35.56: a) There is one half-cycle phase shift, so for constructive interference:
1
2tn
2(380 nm) (1.45) 1102 nm

2t =  m +  0 ⇒ =
=
=
.
1
2 n
(m + 2 )
(m + 12 )
(m + 12 )

Therefore, we have constructive interference at = 441 nm (m = 2), which corresponds
to blue-violet.
b) Beneath the water, looking for maximum intensity means that the reflected
part of the wave at the wavelength must be weak, or have interfered destructively. So:
m 0
2tn 2(380 nm) (1.45) 1102 nm
2t =
⇒ 0=
=
=
.
n
m
m
m
Therefore, the strongest transmitted wavelength (as measured in air) is
= 551 nm (m = 2), which corresponds to green.
22.34:
The α particle feels no force where the net electric field is zero. The fields can
cancel only in regions A and B.
Eline = Esheet
2
0
r
=
2
0
50 µC/m
= 0.16m = 16cm
π (100 µC/m 2 )
The fields cancel 16 cm from the line in regions A and B.
r=
πσ =
22.34:
The α particle feels no force where the net electric field is zero. The fields can
cancel only in regions A and B.
Eline = Esheet
2
0
r
=
2
0
50 µC/m
= 0.16m = 16cm
π (100 µC/m 2 )
The fields cancel 16 cm from the line in regions A and B.
r=
πσ =
22.36: a) For r < a, E = 0, since no charge is enclosed.
For a < r < b, E =
q
1
4
0
r2
, since there is +q inside a radius r.
For b < r < c, E = 0, since now the –q cancels the inner +q.
For r > c, E =
q
1
4
0
r2
, since again the total charge enclosed is +q.
b)
c) Charge on inner shell surface is –q.
d) Charge on outer shell surface is +q.
e)
22.45: a) a < r < b, E =
b) r > c, E =
1
4
0
1 2
, radially outward, as in 22.48 (b).
4πε 0 r
2
r , radially outward, since again the charge enclosed is the
same as in part (a).
c)
d) The inner and outer surfaces of the outer cylinder must have the same amount
of charge on them: l = − inner l ⇒ inner = − , and outer = .
22.46: a) (i) r < a, E (2 rl ) =
q
=
l
⇒E=
.
2 0r
(ii) a < r < b, there is no net charge enclosed, so the electric field is zero.
0
(iii)
0
r > b, E (2 rl ) =
q
0
=
2 l
0
⇒E=
0
r
.
b) (i) Inner charge per unit length is − . (ii) Outer charge per length is + 2α .
22.47: a) (i) r < a, E (2 rl ) =
q
0
=
l
0
⇒E=
2
0r
, radially outward.
(ii) a < r < b, there is not net charge enclosed, so the electric field is zero.
(iii) r > b, there is no net charge enclosed, so the electric field is zero.
b) (i) Inner charge per unit length is − α .
(ii) Outer charge per length is ZERO.
23.18: Initial energy equals final energy:
Ei = E f ⇒ −
keq1 keq2
keq1 keq2 1
2
−
=−
−
+ me v f
2
r1i
r2i
r1 f
r2 f
 (3.00 × 10 −9 C) (2.00 × 10 −9 C) 
 = − 2.88 × 10 −17 J
+
Ei = k ( − 1.60 × 10 −19 C) 
0.25
m
0.25
m


−9
−9
 (3.00 × 10 C) (2.00 × 10 C  1
 + me v 2f
+
E f = k ( − 1.60 × 10 −19 C) 
0.10 m
0.40 m
 2

