be concave .

b. Using the mirror equation (Equation 25.3), we find for the focal length that

11

o i

1 1 1 1 1 10.32 cm or 3.1cm2.0 cm 5.6 cm 0.32 cm

ff d d

−−= + = + = = = +

−

Note that the value of the image distance is −5.6 cm, which is negative because the image is

behind the mirror (virtual). c. Using the magnification equation (Equation 25.4), we find for the magnification that

i

o

–5.6 cm– – +2.82.0 cm

dm

d= = =

d. Since m is positive, the image is upright relative to the object.

31. REASONING

a. You need a concave mirror to make the measurement. Since you must measure the image distance and image height of the tree, the image must be a real image. Only a concave mirror can produce a real image. The image distance of the sun is equal to the focal length of the mirror. The sun is extremely far away, so the light rays from it are nearly parallel when they reach the mirror. According to the discussion in Section 25.5, light rays near and parallel to the principal axis are reflected from a concave mirror and converge at the focal point.

With numerical values for the focal length f of the mirror and the image distance di, the

mirror equation (Equation 25.3) can be used to find the object distance do:

o i

1 1 1d f d

= −


b. The height ho of the tree is related to the height hi of its image by the magnification m; ho = hi/m (Equation 25.4). However, the magnification is given by the magnification equation (Equation 25.4) as m = −di/do, where di is the image distance and do is the object distance, both of which we know.

SOLUTION a. The distance to the tree is given by the mirror equation as

oo i

1 1 1 1 1 so 82 m0.9000 m 0.9100 m

dd f d

= − = − =

b. Since ho = hi/m and m = −di/do, we have that

oi i

o ii i

o

dh hh h

dm dd

⎛ ⎞= = = −⎜ ⎟⎛ ⎞ ⎝ ⎠−⎜ ⎟

⎝ ⎠

Now hi = −0.12 m, where the minus sign has been used since the image is inverted relative to the tree (see Figure 25.18b). Thus, the height of the tree is

( )oo i

i

82 m0.12 m 11 m0.9100 m

dh h

d⎛ ⎞ ⎛ ⎞= − = − − =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

32. REASONING The mirror equation, which is o i

1 1 1d d f

+ = (Equation 25.3), and the

magnification equation, which is i

o–d

md

= (Equation 25.4), apply to both types of mirrors.

In these equations do is the object distance, di is the image distance, f is the focal length, and

m is the magnification. The focal lengths of the two types of mirrors are 1concave 2f R=

(Equation 25.1) and 1convex 2f R= − (Equation 25.2). We will apply these equations to both

types of mirrors, taking particular advantage of the facts that the radius R and the object distance do are the same for each type.

SOLUTION We are given information about the magnification m. In addition, we know

that the object distance do is the same for each type of mirror. We also know that the radius

R is the same for each type of mirror, so that we can substitute 1concave 2f R= (Equation 25.1)

into 1convex 2f R= − (Equation 25.2) and see that


1convex concave2

convex concave

1 1 or f R ff f

= − = − = − (1)

Thus, in the mirror equation and the magnification equation we have information about all of

the variables except the image distance di. To eliminate it from the problem, we solve the magnification equation for di:

ii o

o– or d

m d m dd

= = − (2)

Substituting Equation (2) into the mirror equation gives

( )o i o o o

1 1 1 1 1 1 11f d d d m d d m

⎛ ⎞= + = + = −⎜ ⎟− ⎝ ⎠ (3)

Applying Equation (3) to each type of focal length in Equation (1), we have

1do

1− 1mconvex

⎛

⎝⎜

⎞

⎠⎟

1/ fconvex

= − 1do

1− 1mconcave

⎛

⎝⎜

⎞

⎠⎟

−1/ fconcave

or mconvex =mconcave

2mconcave −1= +2.0

2 +2.0( )−1= +0.67

33. REASONING The radius of curvature R of a concave mirror is related to the focal length f

of the mirror by 12f R= (Equation 25.1), so we have that

2R f= (1)

The mirror’s focal length is given by the mirror equation o i

1 1 1f d d= + (Equation 25.3),

where do = 14 cm is the object distance and di is the image distance. The image distance is

related to the object distance via the magnification equation i

o

dm

d= − (Equation 25.4).

