5.
I SSM
REASONING AND SOLUTION The magnitude of the force can be determined
I
using Equation 21.1, F = Iql vB sin B, where B is the angle between the velocity and the
magnetic field. The direction of the force is determined by using Right-Hand Rule No. 1.
a. F=
Iql
directed
b. F=
Iql
vB sin 30.0° = (8.4 x 10-6 C)(45 m/s)(0.30 T) sin 30.0° = 15.7 X 10-5 NI,
I into
the paper I·
vB sin 90.0° = (8.4 x 10-6 C)(45 m/s)(0.30 T) sin 90.0° =
11.1
x 10-4N
I,
directed I into the paper I·
c. F= Iql vB sin 150° = (8.4 x 10-6 C)(45 m/s)(0.30 T) sin 150° = 15.7 x 10-5 NI,
directed I into the paper I·
6.
REASONING When a charge qo travels at a speed v and its velocity makes an angle Bwith
respect to a magnetic field of magnitude B, the magnetic force acting on the charge has a
magnitude F that is given by F = IqolvBsinB
(Equation 21.1). We will solve this problem
by applying this expression twice, first to the motion of the charge when it moves
perpendicular to the field so that B = 90.0° and then to the motion when B = 38°.
SOLUTION
When the charge moves perpendicular
Equation 21.1 indicates that
to the field so that B= 90.0°,
When the charge moves so that B = 38°, Equation 21.1 shows that
Dividing the second expression by the first expression gives
F380
F90.0o
_
IqolvBsin 38°
-jqolvBsin90.00
F38 ° =F 90.0 o( sin
sin90.0°
38° )=(2.7X10-3
~--N)( sin
sin90.0°
38° )=!1.7X10-3
NI
REASONING
a. The drawing shows the velocity v of the particle at the top
of its path. The magnetic force F, which provides the
centripetal force, must be directed toward the center of the
circular path. Since the directions of v, F, and B are known,
we can use Right-Hand Rule No. 1 (RHR-l) to determine if
the charge is positive or negative.
11.
B (out of paper)
lv
F
~
-'''':,
• ,
\
•
\
\
b. The radius of the circular path followed by a charged
•
•
,
•
particle is given by Equation 21.2 as r = mv / Iql B . The mass
m of the particle can be obtained directly from this relation, since all other variables are
known.
SOLUTION
a. If the particle were positively charged, an application of RHR-l would show that the
force would be directed straight up, opposite to that shown in the drawing. Thus, the charge
on the particle must be Inegativel.
b. Solving Equation 21.2 for the mass of the particle gives
m =_lqIB_r
v
19. ISSMI REASONING
=_(8_.2_X_l0_-4_C_)_(0_.4_8_T_)(_96_0_m_)
=!2.7XlO-3
kg I
140m/s
~.---.
AND SOLUTION
According
to Right-Hand
Rule No. 1, the
magnetic force on the positively charged particle is toward the bottom of the page in the
drawing in the text. If the presence of the electric field is to double the magnitude of the net
force
on·
the
charge,
the
electric
field
must
also
be
I
directed toward the bottom of the page I. Note that this results in the electric field being
perpendicular to the magnetic field, even though the electric force and the magnetic force
are in the same direction.
Furthermore, if the magnitude of the net force on the particle is twice the magnetic force, the
electric force must be equal in magnitude to the magnetic force. In other words, combining
Equations 18.2 and 21.1, we find IqlE = IqlvBsin e, with sin e= sin 90.0° = 1.0. Then,
solving for E
E = vB sine = (270 m / s)(0.52 T)(1.0) = 1140 V/m
------.-- ... _-----. __
._-~------_._'-----------------------
I
21.
REASONING When the proton moves in the magnetic field, its trajectory is a circular path.
The proton will just miss the opposite plate if the distance between the plates is equal to the
radius of the path. The radius is given by Equation 21.2 as r = mv / (Iq\ B). This relation can
be used to find the magnitude B of the magnetic field, since values for all the other variables
are known.
SOLUTION
Solving the relation r = mv / (Iql B) for the magnitude of the magnetic field,
and realizing that the radius is equal to the plate separation, we find that
B= mv = (1.67 x 10-27 kg)(3.5Xl06m/s)
Iqlr
= 016T
I.
(1.60 x 10-19 C)(0.23 m)
I
The values for the mass and the magnitude of the charge (which is the same as that of the
--~-
.--"lpdrr\n)Ju~ve heen.taken from the inside of the front cover.