1
2
= − 5.04 × 10 −17 J + me v f
2
2
(5.04 × 10 −17 J − 2.88 × 10 −17 J)
⇒ vf =
−31
9.11 × 10 kg
= 6.89 × 10 6 m s.
23.56: Fe = mg tan = (1.50 × 10 −3 kg) (9.80 m s ) tan (30°) = 0.0085 N . (Balance
forces in x and y directions.) But also:
Fd (0.0085 N) (0.0500 m)
Vq
Fe = Eq =
⇒V =
=
= 47.8 V.
q
d
8.90 × 10 − 6 C
2
23.57: a) (i) V =
2
(ln(b a) − ln (b b)) =
0
(ii) V =
2
(iii) V = 0.
(ln(b r ) − ln( b b)) =
0
b) Vab = V (a ) − V (b ) =
2
c) Between the cylinders:
V =
2
∴E = −
ln( b r ) =
0
ln( b a).
2
0
2
0
ln(b r ) .
ln( b a).
0
Vab
ln( b r )
ln (b a)
∂V
Vab ∂
Vab 1
=−
(ln(b r )) =
.
∂r
ln( b a ) ∂r
ln( b a) r
d) The potential difference between the two cylinders is identical to that in part
(b) even if the outer cylinder has no charge.
23.58: Using the results of Problem 23.57, we can calculate the potential difference:
Vab 1
E=
⇒ Vab = E ln (b/a)r
ln( b a) r
⇒ Vab = (2.00 × 10 4 N C) (ln (0.018 m 145 × 10 −6 m)) 0.012 m = 1157 V.
23.59: a) F = Eq = (1.10 × 103 V m) (1.60 × 10−19 C) = 1.76 × 10−16 N, downward.
b) a = F me = (1.76 × 10 −16 N) (9.11 × 10 −31 kg) = 1.93 × 1014 m s , downward.
0.060 m
1
1
2
c) t =
= 9.23 × 10 −9 s, y − y0 = at 2 = (1.93 × 1014 m s ) (9.23 × 10 − 9 s) 2
6
6.50 × 10 m s
2
2
2
= 8.22 × 10 −3 m.
d) Angle = arctan(v y vx ) = arctan(at vx ) = arctan(1.78 6.50) = 15.3°.
e) The distance below center of the screen is:
0.120 m
D = d y + v yt = 8.22 × 10− 3 m + (1.78 × 106 m s )
= 0.0411 m.
6.50 × 106 m s
23.60:
(a) Use ∆Vab to get : ∆V = ∫ E ⋅ dl = ∫
=
E=
b
b
a
a
2
dr =
0
2
ln
0
2 0 ∆V
ln b a
2
0
r
=
2
0
∆V ln b a
∆V
=
2 0r
r ln b a
at outer surface of the wire, r = a = 0.1272 mm
850 V
E=
= 2.65 × 10 6 V m
m
00 cm 
( 0.000127
) ln  (10..0127
2
 2 cm ) 
(b) at the inner surface of the cylinder, r = 1.00 cm , which gives
E = 1.68 × 104 V m
b
a
c
kq
kq
dr = .
2
c
∞ r
c
r r b r r kq
kq
b) At r = b : V = − ∫ E ⋅ d r − ∫ E ⋅ d r =
−0= .
c
c
∞
c
23.76: a) At r = c : V = − ∫
a
r r b r r a r r kq
dr
1 1 1 
c) At r = a : V = − ∫ E ⋅ d r − ∫ E ⋅ d r − ∫ E ⋅ d r =
− kq ∫ 2 = kq  − + .
c
r
c b a
∞
c
b
b
1 1 1 
d) At r = 0 : V = kq  − +  since it is inside a metal sphere, and thus at the
c b a 
same potential as its surface.
c
C
2 0
=
L ln (rb ra )
(0.180 m)2 0
C=
= 4.35 × 10−12 F
ln (5.00 0.50)
24.9: a)
b) V = Q / C = (10.0 × 10 −12 C) /(4.35 × 10 −12 F) = 2.30 V
1  rb ra  1  (0.148 m)(0.125 m) 
= 

 = 8.94 × 10 −11 F.
k  rb − ra  k  0.148 m − 0.125 m 
b) The electric field at a distance of 12.6 cm:
kQ kCV k (8.94 × 10 −11 F)(120 V)
E= 2 = 2 =
= 6082 N/C.
r
r
(0.126 m) 2
c) The electric field at a distance of 14.7 cm:
kQ kCV k (8.94 × 10−11 F)(120 V )
E= 2 = 2 =
= 4468 N/C.
r
r
(0.147 m)2
d) For a spherical capacitor, the electric field is not constant between the
surfaces.
24.13: a) C =
24.40: a) E0 = KE = (3.60)(1.20 × 106 V m) = 4.32 × 106 V m ⇒
−5
= 0 E0 =
3.82 × 10 C m .
2
1