Because the image is virtual, the image distance di is negative. The object distance is positive, so we conclude from Equation 25.4 that the magnification is positive: m > 0. The image is twice the size of the object, so we have that the magnification is m = +2.0.

SOLUTION Taking the reciprocal of both sides of o i

1 1 1f d d= + (Equation 25.3) yields

o i

11 1f

d d

=+

. Substituting this result into Equation (1), we find that


o i

22 1 1R f

d d

= =+

(2)

Solving i

o

dm

d= − (Equation 25.4) for di and substituting m = +2.0 yields

i o o o( 2.0) 2d md d d= − = − + = − (3) Substituting Equation (3) into Equation (2), we obtain

( )

( )o

o i o o o

2 2 2 4 4 14 cm 56 cm1 1 1 1 12 2

R d

d d d d d

= = = = = =⎛ ⎞+ + ⎜ ⎟− ⎝ ⎠

34. REASONING AND SOLUTION We know that do − di = 45.0 cm. We also have

1/do + 1/di = 1/f. Solving the first equation for di and substituting the result into the second equation yields,

1do

+1

do − 45.0 cm=

130.0 cm

Cross multiplying gives do

2 − 105 do + 1350 = 0, which we can solve using the quadratic equation to yield two roots, do = (105 ± 75)/2.

a. When the object lies beyond the center of curvature we have

do+ = (1.80 × 102 cm)/2 = + ×9.0 10 cm1 and di+ = +45 cm

b. When the object lies within the focal point

do– = (3.0 × 101 cm)/2 = +15 cm , and di– = − ×3.0 10 cm1

35. SSM REASONING

a. The image of the spacecraft appears a distance di beneath the surface of the moon, which we assume to be a convex spherical mirror of radius R = 1.74×106 m. The image distance di is related to the focal length f of the moon and the object distance do by the mirror equation

o i

1 1 1d d f

+ = (25.3)

The spacecraft is the object, so the object distance do is equal to the altitude of the spacecraft: do = 1.22×105 m. The moon is assumed to be a convex mirror of radius R, so its focal length is given by


12f R= − (25.2)

b. Both the spacecraft and its image would have the same angular speed ω about the center of the moon because both have the same orbital period. The linear speed v of an orbiting body is related to its angular speed ω by v rω= (Equation 8.9), where r is the radius of the orbit. Therefore, the speeds of the spacecraft and its image are, respectively,

o o i i and v r v rω ω= = (1)

We have used vo and ro in Equation (1) to denote the orbital speed and orbital radius of the spacecraft, because it is the object. The symbols vi and ri denote orbital speed and orbital radius of the spacecraft’s image. The orbital radius ro of the spacecraft is the sum of the radius R of the moon and the spacecraft’s altitude do = 1.20×105 m above the moon’s surface: o or R d= + (2)

The “orbital radius” ri of the image is its distance from the center of the moon. The image is below the surface, and the image distance di is negative. Therefore, the distance between the center of the moon and the image is found from

i ir R d= + (3) SOLUTION

a. Solving Equation 25.3 for i

1d

and then taking the reciprocal of both sides, we obtain

o oi

i o o o

1 1 1 or d f d f

dd f d d f d f

−= − = =

− (4)

Substituting Equation 25.2 into Equation (4), we find that

( )( )

( )( )( )

5 611 12o 2 o 5o 2

i 1 5 61 1o o 2o 2 2

1.22 10 m 1.74 10 m1.07 10 m

1.22 10 m 1.74 10 m

d R d Rd fd

d f d Rd R

− × ×− −= = = = = − ×

− +− − × + ×

Therefore, the image of the spacecraft would appear 51.07 10 m× below the surface of the moon.

