Download PHY122 Midterm2 (practice)

FINAL.
1. Shown in the Figure below is a loop of wire that lies in a horizontal plane.
You have a standard bar magnet that you will move either up or down
and in the process either insert or remove it from the loop. This problem
concentrates on knowing which way the current flows in the loop (i.e. all
answers are either “clockwise” or “counter-clockwise”). Which way does
N
W
E
S
the current flow if you:
(a) Insert the North end of the magnet into the loop from below.
ANSWER: clockwise.
(b) Insert the South end of the magnet into the loop from below.
ANSWER: counter-clockwise.
(c) Insert the North end of the magnet into the loop from above.
ANSWER: counter-clockwise.
(d) Insert the South end of the magnet into the loop from above.
ANSWER: clockwise.
(e) Remove the North end of the magnet from the loop from below.
ANSWER: counter-clockwise.
(f) Remove the South end of the magnet from the loop from below.
ANSWER: clockwise.
(g) Remove the North end of the magnet from the loop from above.
ANSWER: clockwise.
(h) Remove the South end of the magnet from the loop from above.
ANSWER: counter-clockwise.
1
2. A coil with 20 turns is placed in a magnetic field whose strength is 0.1
Tesla. The radius of the coil is 0.15 meters and it is turned with a frequency
of f=60 Hz. This device is called a generator and produces an AC voltage.
The output of the generator is feed into a transformer that has 100 turns
on the input side and 400 turns on the output side. Finally, the output of
the transformer is attached to a 15 Ω resistor.
(a) What is the peak voltage produced by the generator?
ANSWER:
V (t) = ωN BA cos ωt
Vpeak = ωN BA
(1)
(2)
Vpeak = (2πf )N BA
(3)
Vpeak = (2πf )N B(πr2 )
Vpeak = 53.30V
(4)
(5)
(b) What is the RMS voltage produced by the generator?
ANSWER:
Vpeak
VRM S = √
2
53.30
VRM S =
1.414
VRM S = 37.69V
(6)
(7)
(8)
(c) What is the peak voltage produced on the output side of the transformer.
ANSWER:
N1
V1
=
V2
N2
N2
V2 =
V1
N1
V2 = 213.18V
(9)
(10)
(11)
(d) What is the RMS voltage produced on the output side of the transformer.
ANSWER:
Vpeak
VRM S = √
2
218.18
VRM S =
1.414
VRM S = 150.74V
2
(12)
(13)
(14)
(e) What is the peak current through the resistor?
ANSWER:
V = IR
VRM S = IRM S R
(15)
(16)
Vpeak = Ipeak R
Vpeak
Ipeak =
R
213.18
Ipeak =
15
Ipeak = 14.21Amps
(17)
(18)
(19)
(20)
(f) What is the RMS current through the resistor?
ANSWER:
Ipeak
IRM S = √
2
14.21
IRM S =
1.414
IRM S = 10.05Amps
(21)
(22)
(23)
(g) What is the AVERAGE power dissipated by the resistor?
ANSWER:
hP i = VRM S IRM S cos φ
(24)
φresistor = 0
(25)
hP i = 150.74 × 10.05 cos 0
hP i = 1515W atts
(26)
(27)
(h) What is the average torque required to keep the loop spinning?
ANSWER: Ah Ha! A non-trivial one. Here we must use the principle of conservation of energy. The power out of the transformer
equals the power input to the system. So, we must be getting 1515
Watts out of the generator due to the applied torque necessary to
spin the loop. In linear dimensions, power is force times velocity. In
rotational terms, it is torque times angular velocity:
Plinear = F · v
Pangular = τ · ω
1515 = τ · 2πf
1515
τ=
2π60
τ = 4.02N ewton · meters
3
(28)
(29)
(30)
(31)
(32)
3. Shown in the Figure below is a rectangular loop of dimensions 0.20 m by
0.30 m. The loop is partly filled with a magnetic field whose strength is
0.25 T and is directed into the page. The loop has a resistance of 3 Ω.
You grab the loop and pull it to the right with a constant speed of 0.4
m/sec.
x
x
x
B=0.25 T
x
x
x
x
x
x
x
x
x
B=0
0.30 m
0.20 m
x
x
x
x
x
x
x
x
x
(a) Which way does the current flow (clockwise or counter-clockwise)?