= 1 −  = (3.82 × 10 −5 C m 2 )(1 − 1 3.60) = 2.76 × 10 − 5 C m 2 .
 K
1
c) U = 2 CV 2 = uAd = 12 0 E 2 Ad
b)
i
⇒ U = 12 (3.60) 0 (1.20 × 106 V m) 2 (0.0018 m)(2.5 × 10−4 m 2 ) = 1.03 × 10 −5 J.
24.44: a) ∆Q = Q − Q0 = ( K − 1)Q0 = ( K − 1)C0V0 = (2.1)(2.5 × 10−7 F)(12 V ) =
6.3 × 10 −6 C.
b) Qi = Q (1 − K1 ) = (9.3 × 10−6 C)(1 − 1 3.1) = 6.3 × 10−6 C.
c) The addition of the mylar doesn’t affect the electric field since the induced
charge cancels the additional charge drawn to the plates.
25.76: a) Rsteel =
RCu =
ρ L (2.0 × 10 −7 Ω ⋅ m) (2.0 m)
=
= 1.57 × 10 −3 Ω
2
A
(π 4) (0.018 m)
L (1.72 × 10 −8 Ω ⋅ m) (35 m)
=
= 0.012 Ω
A
( 4) (0.008 m) 2
⇒ V = IR = I ( Rsteel + RCu ) = (15000 A ) (1.57 × 10 −3 Ω + 0.012 Ω) = 204 V.
b) E = Pt = I 2 Rt = (15000 A) 2 (0.0136 Ω) (65 × 10 −6 s) = 199 J.
26.69: a) When the switch is open, only the outer resistances have current through them.
So the equivalent resistance of them is:
−1


36.0 V
1
1
V
 = 4.50 Ω ⇒ I =
=
= 8.00 A
+
Req = 
6
3
3
6
4.50
Ω
Ω
+
Ω
Ω
+
Ω
R
eq


1

1

⇒ Vab =  8.00 A  (3.00 Ω ) −  8.00 A  (6.00 Ω) = −12.0 V.
2

2

b) If the switch is closed, the circuit geometry and resistance ratios become identical
to that of Problem 26.60 and the same analysis can be carried out. However, we can also
use symmetry to infer the following:
I 6Ω = 23 I 3Ω , and I switch = 13 I 3Ω . From the left loop as in Problem 26.60:
1
2