b. Solving the first of Equations (1) for ω yields o

o

vr

ω = . Substituting this result into the

second of Equations (1), we find that

o ii i i o

o o

v rv r r v

r rω

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(5)

Substituting Equations (2) and (3) into Equation (5) yields


( ) ( )

6 5i i

i o o 6 5o o

1.74 10 m 1.07 10 m1620 m/s 1420 m/s

1.74 10 m 1.22 10 mr R d

v v vr R d

⎡ ⎤× + − ×⎛ ⎞ ⎛ ⎞+ ⎢ ⎥= = = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎢ ⎥+ × + ×⎝ ⎠ ⎝ ⎠ ⎣ ⎦

36. REASONING We will start by drawing the two situations in which the object is 25 cm and

5 cm from the mirror, making sure that all distances (including the radius of curvature of the mirror) and heights are to scale. For each location of the object, we will draw several rays to locate the image (see the Reasoning Strategy for convex mirrors in Section 25.5). Once the images have been located, we can readily answer the questions regarding their positions and heights.

SOLUTION The following two ray diagrams illustrate the situations where the objects are at

different distances from the convex mirror.

a. As the object moves closer to the mirror, it can be seen that the magnitude of the image

distance becomes smaller . b. As the object moves closer to the mirror, the magnitude of the image height becomes

larger .

F C Object

25 cm

Image

3 11

1 11

F C Object

Image

5 cm

1 11

3 11


c. By measuring the image heights, we find that the ratio of the image height when the object distance is 5 cm to that when the object distance is 25 cm is 3 .

37. SSM REASONING The mirror equation relates the object and image distances to the

focal length. Thus, we can apply the mirror equation once with the given object and image distances to determine the focal length. Then, we can use the mirror equation again with the focal length and the second object distance to determine the unknown image distance.

SOLUTION According to the mirror equation, we have

1 1 1 1 1 1d d f d d fo1 i1

First position of object

o2 i2

Second position of object

and + = + =

Since the focal length is the same in both cases, it follows that

1do1

+ 1di1

= 1do2

+ 1di2

1di2

= 1do1

+ 1di1

− 1do2

= 125 cm

+ 1−17 cm( ) −

119 cm

= −0.071 cm−1

di2 = −14 cm

The negative value for di2 indicates that the image is located 14 cm behind the mirror. 38. REASONING The object distance do is the distance between the object and the mirror. It is

found from o i

1 1 1d d f

+ = (Equation 25.3), where di is the distance between the mirror and the

image, and f is the focal length of the mirror. We are told that the image appears in front of the mirror, so, according to the sign conventions for spherical mirrors, the image distance must be positive: di = +97 cm.

SOLUTION Solving o i

1 1 1d d f

+ = (Equation 25.3) for do yields

( ) ( )o

o i

i

1 1 1 1 1 or 74 cm1 1 1 142 cm 97 cm

dd f d

f d

= − = = =− −

+ +


39. REASONING AND SOLUTION a. The height of the shortest mirror would be one-half the height of the person. Therefore,

h = H/2 = (1.70 m + 0.12 m)/2 = 0.91 m b. The bottom edge of the mirror should be above the floor by

h' = (1.70 m)/2 = 0.85 m 40. REASONING The focal length f of the water drop is given by the mirror equation

o i

1 1 1f d d= + (Equation 25.3), where do and di are, respectively, the object distance and the

image distance. The object distance (do = 3.0 cm) is given, and we will determine the image

distance di from the magnification equation i i

o o

h dm

h d= = − (Equation 25.4), where ho is the

diameter of the flower and hi is the diameter of its image. The water drop acts as a convex spherical mirror, so the image is upright. Therefore, the image height hi is positive, and we expect the focal length f to be negative.