ANSWER: The loop contains magnetic field pointer down. When
you slide it to the right, the downward fluc would be reduced. The
loop responds by adding additional downward field lines by making
current in the clockwise direction.
(b) What is the voltage generated in the loop?
ANSWER: Since you already answered the direction of the current,
you do not need to specify the direction of the voltage (its the same).
So all we need to do is calculate the amount.
dΦB
dt
dΦB
|V | =
dt
ΦB = BA = Blw
dΦB
d(Blw)
dw
=
= Bl
= BLv
dt
dt
dt
dΦB
= 0.25 · 0.20 · 0.4 = 0.02V olts
dt
|V | = 0.02V olts
V =−
(c) What is the current in the loop?
4
(33)
(34)
(35)
(36)
(37)
(38)
(39)
ANSWER:
V = IR
V
I=
R
I=
0.02
= 0.0066Amps = 6.6mA
3
(40)
(41)
(42)
(d) How much force is required to keep the loop moving at 0.4 m/s?
F = ILB = 0.0066 · 0.20 · 0.25 = 0.000333N ewtons
5
(43)
4. Shown in the figure below is the view from above of a sliding bar of resistance R and length L. The bar is free to slide along the rails without
friction. The system is immersed in a magnetic field pointed out of the
page (toward you). The bar is given an initial velocity v0 to the right.
Answer all the following:
L
y
Sliding Bar
with resistance R
v
(a) Write an expression for the flux through the loop formed by the bar
and wire slide for the moment when the bar’s position is “x”.
ANSWER:
ΦB = BA = BLx
(44)
(b) Using your result from (a) determine an expression for the voltage
generated in the loop as a function of the bar’s velocity.
ANSWER:
dΦB
dt
(45)
dx
d(Blv)
= −Bl
= −BLv
dt
dt
(46)
V =−
V =−
(c) Determine the current through the bar.
ANSWER:
V = IR
V
I=
R
BLv
I =−
R
6
(47)
(48)
(49)
(d) Determine the force on the bar as a function of velocity.
ANSWER:
F = ILB
BLv
F =−
LB
R
B 2 L2
v
F =−
R
(e) What direction is this force (justify your response).
ANSWER:The force is to the left (opposing the change).
7
(50)
(51)
(52)
5. A wire with radius a carries a current I. The current is uniformly distributed throughout the wire.
I
a
i
ii
(a) Use Ampere’s Law to calculate the magnetic field in region i.
ANSWER: Our Ampere’s path will be a circular loop whose length
is 2πr. This path contains all of the current.
I
−
~ ·→
B
dl = µ0 Iin
(53)
Iin = I
(54)
BL = µ0 I
µ0 I
B=
2πr
(55)
(56)
(b) Use Ampere’s Law to calculate the magnetic field in region ii.
ANSWER: Our Ampere’s path will be a circular loop whose length
is 2πr. This path only a fraction of the current.
I
−
~ ·→
B
dl = µ0 Iin
(57)
Iin = I
r2
πr2
=
πa2
a2
(58)
(59)
2
BL = µ0 I
8
r
a2
(60)
B=
µ0 I
µ0 Ir2
=
r
2
2πra
2πa2
9
(61)
6. Shown in the figure below is a solenoid carrying a current I.
(a) Assume the solenoid is very long and then use Ampere’s law to determine the magnetic field inside the solenoid.
ANSWER: Our Ampere’s path will be a rectangular loop of length
L. This loop makes no contribution outside the coil (long coil approximation) and no contribution on the sides (field perpendicular
to path). So, the only contribution is BL from the side inside the
coil.
Also, if the solenoid carries a current I, the loop catches MORE than
I since it contains more than one loop of the coil. We’ll assume that
the ampere’s path has N loops that cut through it.