36 V −  I 3Ω (6 Ω) − I 3Ω (3 Ω) = 0 ⇒ I 3Ω = 5.14 A ⇒ I switch = I 3Ω = 1.71 A.
3
3

36.0 V
2
5
ε
=
= 4.20 Ω.
(c) I battery = I 3Ω+ I 3Ω = I 3Ω = 8.57 A ⇒ Req
3
3
I battery 8.57 A
26.70: a) With an open switch: Vab = ε = 18.0 V, since equilibrium has been reached.
b) Point “a” is at a higher potential since it is directly connected to the positive
terminal of the battery.
c) When the switch is closed:
18.0 V = I (6.00 Ω + 3.00 Ω) ⇒ I = 2.00 A ⇒ Vb = (2.00 A )(3.00 Ω) = 6.00 V.
d) Initially the capacitor’s charges were:
Q3 = CV = (3.00 × 10 −6 F)(18.0 V) = 5.40 × 10 −5 C.
Q6 = CV = (6.00 × 10 −6 F)(18.0 V) = 1.08 × 10 − 4 C.
After the switch is closed:
Q3 = CV = (3.00 × 10 −6 F)(18.0 V − 12.0 V) = 1.80 × 10 −5 C.
Q6 = CV = (6.00 × 10 −6 F)(18.0 V − 6.0 V) = 7.20 × 10 −5 C.
So both capacitors lose 3.60 × 10 −5 C.
26.82: For a charged capacitor, connected into a circuit:
Q
I 0 = 0 ⇒ Q0 = I 0 RC = (0.620 A )(5.88 kΩ)(8.55 × 10 −10 F) = 3.12 × 10 −6 C.
RC
110 V
ε
=
= 1.69 × 10 6 Ω ⇒
−5
I 0 6.5 × 10 A
6.2 s
τ
C= =
= 3.67 × 10 −6 F.
6
R 1.69 × 10 Ω
26.83: ε = I 0 R ⇒ R =
27.17: K 1 + U 1 = K 2 + U 2
U1 = K 2 = 0, so K 1 = U 2 ;
v=e
b)
1 2
mv = ke 2 r
2
2k
= (1.602 × 10 −19 C)
mr
∑ F = ma gives qvB = mv
−27
(3.34 × 10
2
− 27
2k
= 1.2 × 10 7 m s
kg)(1.0 × 10 −15 m)
r
mv (3.34 × 10 kg)(1.2 × 10 7 m/s)
B=
=
= 0.10 T
qr
(1.602 × 10 −19 C)(2.50 m)
27.19: q = (4.00 × 10 8 )( − 1.602 × 10 −19 C) = 6.408 × 10 −11 C
speed at bottom of shaft:
1
2
mv 2 = mgy; v = 2 gy = 49.5 m/s
v is downward and B is west, so v × B is north. Since q < 0, F is south.
F = qvB sin = (6.408 × 10 −11 C)(49.5 m s )(0.250 T) sin 90° = 7.93 × 10 −10 N
27.34: a) F = Ilb = (1.20 A) (0.0100 m) (0.588 T) = 7.06 × 10 −3 N, and by the righthand
rule, the easterly magnetic field results in a southerly force.
b) If the field is southerly, then the force is to the west, and of the same magnitude as
part (a), F = 7.06 × 10 −3 N.
c) If the field is 30° south of west, the force is 30° west of north ( 90°
counterclockwise from the field) and still of the same magnitude, F = 7.60 × 10 −6 N.
27.39: a) FI = mg when bar is just ready to levitate.
2
mg (0.750 kg)(9.80m s )
IlB = mg, I =
=
= 32.67 A
lB
(0.500 m)(0.450 T)
= IR = (32.67 A)(25.0 ) = 817 V
b) R = 2.0 Ω, I = ε R = (816.7 V) (2.0 Ω) = 408 A
FI = IlB = 92 N
a = ( FI − mg ) a = 113 m s
2
→
→
27.55: The direction of E is horizontal and perpendicular to v , as shown in the sketch:
FB = qvB, FE = qE
FB = FE for no deflection , so qvB = qE
E = vB = (14.0 m s)(0.500 T) = 7.00 V m
We ignored the gravity force. If the target is 5.0 m from the rifle, it takes the
bullet 0.36 s to reach the target and during this time the bullet moves downward
y − y 0 = 12 a y t 2 = 0.62 m. The magnetic and electric forces we considered are horizontal.
A vertical electric field of E = mg q = 0.038 V m would be required to cancel the
gravity force. Air resistance has also been neglected.
28.15: a) B =
I
(800 A)
= 0
= 2.90 × 10 − 5 T, to the east.
2 r 2 (5.50 m)
0
b) Since the magnitude of the earth’s magnetic filed is 5.00 × 10 −5 T, to the north,
the total magnetic field is now 30 o east of north with a magnitude of 5.78 × 10 −5 T. This
could be a problem!
µ 0 I1 I 2 L µ 0 (5.00 A ) (2.00 A) (1.20 m)
=