--
_~
- ..-....~--~
..,,--..,,-- ~
~--.-....
_
..,~ ... - ,
"".....~.__ .The magnitude of the magnetic force exerted on a long straight wire is given
by Equation 21. 3 as F = ILB sin 8. The direction of the magnetic force is predicted by
Right-Hand Rule No. 1. The net force on the triangular loop is the vector sum of the forces
on the three sides.
...
30. REASONING
•......
•.
.•..
..
..•..
SOLUTION
a. The direction of the current in side AB is opposite to the direction of the magnetic field,
so the angle 8 between them is 8 = 180°. The magnitude of the magnetic force is
FAB
= ILB sin 8 = ILB sin 180° = ~
For the side BC, the angle is 8= 55.0°, and the length of the side is
L = 2.00 m
cos 55.00 = 3.49 m
The magnetic force is
FBC
=ILBsin8
= (4.70A)(3.49m)(1.80T)sin55.00
=
I
24.2N
,
An application of Right-Hand No. I shows that the magnetic force on side BC is directed
I
perpendicularly out of the paper
I, toward
the reader.
For the side AC, the angle is 8= 90.0°. We see that the length of the side is
L = (2.00 m) tan 55.0° = 2.86 m
The magnetic
FAC
fDTce
is
=ILBsin8
= (4.70A)(2.86
m)(1.80T) sin 90.0° = !24.2N
I
An application of Right-Hand No. I shows that the magnetic force on side AC is directed
I perpendicularly into the paper I, away from the reader.
b. The net force is the vector sum of the forces on the three sides. Taking the positive
direction as being out of the paper, the net force is
IF
= ON+24.2N+(-24.2N)
= ~
---"'.-------
34. REASONING Since the rod does not rotate about the axis at P, the net torque relative to
that axis must be zero; L1" = -0 (Equation 9.2). There are two torques that must be
considered, one due to the magnetic force and another due to the weight of the rod. We
consider both of these to act at the rod's center of gravity, which is at the geometrical center
of the rod (length = L), because the rod is uniform. According to Right-Hand Rule No. 1, the
magnetic force acts perpendicular to the rod and is directed up and to the left in the drawing.
Therefore, the magnetic torque is a counterc1ockwise (positive) torque. Equation 21.3 gives
the magnitude F of the magnetic force as F =!LB sin 90.0°, since the current is
perpendicular to the magnetic field. The weight is mg and acts downward, producing a
clockwise (negative) torque. The magnitude of each torque is the magnitude of the force
times the lever arm (Equation 9.1). Thus, we have for the torques:
1"magnetic
= +~
(ILB) (L
/2)
'-v----'
force
and
1"weight
= - ~[
lever arm
(L / 2) cos BJ
force
leve; arm
'
Setting the sum of these torques equal to zero will enable us to find the angle Bthat the rod
makes with the ground.
SOLUTION Setting the sum of the torques equal to zero gives L1"=
and we have
or
+ ( ILB) (L /2) - ( mg ) [ (L /2) cos BJ = 0
B
----..-.
---
41.
+
1"weight
= 0,
cos B = ILB
mg
(0.094 kg) (29.80 m/s )
= cos -l(4.1A)(0.4?m)(0.36T)l=
I
1440 I
.
...•.........
I SSMII
1"magnetic
wwwl
REASONING
,.~--
...••.
--
The torque
on the loop is given by Equation
21.4,
= NIABsin r/J. From the drawing in the text, we see that the angle r/J between the nonnal
to the plane of the loop and the magnetic field is 90° - 35° = 55°. The area of the loop is
0.70 m x 0.50 m = 0.35 m2.
1"
SOLUTION
a. The magnitude of the net ~orque exerted on the loop is
1" =
NIAB sin r/J = (75)(4.4 A)(0.35
m2
)(1.8 T) sin 55°= 1170 N· m I
b. As discussed in the text, when a current-carrying loop is placed in a magnetic field, the
loop tends to rotate such that its normal becomes aligned with the magnetic field. The
normal to the loop makes an angle of 55° with respect to the magnetic field. Since this angle
decreases as the loop rotates, the I 35° angle increases I.
54. REASONING
Theon
nettheforce
the wire
loop is a sum
of the
onl;,
each segment ofAND
the SOLUTION
loop. The forces
two on
segments
perpendicular
to the
longforces
straight
wire cancel each other out. The net force on the loop is therefore the sum of the forces
the parallel segments (near and far). These are:!