I
−
~ ·→
B
dl = µ0 Iin
(62)
Iin = N I
(63)
BL = µ0 N I
N
B = µ0 I
L
(64)
(65)
(b) Calculate the field if this solenoid has 500 coils, is 0.2 meters long,
and carries a current of 1 Ampere.
ANSWER:
N
I
L
500
B = 4π × 10−7
1
0.2
B = 0.00314T esla
B = µ0
(66)
(67)
(68)
(c) Calculate the energy stored in the solenoid if its radius is 0.01 meters.
10
ANSWER:There are TWO ways to calculate this result. The first
way to calculate it would be to use the energy per unit volume, u:
U
1 B2
=
V
2 µ0
2
1B
U = uV =
× πr2 L
2 µ0
1 µ0 N 2 I 2
U = uV =
× πr2 L
2 L2
1 µ0 N 2 πr2 2
I
U = uV =
2
L
u=
(69)
(70)
(71)
(72)
However, we can ALSO calculate this using the formula for the inductance of such a coil:
1 2
LI
2
2
2
µ0 N A
µ0 N πr2
L=
=
Length
Length
1 µ0 N 2 πr2 2
I
U=
2 Length
U=
(73)
(74)
(75)
Of course, these two method produce precisely the same result! We
can simply plug in the numbers and get the final result:
U = 0.000247Joules
11
(76)
7. The circuit below begins with the switch in the open position and the
capacitor carrying zero charge. The battery has VB = 10V , the resistor
has R = 5kΩ, and the capacitor has C = 12µF . At time=0, the switch in
the circuit is closed.
(a) Determine the time constant of this circuit.
ANSWER: τ = RC = 5kΩ12µF = 60msec
(b) Draw sketches of the time dependence of each of the following:
i. The voltage on the capacitor as a function of time, VC (t).
VC
10 V
t
ii. The charge on the capacitor as a function of time, QC (t).
QC
120 µ C
t
iii. The voltage on the resistor as a function of time, VR (t).
iv. The current through the resistor as a function of time.
To be counted for full credit the vertical axis of each sketch should
be labelled with a numerical value indicating either the initial or
asymptotic value of the quantity plotted.
12
VR
10 V
t
I R
2 mA
t
(c) Write an equation for each of the following:
i. The voltage on the capacitor as a function of time, VC (t).
ANSWER:The voltage on the capacitor is a saturating exponential whose eventual
is the full 10 V of the battery.
voltage
t
− 60msec
Thus, VC (t) = 10V 1 − e
ii. The charge on the capacitor as a function of time, QC (t).
ANSWER:Since the asymptotic voltage on the capacitor is 10
V, and V = Q
C or Q = CV , the asymptotic
charge on the
capact
itor is Qf inal = 120µC. QC (t) = 120µC 1 − e− 60msec
iii. The voltage on the resistor as a function of time, VR (t).
ANSWER:The resistor takes up the balance of the battery voltage that is not across the capacitor. This is a decaying exponent
tial: VR (t) = 10V e− 60msec
iv. The current through the resistor as a function of time.
ANSWER:Well, V = IR implies that I = VR . Since that starting voltage on the r esistor is 10 V, the starting I = 10V
5kΩ = 2mA.
t
− 60msec
IR (t) = 2mAe
(d) At what time does the voltage on the capacitor reach 3 V?
ANSWER:Using the capacitor voltage formula, we plug in 3 V for
the voltage in the VC (t) formula. Then we have:
t
3V = 10V 1 − e− 60msec
13
(77)
t
0.3 = 1 − e− 60msec
0.7 = e
t
− 60msec
t
ln 0.7 = −
60msec
t = −60msec ∗ ln 0.7 = 21.4msec
14
(78)
(79)
(80)
(81)
8. The circuit below begins with the switch in the open position. The battery
has VB = 8V , the resistor has R = 2kΩ, and the inductor has L = 3mH.
At time=0, the switch in the circuit is closed.
(a) Determine the time constant of this circuit.
ANSWER:
τ=
L
3mH
=
= 1.5µsec
R
2kΩ
(82)
(b) Draw sketches of the time dependence of each of the following:
i. The voltage on the inductor as a function of time, VL (t).