= 6.00 × 10 −6 N, and the force is
2πr
2π (0.400 m)
repulsive since the currents are in opposite directions.
b) Doubling the currents makes the force increase by a factor of four to
F = 2.40 × 10 −5 N.
28.22: a) F =
28.32: Consider a coaxial cable where the currents run in OPPOSITE directions.
r
I
a) For a < r < b, I encl = I ⇒ ∫ B ⋅ dl = 0 I ⇒ B 2 r = 0 I ⇒ B = 0 .
2 r
b) For r > c, the enclosed current is zero, so the magnetic field is also
zero.
28.37:
Br
I
, so I =
= 3.72 × 106 A
2 r
(µ0 2 )
NI
2aBx
b) Bx = 0 , so I =
= 2.49 × 105 A
2a
0N
c) B = 0 nI = µ 0 ( N L) I , so I = BL 0 N = 237 A
a) B =
0
28.73:
Apply Ampere’s law to a circular path of radius 2a.
B (2 r ) = 0 I encl
 ( 2a ) 2 − a 2 
 = 3I 8
I encl = I 
2
2 
 (3a) − a 
3 0I
; this is the magnetic field inside the metal at a distance of 2a from
16 2 a
the cylinder axis.
I
Outside the cylinder, B = 0 . The value of r where these two fields are equal is
2 r
given by 1 r = 3 (16a) and r = 16a 3.
B=
29.10: According to Faraday’s law (assuming that the area vector points in the positive zdirection)
0 − (1.5 T)π (0.120 m) 2
∆Φ
ε =−
=−
= + 34 V(counterclockwise )
2.0 × 10 −3 s
∆t
29.12: a)
dΦ B d
ε =
= ( NBA cos ωt ) = NBA ω sin t and 1200 rev min = 20 rev s, so :
dt
dt
⇒ ε max = NBAω = (150)(0.060 T )π (0.025 m) 2 (440 rev min)(1 min 60sec)( 2π rad rev) = 0.814 V.
b) Average ε =
2
2
ε max = 0.814 V = 0.518 V.
π
π
∆Φ B
π (0.0650 2 m) 2
∆A
− πr 2
29.53: a) ε = −
=−B
=−B
= (0.950 T)
= 0.0126 V.
∆t
∆t
∆t
0.250 s
b) Since the flux through the loop is decreasing, the induced current must produce a
field that goes into the page. Therefore the current flows from point a through the
resistor to point b .
29.56: The bar will experience a magnetic force due to the induced current in the loop.
According to Example 29.6, the induced voltage in the loop has a magnitude BLv, which
opposes the voltage of the battery, ε . Thus, the net current in the loop is I = ε −RBLv . The
acceleration of the bar is a =
F
m
=
ILB sin(90 ° )
m
=
( ε − BLv ) LB
mR
.
) LB
a) To find v(t ), set dv
and solve for v using the method of separation
= a = (ε − BLv
dt
mR
of variables:
2 2
v
t LB
− 3.1t s
dv
ε
− BmRL t
dt
v
(
1
e
)
(
10
m
s
)
(1
e
).
=
→
=
−
=
−
∫0 (ε − BLv) ∫0 mR
BL
Note that the graph of this function is similar in appearance to that of a charging
capacitor.
2
b) I = R = 2.4 A; F = ILB = 2.88 N; a = F m = 3.2 m s
c) When
[12 V − (1.5 T) (0.8 m) (2.0 m s)] (0.8 m) (1.5 T)
2
v = 2.0 m s , a =
= 2.6 m s
(0.90 kg) (5.0Ω)
d) Note that as the velocity increases, the acceleration decreases. The velocity will
V
asymptotically approach the terminal velocity BLε = (1.5 T)12 (0.8
m) = 10 m s , which makes the
acceleration zero.
30.65: a) Just after the switch is closed the voltage V5 across the capacitor is zero and there
is also no current through the inductor, so V3 = 0. V2 + V3 = V4 = V5 , and since
V5 = 0 and V3 = 0, V4 and V2 are also zero. V4 = 0 means V3 reads zero.
V1 then must equal 40.0 V, and this means the current read by A1 is
(40.0 V) / (50.0 Ω) = 0.800 A.
A2 + A3 + A4 = A1 , but A2 = A3 = 0 so A4 = A1 = 0.800 A.
A1 = A4 = 0.800 A; all other ammeters read zero.
V1 = 40.0 V and all other voltmeters read zero.