On "
y
Fnear
Ffar
::;
-4
= flil2L/(2J[dnear)
= flo(l2 A)(25 A)(0.50 m)/[2J[ (0.11 m)] = 2.7 x ION
= flil2L/(2J[dfa)
= flo(l2 A)(25 A)(0.50 m)/[2J[ (0.26 m)] = 1.2 x 10 N
}
-4
Note: F near is a force of attraction, while FeJar is a repulsive one. The magnitude of the net
force is, therefore,
F=F
-Fe =2.7x
near
lar.
,.-
'.~
-t.·
L
-.
1-----.
10-4N-1.2x
~LA
1.\.; d-
/1···.
[§sMJ Iwwwl REASONING AND SOLUTION
have the following:
Rod A:
1O-4N = !1.5X10-4N
I
----'.-.-----',,"
_
For the three rods in the drawing, we
The motional emf isl zero I, because the velocity of the rod is parallel to the
direction of the magnetic field, and the charges do not experience a magnetic force.
Rod B: The motional emf ~ is, according to Equation 22.1,
~ = vEL = (2,7 m/s)(0.45 T)(1.3 m) = 11.6 V ,
The positive end of Rod B is I end 2 I.
Rod C:
The motional emf is I zero [, because the magnetic force F on each charge is
directed perpendicular to the length of the rod. For the ends of the rod to become charged,
the magnetic force must be directed parallel to the length of the rod.
16. REASONING
SOLUTION
The change in flux L\<P is given by
where M is the area of the loop that leaves the region of the magnetic
field in a time L\t. This area is the product of the width of the rectangle (0.080 m) and the
length v L\t of the side that leaves the magnetic field, M = (0.080 m) v M.
L\<P
= B M cos
L\<P
AND
rp,
= E M cos rp= (2.4 T) (0.080 m)(0.020 m/s)(2.0 s) cos 0.00 = /7.7 X 10-3 Wb
,.--
------_._--,--,-
---'~-------'-~-"------' -~--_.
I
-------------"-_._----~------~
'-""-...._
./
18.
---
REASONING The magnitude I~Iof the emf induced in the loop can be found using
Faraday's law of electromagnetic induction:
(22.3)
where N is the number of turns,
t -
to
<I>
and
<1>0
are, respectively, the final and initial fluxes, and
is the elapsed time. The magnetic flux is given by
<I>
= BA cos r/J (Equation 22.2),
where B is the magnitude of the magnetic field, A is the area of the surface, and
.angle between the'direction of the magnetic field and the normal to the surface.
r/J
is the
SOLUTION Setting N = 1 since there is only one turn, noting that the final area is A = 0 m2
and the initial area is Ao = 0.20 m x 0.35 m, and noting that the angle r/J between the
magnetic field and the normal to the surface is 0°, we find that the magnitude of the emf
induced in the coil is
I~I= 1_ N
BA
cos r/J-
t-t BAo
o
cos r/J
= 1-(1) (0.65 T) (0 m2 )cosOo-
(0.65
mxO.35 m)cosOOj = 10.25 vi
0.18 T)(0.20
s
30. REASONING According to Lenz's law, the induced current in the triangular loop flows in
such a direc;tion so as to create an induced magnetic field that opposes the ·original flux
change.
SOLUTION
a. As the triangle is crossing the +y axis, the magnetic flux down into the plane of the paper
is increasing, since the field now begins to penetrate the loop. To offset this increase, an
induced magnetic field directed up and out of the plane of the paper is needed. By applying
RHR-2 it can be seen that such an induced magnetic field will be created within the loop by a
I counterclockwise
induced current I·
b. As the triangle is crossing the -x axis, there is no flux change, since all parts of the
triangle remain in the ma netic field, which remains constant. Therefore, there is no induced
magnetic field, and no induced current aDDears .
c. As the triangle is crossing the -y axis, the magnetic flux down into the plane of the paper
is decreasing, since the loop now begins to leave the field region. To offset this decrease, an
induced magnetic field directed down and into the plane of the paper is needed. By applying
RHR-2 it can be seen that such an induced magnetic field will be created within the loop by a
I clockwise induced current I.