VL
8V
t
ii. The voltage on the resistor as a function of time, VR (t).
VR
8V
t
iii. The current through the resistor as a function of time, IR (t).
To be counted for full credit the vertical axis of each sketch should
be labelled with a numerical value indicating either the initial or
asymptotic value of the quantity plotted.
(c) Write an equation for each of the following:
ANSWER:
15
I R
4 mA
t
i. The voltage on the inductor as a function of time, VL (t).
t
(83)
VL (t) = 8V e− 1.5µsec
ii. The voltage on the resistor as a function of time, VR (t).
t
VR (t) = 8V 1 − e− 1.5µsec
(84)
iii. The current through the resistor as a function of time, IR (t).
t
IR (t) = 4mA 1 − e− 1.5µsec
(85)
(d) At what time is the voltage on the inductor 3 V?
t
3V = 8V e− 1.5µsec
0.375 = e
t
− 1.5µsec
t
ln 0.375 = −
1.5µsec
t = −1.5µsec × ln 0.375 = 1.47µsec
16
(86)
(87)
(88)
(89)
9. Shown below is an RL circuit driven by an AC power source.
R
L
Vp
(a) Draw a phasor disgram representing this circuit.
Z
ωL
φz
R
(b) Analyze your phasor diagram to determine the magnitude of the
impedance, |Z|.
ANSWER:Z is the vector sum of the resistor and inductor impedance.
|Z| is the magnitude of the impedance vector. The x-component of
the impedance vector is R and the y-component is ωL. Thus, the
magntude of the impdance vector is
q
2
|Z| = R2 + (ωL)
(c) Analyze your phasor diagram to determine the phase of the impedance,
φZ .
ANSWER:The phase of the impedance is the same as the direction
of the impendance vector. The direction of a vector is obtained using
the relation θ = tan−1 xy . In the case of our circuit, this yields
φZ = tan−1 ωL
R .
(d) Explain in a single sentence what the phase of the impendance means.
ANSWER:The phase of the impendance tells you the phase difference between the total voltage applied to the circuit and the total
current flowing through the circuit.
(e) Let the resistance be R = 5kΩ, the inductance be L = 3mH and the
voltage source have a peak voltage, Vpeak = 10V , at a frequency of
f = 5kHz. Determine the peak current through the circuit.
17
ANSWER:
Vpeak = Ipeak |Z|
q
2
|Z| = R2 + (ωL)
ω = 2πf = 2π5000Hz = 10000π = 31415rad/s
ωL = 94.2Ω
q
|Z| = R2 + (ωL)2 = 5001Ω
Ipeak = 1.9996mA
(90)
(91)
(92)
(93)
(94)
(95)
(f) Determine the peak voltage across the inductor.
ANSWER:
Vpeak(inductor) = Ipeak |Zinductor |
Vpeak(inductor) = Ipeak ωL
(96)
(97)
Vpeak(inductor) = 1.9996mA94.2Ω = 188mV
(98)
(g) If the voltage across the inductor is used to power another circuit,
would this system be considered a high-pass or a low-pass filter?
Explain.
ANSWER:The impedance of an inductor is high when the frequency is high. Thus, it takes the most voltage at the highest frequencies. This circuit would then be a high pass filter.
18
10. Shown below is an RC circuit driven by an AC power source.
(a) Draw a phasor diagram representing this circuit.
φz
R
1
ωC
Z
~
(b) Determine the magnitude of the impedance, |Z|.
ANSWER:
s
2
1
2
~
|Z| = R +
ωC
(c) Determine the phase of the impedance, φZ .
ANSWER:
−1 −1
φZ = tan−1 ωC = tan−1
R
ωCR
(99)
(100)
(d) Explain in a single sentence what the phase of the impedance means.
ANSWER:The phase of the impedance equals the phase difference
between the total voltage and total current.
(e) Let the resistance be R = 1.0kΩ, the capacitance be C = 0.5µF
and the voltage source have an RMS voltage, VRM S = 12V , at a
frequency of f = 300Hz. Determine the RMS current through the
circuit.