SOLUTION Solving i i

o o

h dh d

= − (Equation 25.4) for di and taking the reciprocal, we obtain

o i oi

o i o i

1 or d h h

dh d d h

= − = − (1)

Substituting Equation (1) into o i

1 1 1f d d= + (Equation 25.3) yields

o o

o i o o i o i

1 1 1 1 1 1h h

f d d d d h d h⎛ ⎞

= + = − = −⎜ ⎟⎜ ⎟⎝ ⎠ (2)

Taking the reciprocal of Equation (2), we find that the focal length of the water drop is

o

oo o

i i

1 3.0 cm 0.16 cm2.0 cm11 10.10 cm

df d h h

h h

⎛ ⎞= = = = −⎜ ⎟−− −⎜ ⎟⎜ ⎟⎝ ⎠


41. SSM REASONING This problem can be solved by using the mirror equation, Equation 25.3, and the magnification equation, Equation 25.4.

SOLUTION a. Using the mirror equation with d di o= and f R= / 2 , we have

1 1 1 2d f d R d d Ro i o o

1/ 2

– 1 or 2= = =–

Therefore, we find that d Ro = .

b. According to the magnification equation, the magnification is

mdd

dd

= = =– –i

o

o

o

–1

c. Since the magnification m is negative, the image is inverted . 42. REASONING The mirror equation relates the object and image distances to the focal length.

The magnification equation relates the magnification to the object and image distances. The problem neither gives nor asks for information about the image distance. Therefore, we can solve the magnification equation for the image distance and substitute the result into the mirror equation to obtain an expression relating the object distance, the magnification, and the focal length. This expression can be applied to both mirrors A and B to obtain the ratio of the focal lengths.

SOLUTION The magnification equation gives the magnification as m = –di/do. Solving for

di, we obtain di = –mdo. Substituting this result into the mirror equation, we obtain

1do

+ 1di

= 1do

+ 1−mdo( ) =

1f

or 1do

1− 1m

⎛⎝⎜

⎞⎠⎟= 1

f or f =

domm−1

Applying this result for the focal length f to each mirror gives

fd mm

fd mmA

o A

AB

o B

B

and =−

=−1 1

Dividing the expression for fA by the expression for fB, we find


fAfB

=d0mA / mA −1( )d0mB / mB −1( ) =

mA mB −1( )mB mA −1( ) =

4.0 2.0−1( )2.0 4.0−1( ) = 0.67

43. REASONING According to the discussion about relative velocity in Section 3.4, the

velocity vIY of your image relative to you is the vector sum of the velocity vIM of your image relative to the mirror and the velocity vMY of the mirror relative to you:

vIY = vIM + vMY

As you walk toward the stationary mirror, you perceive the mirror moving toward you in the

opposite direction. Thus, vMY = –vYM. The velocity vYM has the components vYM,x and vYM,y, while the velocity vIM has the components vIM,x and vIM,y. The x direction is perpendicular to the mirror, and the two x components have the same magnitude. This is because the image in a plane mirror is always just as far behind the mirror as the object is in front of it. For instance, if an object moves 1 meter perpendicularly toward the mirror in 1 second, the magnitude of its velocity relative to the mirror is 1 m/s. But the image also moves 1 meter toward the mirror in the same time interval, so that the magnitude of its velocity relative to the mirror is also 1 m/s. The two x components, however, have opposite directions. The two y components have the same magnitude and the same direction. This is because an object moving parallel to a plane mirror has an image that remains at the same distance behind the mirror as the object is in front of it and moves in the same direction as the object.

SOLUTION According to the discussion in the REASONING,

vIY = vIM + vMY = vIM – vYM This vector equation is equivalent to two equations, one for the x components and one for the

y components. For the x direction, we note that vYM,x = –vIM,x.

( ) ( )IY, IM, YM,

0.90 m/s cos 50.0 0.90 m/s cos 50.0 1.2 m/s

x x xv v v= −

= − °− ° = −

For the y direction, we note that vYM,y = vIM,y.