23. REASONING
I~I= I-N
The magnitude
I~Iof
the average emf induced in the triangle is given by
~~I
(see Equation 22.3), which is Faraday's law. This expression can be used
directly to calculate the magnitude of the average emf. Since the triangle is a single-turn coil,
the number of turns is N = 1. According to Equation 22.2, the magnetic flux <Dis
(1)
<D= BA cos fjJ = BA cos 0° = BA
where B is the magnitude of the field, A is the area of the triangle, and fjJ = 0° is the angle
between the field and the normal to the plane of the triangle (the magnetic field is
perpendicular to the plane of the triangle). It is the change ~<P in the flux that appears in
Faraday's law, so that we use Equation (1) as follows:
~<P
= BA-BAa
= BA
where Aa = 0 m2 is the initial area of the triangle just as the bar passes point A, and A is the
area after the time interval ~t has elapsed. The area of a triangle is one-half the base (d AC)
times the height (dCB) of the triangle. Thus, the change in flux is
The base and the height of the triangle are related, according to dCB = d AC tan e, where
= 19°. Furthermore, the base of the triangle becomes longer as the rod moves. Since the
rod moves with a speed v during the time interval ""t, the base is d AC = vM. With these
(J
substitutions the change in flux becomes
(2)
SOLUTION Substituting Equation (2) for the change in flux into Faraday's law, we find
that the magnitude of the induced emf is
I~I= -N-""<PI
2
~t = I-N l.B(v""t)2tanBI
~t
I
=(1)1(0.38
=
Nl.Bv2~ttanB
2
T)(0.60 m/s)2 (6.0 s)tan19°=10.14
vi
31. ~SMJ REASONING
In solving this problem, we apply Lenz's law, which essentially says
that the change in magnetic flux must be opposed by the induced magnetic field.
SOLUTION
a. The magnetic field due to the wire in the vicinity of position 1 is directed out of the
paper. The coil is moving closer to the wire into a region of higher magnetic field, so the
flux through the coil is increasing. Lenz's law demands that the induc~d field counteract this
increase. The direction of the induced field, therefore, must be into the paper. The current
in the coil must be I clockwise I·
b. At position 2 the magnetic field is directed into the paper and is decreasing as the coil
moves away from the wire. The induced magnetic field, therefore, must be directed into the'
paper, so the current in the coil must be I clockwise
-------------_._-~--".
-_.'
.. ~"""
•.....•
-.,. .'. --
--.".
----.
- -
---
--
I.
--
.•
...
69. REASONING
Using Equation 22.3 (Faraday's law) and recognizing that N = 1, we can
write the magnitude of the emfs for parts a and b as follows:
and
(1)
(2)
To solve this problem, we need to consider the change in flux
for both parts of the drawing.
SOLUTION
Ll<1>
and the time interval
/!"t
The change in flux is the same for both parts of the drawing and is given by
( Ll<1> )
a
= (Ll<1»b
= <1>inside
-
<1>outside
(3)
= <1>inside = BA
In Equation (3) we have used the fact that initially the coil is outside the field region, so that
= 0 Wb for both cases. Moreover, the field is perpendicular to the plane of the coil
and has the same magnitude B over the entire area A of the coil, once it has completely
<1>ou tSI'de
entered the field region. Thus,
<1>.InS1'de
= BA in both cases, according to Equation 22.2.
The time interval required for the coil to entire the field region completely can be expressed
as the distance the coil travels divided by the speed at which it is pushed. In part a of the
drawing the distance traveled is W, while in part b it is 1. Thus, we have
(5)
(4)
Substituting Equations (3), (4), and (5) into Equations (1) and (2), we find
(7)
(6)
Dividing Equation (6) by Equation (7) gives
BA
Iqal_ W/v
~-~
L/v
=~=3.0
W
or
Iqbl--1tl=
3.0
0.15
3.0V =10.050 V I
4.
REASONING
The two capacitors, each of capacitance C and wired in parallel, are
equivalent to a single capacitor Cp that is the sum of the capacitances; Cp = C + C = 2C
(Equation 20.1S). The equivalent capacitance is related to the capacitive reactance Xc of the
circuit by Equation 23.2; Cp = 1/ (2.nI Xc).
Xc = Vnns
/
We can determine Xc by using Equation 23.1,
Inns' since the voltage and current are known.