ANSWER:
s
2 s
2
1
1
2
2
~
= R +
(101)
|Z| = R +
ωC
2πf C
19
1
1
=
= 1061Ω
2πf C
2π × 300 × 0.5 × 10−6
p
~ = 10002 + 10612 = 1458Ω
|Z|
12
VRM S
=
IRM S =
= 8.23mA
~
1458
|Z|
(102)
(103)
(104)
(f) Determine the RMS voltage across the capacitor.
ANSWER:
VRM S(cap)
~ (cap)
VRM S(cap) = IRM S |Z|
1
VRM S(cap) = IRM S
2πf C
= 8.23mA × 1061Ω = 8.73V
(105)
(106)
(107)
(g) What circuit element would you add to this circuit (i.e. making a
new circuit) such that the result was a resonant circuit with resonant
frequency f = 30kHz? Please specify the type of circuit element (resistor, capacitor, or inductor) and the value (resistance, capacitance,
or inductance) required.
ANSWER:Add an inductor such that:
1
LC
√
1
L= √
ω C
ω=√
L=
1
ω2 C
=
1
= 56.3µH
(2πf )2 C
20
(108)
(109)
(110)
11. The “middle” of the AM radio band is at roughly f=1000 kHz. The
“middle” of the FM radio band is at roughly f=100 MHz. Determine the
wavelength of each of these two radio stations.
ANSWER: OK, whew! A very easy one:
c = λf
f
λ=
c
1000 × 103
= 0.00333m
3 × 108
100 × 106
λ2 =
= 0.333m
3 × 108
λ1 =
21
(111)
(112)
(113)
(114)
(115)
12. You were outside your family spacecraft taking a leasurely spacewalk and
your impetulant kid speed off into the cosmos leaving you just floating
there, about 150 billion meters from the sun. At your location, the inW
tensity of sunlight is roughly 1000 m
2 and the gravity is not completely
zero, so you start to fall toward the sun. Even though it would take a long
time to reach the sun, you are concerned that if your position changes alot
you’ll never be found.
So you take action:
(a) You calculate the average electric field of the light.
W
ANSWER: OK, now you should remember that the intensity ( m
2)
U
2 J
is related to the energy density (u = V = 0 E ; m3 ) by the relation
I = uc. Thus:
I = 0 E 2 × c
r
I
E=
0 c
r
1000
E=
8.85 × 10−12 · 3 × 108
V
E = 614
m
(116)
(117)
(118)
(119)
(b) You calculate the average magnetic field of the light.
W
ANSWER: OK, now you should remember that the intensity ( m
2)
B2 J
U
is related to the energy density (u = V = µ0 ; m3 ) by the relation
I = uc. Thus:
B2
×c
µ0
r
µ0 I
B=
c
r
−7
4π × 10 · 1000
B=
3 × 108
B = 0.00000205T esla
I=
(120)
(121)
(122)
(123)
WAYYY-DA-MINNIT. Aren’t the magnetic field and electric field in
an EM wave proportional via:
E = cB
E
B=
c
B=
614
= 0.00000205T esla
3 × 108
Um, yup :)
22
(124)
(125)
(126)
(c) You calculate the pressure applied to a reflective solar sail.
ANSWER: The pressure applied by EM waves varies from P = Ic
to P = 2I
c as the substance to which the light is applied goes from
absorbative to reflective. Yours is reflective:
2I
c
2000
P =
3 × 108
N
P = 0.00000666P a( 2 )
m
P =
(127)
(128)
(129)
(d) You calculate the size of a solar sail necessary to exactly balance the
sun’s gravity.
ANSWER: Well, due to gravity you get F = G mM
r 2 . Due to the
solar wind, you get F = P ressure × Area. Set these equal to avoid
falling into the sun!
mM
= PA
r2
GmM
A=
P r2
−11
6.67 × 10
· 100 · 2 × 1030
A=
0.00000666 · (150 × 109 )2
G
A = 88942m2
(130)
(131)
(132)
(133)
As a square, this would be 300 meters on each side. See how bad
the Star Wars Part-II Solar Sail was...WAY WAY too small.
(e) What do you write on the sail?