( ) ( )IY, IM, YM,

0.90 m/s sin 50.0 0.90 m/s sin 50.0 0 m/s

y y yv v v= −

= °− ° =


Since the y component of the velocity vIY is zero, the velocity of your image relative to you points in the –x direction and has a magnitude of 1.2 m/s .

44. REASONING The focal length f of the convex mirror determines the first image distance di1

via o i1

1 1 1f d d= + (Equation 25.3), where do = 15.0 cm is the distance between the candle and

the convex mirror. The second image, formed by the plane mirror, is located as far behind the mirror as the object is in front of the mirror. Therefore, when the plane mirror replaces the convex mirror, the second image distance di2 is given by i2 o 15.0 cmd d= − = − (1) The negative sign in Equation (1) arises because the image lies behind the plane mirror and, therefore, is a virtual image. When the plane mirror replaces the convex mirror, the image moves a distance x farther away from the mirror, so the initial and final image distances are related by i2 i1d d x= − (2) The negative sign in Equation (2) occurs because the image moves farther from the mirror, thus making the second image distance di2 a negative number of greater magnitude than the first image distance di1. SOLUTION Substituting Equation (2) into Equation (1) and solving for di1 yields i1 o i1 o or d x d d x d− = − = − (3)

Substituting Equation (3) into o i1

1 1 1f d d= + (Equation 25.3), we obtain

o i1 o o

1 1 1 1 1f d d d x d= + = +

− (4)

Taking the reciprocal of both sides of Equation (4), we find that the focal length of the convex mirror is

o o

1 1 17 cm1 1 1 115.0 cm 7.0 cm 15.0 cm

f

d x d

= = = −+ +

− −


45. SSM REASONING Since the size hi of the image is one-fourth the size ho of the object,

we know from the magnification equation that

m14

=hiho

= −dido

so that di = − 14 do (25.4)

We can substitute this relation into the mirror equation to find the ratio do/f. SOLUTION Substituting 1

i o4d d= − into the mirror equation gives

1o i o 0 o4

1 1 1 1 1 1 3 1or or d d f d d f d f

−+ = + = =−

(25.3 )

Solving this equation for the ratio do/f yields do/f = 3− . 46. REASONING In both cases, the image and object distances (di and do) are related to the

focal length f of the mirror by the mirror equation o i

1 1 1d d f

+ = (Equation 25.3). Letting the

subscript 1 denote the first situation (convex side of the mirror) and the subscript 2 denote the second situation (concave side of the mirror), we have that

o i2 2 o i1 1

1 1 1 1 1 1 and d d f d d f

+ = + = (1)

In Equations (1), we have made use of the fact that the man maintains the same distance of do = 45 cm between his face and the mirror in both cases. Because the mirror is double-sided, the focal lengths of each side differ only by sign: that of the convex side is negative, and that of the concave side is positive. Therefore, we have that f2 = −f1, where f1 is the focal length of the convex side and f2 is the focal length of the concave side. Given this relation, we can rewrite the first of Equations (1) as

o i2 1

1 1 1d d f

+ = − (2)


We will determine the image distance di1 from the magnification m1 of the first image by

means of i

o

dm

d= − (Equation 25.4), the magnification equation:

i11

o

dm

d= − (3)

After using Equation 25.4 to determine di1, we will use mirror equation to determine the image distance di2 when the man looks into the concave side of the mirror. If the image distance is positive, it will appear in front of the mirror; if the image distance is negative, the image will appear behind the mirror.

SOLUTION Solving Equation (2) for i2

1d

, we obtain

i2 1 o 1 o

1 1 1 1 1d f d f d

⎛ ⎞= − − = − +⎜ ⎟⎜ ⎟⎝ ⎠

(4)

In this result, we can replace the term 1/f1 by using the second of Equations (1). In this way we find that

i2 1 o o i1 o o i1

1 1 1 1 1 1 2 1d f d d d d d d

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + = − + + = − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(5)

Taking the reciprocal of Equation (5), we obtain

i2

o i1

12 1

d

d d

= −⎛ ⎞

+⎜ ⎟⎜ ⎟⎝ ⎠

(6)

Solving Equation (3) for di1 yields i1 1 o o0.20d m d d= − = − . Substituting this result into Equation (6) yields

( )i2

oo i1 o o

1 1 112 1 2 1 2 5.0

0.20

d

dd d d d

= − = − = − = −⎛ ⎞ ⎛ ⎞ −+ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

od−

45 cm 15 cm3.03.0

+= = +

In the preceding equation, we have used the fact that 1 5.00.20

= .