b) After a long time the capacitor is fully charged so A4 = 0. The current through
the inductor isn’t changing, so V2 = 0. The currents can be calculated from the equivalent
circuit that replaces the inductor by a short-circuit.:
I = (40.0 V) / (83.33 Ω) = 0.480 A; A1 reads 0.480 A
V1 = I (50.0 Ω) = 24.0 V
The voltage across each parallel branch is 40.0 V − 24.0 V = 16.0 V
V2 = 0, V3 = V4 = V5 = 16.0 V
V3 = 16.0 V means A2 reads 0.160 A. V4 = 16.0 V means A3 reads 0.320 A. A4 reads zero.
Note that A2 + A3 = A1 .
c) V5 = 16.0 V so Q = CV = (12.0 µF)(16.0 V) = 192 µC
d) At t = 0 and t → ∞, V2 = 0. As the current in this branch increases from zero to
0.160 A the voltage V2 reflects the rate of change of current.
30.66: (a) Initially the capacitor behaves like a short circuit and the inductor like an open
circuit. The simplified circuit becomes
i=
75 V
ε
=
= 0.500 A
R 150 Ω
V1 = Ri = (50 Ω)(0.50 A) = 25.0 V
V3 = 0, V4 = (100 Ω)(0.50 A) = 50.0 V
V2 = V4 (in parallel) = 50.0 V
A1 = A3 = 0.500 A, A2 = 0
(b) Long after S is closed, capacitor stops all current. Circuit becomes
V3 = 75.0 V and all other meters read zero.
(c) q = CV = (75 nF)(75 V) = 5630 nC, long after S is closed.
30.67: a) Just after the switch is closed there is no current through either inductor and
they act like breaks in the circuit. The current is the same through the 40.0 Ω and 15.0 Ω
resistors and is equal to (25.0 V) (40.0 Ω + 15.0 Ω) = 0.455 A. A1 = A4 = 0.455 A;
A2 = A3 = 0.
b) After a long time the currents are constant, there is no voltage across either inductor,
and each inductor can be treated as a short-circuit . The circuit is equivalent to:
I = (25.0 V) (42.73 Ω) = 0.585 A
A1 reads 0.585 A. The voltage across each parallel branch is 25.0 V − (0.585 A)(40.0 Ω) =
1.60 V. A2 reads (1.60 V) (5.0 Ω) = 0.320 A. A3 reads (1.60 V) 10.0 Ω) = 0.160 A. A4
reads (1.60 V) (15.0 Ω) = 0.107 A.
c 3.00 × 108 m s
=
= 361 m.
8.30 × 10 5 Hz
f
2
2
=
= 0.0174 m −1
b) k =
361 m
32.7: a)
c)
=
= 2π f = 2 (8.30 × 10 5 Hz) = 5.21 × 10 6 rad s.
E max = cBmax = (3.00 × 10 8 m s ) (4.82 × 10 −11 T ) = 0.0145 V m.
=
32.11: a)
b)
0
=
v 2.17 × 108 m s
=
= 3.81 × 10 −7 m.
14
5.70 × 10 Hz
f
c 3.00 × 108 m s
=
= 5.26 × 10 −7 m.
14
f 5.70 × 10 Hz
c 3.00 × 10 8 m s
c) n = =
= 1.38.
v 2.17 × 10 8 m s
d) v =
c
KE
⇒ KE =
c2
= n 2 = (1.38) 2 = 1.90.
2
v
ρI
ρI
= 2 , in the direction of the current.
A πa
µ I
b) ∫ B ⋅ d l = µ 0 I ⇒ B = 0 , counterclockwise when looking into the current.
2πa
c) The direction of the Poynting vector S = E × B = k × φ = − , where we have used
cylindrical coordinates, with the current in the z-direction.
EB
1 I µ0 I
I2
Its magnitude is S =
.
=
=
µ 0 µ 0 πa 2 2πa 2π 2 a 3
32.45: a) E = ρJ =
d) Over a length l, the rate of energy flowing in is SA =
ρI 2
ρlI 2
al
2
π
=
.
2π 2 a 3
πa 2
ρl ρlI 2
=
, which exactly equals the flow of
The thermal power loss is I R = I
A πa 2
electromagnetic energy.
2
2
33.8
(a)
Apply Snell’s law at both interfaces.
At the air-glass interface:
(1.00) sin 41.3° = nglass sin α
At the glass-methanol interface:
Combine (1) and (2):
nglass sin α = (1.329) sin θ
sin 41.3° = 1.329 sin
= 29.8°
(b) Same figure as for (a), except
= 20.2°.
(1.00) sin 41.3° = n sin 20.2°
n = 1.91
(2)
n 
33.20 : θ crit = arcsin  b  = arcsin
 na 
 1.00 