ANSWER: No allowance for the next 3 centons. QUICK: Which
science fiction program measures time in centons???
Oh, you and your space suit have a total mass of 100 kg and you happen
to remember that the mass of the sun is 2 × 1030 kg.
23
13. Shown in the figure below is a system containing an object, and two lenses.
Use the shapes of the lenses in the figure to decide whether they are
converging or diverging optical elements.
|f|=8 cm
15 cm
|f|=30 cm
4 cm
(a) Find the image location and magnification of the first lens (assuming
that only this lens exists). Specify this image location, di1 , as some
number of centimeters to the left or to the right of this lens.
ANSWER:
do1 = 15cm
f1 = 8cm
1
1
1
+
=
15 di1
8
di1 = 17.14cm
17.14cm
m1 = −
= −1.143
15cm
(134)
(135)
(136)
(137)
(right of lens)
(b) The image of the first lens acts as the object for the second. Find
the location and magnification of the image produced by the second
lens. Specify this image location, di2 , as some number of centimeters
to the left or to the right of this lens.
ANSWER:
do2 = −13.14cm
f2 = −30cm
1
1
1
+
=
−13.14 di2
−30
di1 = 23.38cm
23.38cm
m1 = −
= 1.779
−13.14cm
(right of lens)
24
(138)
(139)
(140)
(141)
(c) Calculate the total magnification of this entire system.
ANSWER:
mT OT = m1 × m2 = −2.034
(d) Is the final image real or virtual?
ANSWER: real
25
(142)
14. Shown in the figure below is a system containing an object, a lens, and a
mirror. This is a 3-pass system in that light goes from the object, through
the lens, off the mirror, and back through the lens a second time.
Use the shape of the objects in the figure to decide whether they are
converging or diverging optical elements.
|f|=5 cm
20 cm
R = 10 cm
8 cm
(a) Find the image location and magnification of the first lens (assuming
that only this lens exists).
ANSWER: Because the object is on the incoming side, do1 is positive, i.e. do1 = +20cm. Because the lens is diverging, the focal length
is negative, i.e. f1 = −5cm. Using d1o1 + d1i1 = f11 , we have
m1 = −
di1
do1
1
1
1
+
=
20 di1
−5
1
1
1
=
−
di1
−5 20
1
= −0.25
di1
di1 = −4cm
−4
=−
= +0.2
20
(143)
(144)
(145)
(146)
(147)
This means the image is NOT on the outgoing side, so the image is
4 cm to the left of the first lens.
(b) The image of the first lens acts as the object for the mirror. Find the
location and magnification of the image produced by the mirror.
ANSWER: Since the image of the first lens is 4 cm to the left of the
lens itself, the object distance for the mirror is 8cm + 4cm = 12cm.
This distance is positive since this “object” is on the incoming side,
i.e. do2 = +12cm. The focal length of a curved mirror is 21 of the
26
radius. Also, since this mirror is a converging device, the focal length
is positive. Thus, f2 = +5cm. Plugging these in:
di2
m2 = −
do2
1
1
1
+
=
12 di2
5
1
1
1
= −
di2
5 12
1
= 0.11666
di2
di2 = 8.5714cm
8.5714
=−
= −0.714
12
(148)
(149)
(150)
(151)
(152)
Since this image distance is positive, the image is on the outgoing
side of the mirror at a distance of 8.57 cm. The outgoing side of the
mirror is to the left of the mirror.
(c) Lastly, the image from the mirror acts as the object for the light’s
second pass through the lens. Find the location and magnification of
the image produced in this step.
ANSWER: This one is a bit tricky. Since the image of the mirror
(object for the lens) is 8.5714 cm away from the mirror and the
lens is only 8 cm from the mirror, the object for the lens NOT
on the incoming side. Thus, the object distance will be negative,
do3 = −0.5714cm. Again, the lens is diverging, so that the focal
length is f = −5cm. OK, putting this in:
di3
m3 = −
do3
1
1
1
+
=
−0.5714 di3
−5
1
1
1
=
−
di3
−5 −0.5714
1
= 1.55
di3
di3 = 0.645cm
0.645
=−
= 1.129
−0.5714
(153)
(154)
(155)
(156)
(157)
This image is on the outgoing side meaning to the left of the lens.