SOLUTION Since Cp = 1/ (21Cf Xc)
and we know that Cp =2C, the capacitance of each
capacitor can be obtained as follows:
1
1
or
C = 41Cf Xc
Cp = 2C = 21Cf Xc
Since the capacitive reactance is related to the voltage and current by Xc = Vnns / Inns' we
have that
C=
1
_
1
41Cf
J=
---
0.16 A
(24V)=ls.7XI0-7FI
The nns voltage across and the nns current in the inductor are related
according to Vnns = InnsXL
XL
41C(610 Hz) --
nns
(VInns
41CfXc-
12. REASONING
1
(Equation 23.3), where the inductive reactance XL is given by
= 21Cf L (Equation 23.4). The frequency is denoted by f and the inductance by 1.
Since the voltage, current, and frequency are given, we can use these two expressions to
detennine the inductance.
SOLUTION According to Equation 23.3 the nns voltage and current are related by
Substituting XL = 21Cf L (Equation 23.4) for the inductive reactance gives
Solving for the inductance L, we find that
L=
Vnns
Inns21Cf
19. I SSMI REASONING
-----------=
39V
(42Xl0-3 A)21C(7.5Xl03 Hz)
I 0.020H ,
We can use the equations for a series RCL circuit to solve this
problem provided that we set Xc = 0 since there is no capacitor in the circuit. The current
in the circuit can be found from Equation 23.6, Vnns = I nns Z, once the impedance of the
Equation 23.8, tanqJ=(XL -Xc)/R,
circuit has been obtained.
can then be used (with
Xc = 0 Q) to find the phase angle between the current and the voltage.
SOLUTION
The inductive reactance is (Equation 23.4)
XL = 2n
f L = 2n(1 06 Hz) (0.200 H) = 133 Q
The impedance of the circuit is
a. The current through each circuit element is, using Equation 23.6,
234 Q
V =10.925 A I
1rms _- Vrms
Z = 253
b. The phase angle between the current and the voltage is, according to Equation 23.8 (with
Xc = 0 Q),
tanqJ = X-X
L
R
C
= ~X
R
= --133Q = 0.619
215 Q
or
qJ=tan-1(0.619)=
~31.80
24. REASONING AND SOLUTION
At very low frequencies the capacitors behave as if they
were cut out of the circuit, while the inductors behave as if they were replaced with wires
that have zero resistance. The circuit behaves as shown in drawing A, and the current
delivered by the generator in the limit of very low frequency is 11ow fr equency = V / Rj .
A: Low frequency
B: High frequency
At very high frequencies the capacitors behave as if they were replaced with wires that have
zero resistance, while the inductors behave as if they were cut out of the circuit. The circuit·
now behaves as in drawing B, and the two resistors are in series. The current at very high
frequencies is 1 h·rg h frequency = V / (Rj + Rz ) .
The ratio ofthe currents is known, so that we can obtain the ratio of the resistances:
1
low frequency
____
_=_
_
1 high frequency
-
V / RI
V / (Rj
+ Rz
or
)
8.
f
REASONING
The wavelength A of a wave is related to its speed v and frequency
by
A = v/
(Equation 16.1). Since blue light and orange light are electromagnetic waves, they
travel through a vacuum at the speed of light e; thus, v = e.
f
SOLUTION
a. The wavelength of the blue light is
A=~=
f
3.00x108 m/s
Hz = 4.73x10-7 m
6.34x1014
Since 1 nm = 10-9 m,
b. In a similar manner, we find that the wavelength of the orange light is
A
=
e
f
15. I SSMI REASONING
3.00x108 m/s = 6.06x10-7 m = 1606 nml
= 4.95x1014 Hz
We proceed by first finding the time t for sound waves to travel
between the astronauts. Since this is the same time it takes for the electromagnetic waves to
travel to earth, the distance between earth and the spaceship is dearth-ship = et .
SOLUTION The time it takes for sound waves to travel at 343 m/s through the air between
the astronauts is
d
t = astronaut _ 1.5 m
vsound
- 343 m/s = 4.4 x 10-3 s
Therefore, the distance between the earth and the spaceship is
dearth-ship
=et= (3.0x108
m/s)(4.4x10-3
s)=11.3X106
m I
4.