 = 24.4°.
 2.42 
n
33.35: θ b = 90° − arcsin  a
 nb

 1.00 
 = 90° − arcsin 
 = 43.6°.
 1.38 

 n sin θ b
But na sin θ a = nb sin θ b ⇒ θ a = arcsin  b
 na

 1.38 sin( 43.6°) 
 = arcsin 
 = 72.1°.
1.00



 4.0 cm 
 8.0 cm 
 = 14°.
 = 27° and θ b = arctan
33.41 θ a = arctan
 16.0 cm 
 16.0 cm 
 n sin θ a   1.00 sin 27° 
 = 
So, na sin θ a = nb sin θ b ⇒ nb =  a
 = 1. 8.
sin
θ
 sin 14° 
b 

33.42: The beam of light will emerge at the same angle as it entered the fluid as seen by
following what happens via Snell’s Law at each of the interfaces. That is, the emergent
beam is at 42.5° from the normal.
 n sin 90° 
 1.000 
 = arcsin 
33.43: a) θ i = arcsin  a
 = 48.61°.
nw
 1.333 


The ice does not come into the calculation since nair sin 90° = nice sin θ c = nw sin θ i .
b) Same as part (a).
35.8: a) For the number of antinodes we have:
m
mc
m(3.00 × 10 8 m s )
=
=
sin θ =
= 0.2317 m, so, setting θ = 90°,
(12.0 m) (1.079 × 108 Hz)
d
df
the maximum integer value is four. The angles are ± 13.4°, ± 27.6°, ± 44.0°, and ± 67.9°
for m = 0, ± 1, ± 2, ± 3, ± 4.
( m + 1 2)
b) The nodes are given by sin θ =
= 0.2317 (m + 1 2). So the angles are
d
± 6.65°, ± 20.3°, ± 35.4°, 54.2° for m = 0, ± 1, ± 2, ± 3.
35.10: For bright fringes:
(1.20 m) (20)(5.02 × 10 −7 m)
Rm
=
= 1.14 × 10 −3 m = 1.14 mm.
d=
0.0106 m
ym
35.14: Using Eq.35.6 for small angles,
m
,
d
we see that the distance between corresponding bright fringes is
(5.00 m) (1)
Rm
(660 − 470) × (10 −9 m) = 3.17 mm.
∆y =
∆ =
(0.300 × 10 −3 m)
d
ym = R
35.36: a) Since there is a half-cycle phase shift at just one of the interfaces, the
minimum thickness for constructive interference is:
550 nm
t= = 0 =
= 74.3 nm.
4 4n 4(1.85)
b) The next smallest thickness for constructive interference is with another half
3
3
3(550 nm )
= 0 =
= 223 nm.
wavelength thickness added: t =
4
4n
4(1.85)
35.41: a) Hearing minimum intensity sound means that the path lengths from the
individual speakers to you differ by a half-cycle, and are hence out of phase by 180° at
that position.
b) By moving the speakers toward you by 0.398 m, a maximum is heard, which means
that you moved the speakers one-half wavelength from the min and the signals are back
in phase. Therefore the wavelength of the signals is 0.796 m, and the frequency is
v 340 m/s
f = =
=427 Hz.
0.796 m
c) To reach the next maximum, one must move an additional distance of one
wavelength, a distance of 0.796 m.
35.55: a) Intensified reflected light means we have constructive interference. There is
one half-cycle phase shift, so:
1
2tn
2(485 nm) (1.53) 1484 nm

2t =  m +  ⇒ =
=
=
.
1
2 n
(m + 2 )
(m + 12 )
(m + 12 )

⇒ = 593 nm( m = 2), and = 424 nm (m = 3).
b) Intensified transmitted light means we have destructive interference at the upper
surface. There is still a one half-cycle phase shift, so:
m
2tn 2(485 nm) (1.53) 1484 nm
⇒ =
=
=
2t =
.
n
m
m
m
⇒ = 495 nm (m = 3)
is the only wavelength of visible light that is intensified. We could also think of this as
the result of internal reflections interfering with the outgoing ray without any extra phase
shifts.