(d) Calculate the total magnification of this entire system. Is the final
image real or virtual?
ANSWER: mT OT = m1 ×m2 ×m3 = 0.2×−0.714×1.129 = −0.161
This image is real since it is on the outgoing side of the last optical
element.
27
15. A particular person’s nearsighted eye has a near point of 12 cm and a far
point of 17 cm. You will place a lens 2 cm in front of their eye. Answer
all the following:
(a) Do you use a converging or diverging lens?
ANSWER: This near-sighted person should have a diverging lens.
(b) What is the focal length of the chosen lens?
ANSWER: Well, this person has a far point of 17 cm. Normally,
the eye should have a far point of infinity. So, what we do with the
lens is we take objects at infinity, and put the image 17 cm in front
of this person’s eye.
Now, since the lens is not touching the eye (these are glasses not
contacts), we need an image that is 17-2=15 cm from the lens. Note
that this image will NOT be on the outgoing side, so we conclude
di = − 15cm when do = ∞.
1
1
1
+
=
do
di
f
1
1
1
+
=
∞ −15
f
1
1
=
−
15
f
f = −15cm
(158)
(159)
(160)
(161)
(c) What is this person’s new nearpoint?
ANSWER: Well, this person will reach the nearpoint when the
image from the glasses is 12 cm in front of the eye, or 10 cm in front
of the lens di = − 10 cm.
1
1
1
+
=
do
di
f
1
1
1
+
=
do
−10
−15
1
1
1
=
−
do
10 15
do = 30cm
(162)
(163)
(164)
(165)
Wait, what if this poor person wants to see an object when it is
closer than 30 cm? Well, they take off their glasses!! How many
times have you seen a person look at something closely by taking off
their glasses? Each such time, you were witnessing a near-sighted
person.
28
16. Shown in the figure below is a water glass at two different moments. In
the first moment (left figure) the water glass is empty and the light ray
strikes the bottom corner as shown. In the second moment, the glass has
been filled with water (n=1.33). As a result, the light ray has been bent.
Light Ray
Light Ray
θ1
θ1
n=1.33
θ2
15 cm
Full water glass
Empty water glass
x
8 cm
(a) Determine the angle θ1 as shown in the left figure.
ANSWER: Examining the figure closely, we see that the angle θ1
is also the upper angle in the triangle formed by the right ray and
the left and bottom sides of the glass (using vertical angles). In this
8cm
. Solving this we find that θ1 = 28.1o .
case, we see that tan θ1 = 15cm
(b) Determine the angle θ2 as shown in the right figure.
ANSWER: Using Snell’s law:
n1 sin θ1 = n2 sin θ2
1 × sin 28.1o = 1.33 × sin θ2
0.354 = sin θ2
θ2 = 20.74o
(166)
(167)
(168)
(169)
(c) Determine the distance x shown in the right figure.
ANSWER: Examining the triangle, we see the following:
x
15cm
x = 15cm × tan θ2
tan θ2 =
o
x = 15cm × tan 20.74
x = 5.68cm
29
(170)
(171)
(172)
(173)
17. As shown in the figure below, a small boy has lost his toy at the bottom
of a pool (nwater = 1.33). When he shines his flashlight as shown, he
can see the toy.
θ1
1m
3.0 m
θ2
1.5 m
x
(a) What is the angle of incidence, θ1 , of his light upon the water?
ANSWER:
3.0
1.0
θ1 = 71.6o
tan θ1 =
(174)
(175)
(b) What is the angle of refraction θ2 ?
ANSWER:
n1 sin θ1 = n2 sin θ2
o
(176)
1.0 sin 71.6 = 1.33 sin θ2
(177)
o
(178)
θ2 = 45.5
(c) What is the distance from the water’s edge, x, to the lost toy?