47. SSM REASONING The time of travel

is proportional to the total distance for each path. Therefore, using the distances identified in the drawing, we have

tt

d dd

LM MP

LP

reflected

direct

=+

(1)

We know that the angle of incidence is

equal to the angle of reflection. We also know that the lamp L is a distance 2s from the mirror, while the person P is a distance s from the mirror.

30.0°30.0°

lamp, L

person, P

mirror, M

d MP

d LP

d LM

2s

s

Therefore, it follows that if d dMP = , then d dLM = 2 . We may use the law of cosines (see

Appendix E) to express the distance d LP as

dLP = 2d( )2 + d2 – 2 2d( ) d( )cos 2 30.0°( )

SOLUTION Substituting the expressions for d MP, d LM , and d LP into Equation (1), we find

that the ratio of the travel times is

treflectedtdirect

= 2d + d

2d( )2 + d2 – 2 2d( ) d( )cos 2 30.0°( )= 3

5 – 4 cos 60.0°= 1.73

48. CONCEPTS

(i) Either type is possible. A concave mirror can form a virtual image if the object is between the mirror’s focal point and the mirror. A convex mirror always forms a virtual image. (ii) They indicate a convex mirror. A concave mirror can produce an image that is smaller than the object if the object is located beyond the center of curvature of the mirror, as in Figure 25.18b. However, the image is real, not virtual. A convex mirror, in contrast, always produces an image that is virtual and smaller than the object. (iii) The focal length is negative because the mirror is convex. A concave mirror has a positive focal length. CALCULATIONS


The virtual image is located behind the mirror and, therefore, has a negative image distance,di = − 4.5 cm . Using this value together with the object distance of do = 7.0 cm, we can apply the mirror equation to find the focal length: As expected, the focal length is negative.

______________________________________________________________________________ 49. SSM CONCEPTS

(i) A convex mirror always forms an image that is smaller than the object, as Figure 25.21a shows. Therefore, the mirror must be concave. (ii) There are two places. Figure 25.18a illustrates that one of the places is between the center of curvature and the focal point. The enlarged image is real and inverted. Figure 25.19a shows that another possibility is between the focal point and the mirror, in which case the enlarged image is virtual and upright. (iii) Since the image is inverted in Figure 25.18a, the magnification for this possibility is m = −2 . In Figure 25.19a, however, the image is upright, so the magnification ism = +2 for this option. In either case, the image is twice the size of the ring. CALCULATIONS According to the mirror equation and the magnification equation, we have We can solve the magnification equation for the image distance and obtaindi = −mdo . Substituting this expression into the mirror equation gives Applying this result with the two magnifications (and noting from Equation 25.1 that f = 1

2 R = 12 cm ), we obtain:

1f= 1do

+ 1di

= 17.0 cm

+ 1− 4.5 cm

= −0.079 cm−1 or f = −13 cm

1do

+ 1di

= 1f

Mirror equation

and m = − di

d0

Magnification equation

1do

+ 1−mdo( ) =

1f

or do =f m −1( )m


−2 m =

m = +2

______________________________________________________________________________

do =f m −1( )m

=12 cm( ) −2 −1( )

−2= 18 cm

do =f m −1( )m

=12 cm( ) +2 −1( )

+2= 6.0 cm

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CHAPTER 25 THE REFLECTION OF LIGHT: MIRRORSlwillia2/21/ch25.pdfof light reflecting from the mirror...

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