REASONING
The drawing at the right
shows the laser beam after reflecting from
the plane mirror. The angle of reflection is
a, and it is equal to the angle of incidence,
which is 33.0°. Note that the angles labeled
fJ 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
1.10
ID
A
B
1.80 Ih
D
I
c
Floor
Applying trigonometry to triangle DBC, we see that
tanfJ = dBC
dBC
or
dDC
dDC
= tan
(1)
fJ
The distance dBC can be determined from dBC = 1.80 m - dAB' which can be substituted into
Equation (1) to show that
d
- _d_BC 1._8_0
_m_-_dAB
tanfJ tanfJ
(2)
DC -
The distance dAB can be found by applying trigonometry to triangle LAB, which shows that
d
AB
or
dAB = (1.10 m) tan fJ
tan fJ = 1.10
m
Substituting this result into Equation (2), gives
1.80 m-dAB
dDC =
tan fJ
5.
I SSMI
REASONING
1.80 m-(1.1O m)tanfJ
tanfJ
_1.80 m-(1.10 m)tan33.00
tan33.0°
11.67
ml
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
We can use the law of
the angle of reflection Br equals the angle of incidence Bi
reflection and the properties of triangles to determine the angle B at which the ray leaves
.
M2•
.........
"
:
..•.•.••..
lJ}:
)W
:::: .
fJ.
.:~~W
'::::'
.:::'
,:':~: :0
::Hr
'::~:
:
)?M 2
::::
."iiiiii!i::?
H~mmmmmmwmmwmmmmwmmmm~wmmr:
M1
SOLUTION
From the law of reflection, we know that r/J = 65°. We see from the figure
that r/J+ a = 90°, or a = 90° - r/J= 90° - 65° = 25°. From the figure and the fact that the
sum of the interior angles in any triangle is 180°, we have a + fJ + 120° = 180°. Solving for
fJ, we find that fJ = 180° - (120° + 25°) = 35°.
Therefore, since fJ + r = 90°, we find that
the angle r is given by r = 90° - fJ = 90° - 35° = 55°. Since r is the angle of incidence of
the rav on mirror M_. wekno"!"Jrom.lheJ~.w_ofreflecti()nJhat
I e3C_55~ I.. __..
__
h
..
16. REASONING
Since the image is behind the mirror, the image is virtual, and the image
distance is negative, so that dj = -34.0 cm. The object distance is given as do = 7.50 cm.
f
The mirror equation relates these distances to the focal length 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
1
1
1
-=-+-
f
do
1
f=---=
1
1
-+d
d.
or
dj
o
1
-+--7.50 cm (-34.0
1
1
= 19.62 cml
1
cm)
Since the focal length is positive, the mirror is Iconcavel·
17. I SSMI REASONING
We have seen that a convex mirror always forms a virtual image as
shown in Figure 25.22a of the text, where the image is upright and smaller than the object.
These characteristics should bear out in the results of our calculations.
SOLljTION
The radius of curvature of the convex mirror is 68 cm. Therefore, the focal
= -(l/2)R
= -34 cm. Since the image is virtual, we know
length is, from Equation 25.2,
that d.I = -22 cm.
f
a. With
dj
= -22 cm and
1
-1 --do
f
f
= -34 cm, the mirror equation gives
1
1
----
1
d.I
-34 cm
-22 cm
or
I do
= +62 cm I
b. According to the magnification equation, the magnification is
dj
m=-do
__
-22 cm = I +0.35
62cm
I
c. Since the magnification m is positive, the image is I upright I·
d. Since the magnification m is less than one, the image is
I
smaller
I
than the object.
18.
REASONING The magniftcation of the mirror is related to the image and object distances
via the magnification equation. The image distance is given, 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 dj and the object distance do according to
d
m=-_1
(25.4)
do
According to the mirror equation, the image distance
to the focallengthfas
follows:
1
1
1
-+-=do
f
dj
1
or
dj
1
-=--do f
and the object distance do are related
1
(25.3)
dj
Substituting this expression for lido into Equation 25.4 gives
12 cm
= _(36
cm)+ 1= 1-2.01
----- ------~----------~--- -- ------------ -..-..~=;; ,~, -22.
REASONING
a. We are given that the focal length of the mirror is = 45 cm and that the image distance is
one-third the object distance, or dj = tdo . These two pieces of information, along with the
mirror equation, will allow us to find the object distance.
1
b. Once the object distance has been determined, the image distance is one-third that value,
since we are given that dj = do .
t
SOLUTION
a. The mirror equation indicates that
Substituting
dj
-+-=
-1
d
d
1
1
o
1
1
(25.3)
t
= do and solving for do gives
4_~
or
d -1
o
or
do = 41 = 4(45 cm) = 1180 cm I
b. The image distance is
d1303
=ld
=1(180 cm)=!6.0X101
.
_cm I