ANSWER:
x = 3m + l
l
tan θ2 =
1.5
l = 1.5 tan θ2
l = 1.5 tan 45.5o
(179)
l = 1.53m
x = 4.53m
(183)
(184)
30
(180)
(181)
(182)
18. Shown in the figure below is Young’s two-slit experiment. Monochromatic
light of wavelength λ = 0.633µm comes in from the left and passes through
the two (very narrow) slits. The result is a pattern of bright and dark
bands visible on a screen 5 meters away from the two slits. WARNING:
this drawing is not to scale.
3 cm
d
Dark
Bright
Dark
Bright
5 meters
(a) Determine the distance, d, between the two slits.
ANSWER: The distance labelled in the figure (3 cm) is the distance
to the first bright band. This band follows the relation d sin θ = λ.
The angle to this band can be found as follows:
0.03m
3cm
=
5m
5m
θ = 0.344o
d sin θ = λ
λ
0.633µm
d=
=
= 105.5µm
sin θ
sin 0.344o
(185)
tan θ =
(186)
(187)
(188)
(b) If the light is changed to green light, λ = 0.532µm, the diffraction
pattern will change. What is the location, y, of the first dark band
using this light.
ANSWER: The dark bands follow the relation d sin θ = n + 12 λ
where n = 0, ±1, ±2, .... The first dark band is n = 0 or d sin θ = 21 λ.
Using this with the new wavelength yields:
d sin θ =
1
λ
2
1
× 0.532µm
2
1 0.532µm
sin θ = ×
2 105.5µm
sin θ = 0.00252
105.5µm sin θ =
sin θ = 0.00252
θ = 0.1445o
y
tan θ =
5meters
y = 5meters × tan 0.1445o = 0.00504m = 0.504cm
31
(189)
(190)
(191)
(192)
(193)
(194)
(195)
(196)
(c) Make a sketch of the diffraction pattern resulting from single slit
interference.
Double width bright spot
Single width bright bumps
32
19. A thin film of soap water (n=1.33) appears green when lit from above.
Calculate the thickness of the film if it appears green (λ = 500 nm).
Assume that this is the thinnest film that can produce green light.
ANSWER: OKAY. Each time light is reflected from an interface between
two materials of different index of refraction, there is a chance there will
be a phase change in the light. If the second medium is higher n, the
phase change is ∆φ = π. However, if the second medium is lower n, the
phase change is ∆φ = 0.
So, if you have a film with nearly zero thickness, the reflections from the
two surfaces cancel (watch the top of a soap bubble just before it bursts,
the reflections go away!). Now, we *want* reflections, so the film should
not have zero thickness. Instead the round-trip distance of the light should
be 1/2 wavelength so that the destructive turns to constructive.
Finally, the 1/2 wavelength should be 1/2 of the wavelength while the light
is in the soap film!!
1λ
2n
λ
1λ
=
d=
4n
4n
d = 94nm
2d =
33
(197)
(198)
(199)
20. Assume that three perfect polarizers are placed in sequence and receive
incoming light from one side. The first polarizer has it’s optic axis oriented
vertically. The second is oriented 30 degrees to one side of vertical. The
third is oriented horizontally.
(a) What fraction of the incoming light gets through the first polarizer?
For any polarizer the outgoing intensity is related to the incoming intensity via the formula Iout = Iin cos2 φ. Here φ is the angle between
the polarization of the light and the axis of the polarizer. Oops!
In the first stage the incoming light is unpolarized. This means φ
takes2 all
possible angles. However, we know about the average value:
cos φ = 21 . Thus:
I1 = I0 cos2 φ
1
I1 = I 0
2
1
I1
=
F raction =
I0
2
(200)
(201)
(202)
(b) What fraction of the incoming light gets through both the first and
second polarizer?
In this case the calculation is simpler:
I2 = I1 cos2 φ
√ !2
3
2
o
I2 = I1 cos 30 = I1
2
3
4
13
I2 = I 0
24
I2
13
3
=
=
F raction =
I0
24
8
I2 = I 1
(203)
(204)
(205)
(206)
(207)
(c) What fraction of the incoming light gets through the first, second,
and third polarizers?
Again, a simple case:
I3 = I2 cos2 φ
2
1
I3 = I2 cos2 60o = I2
2
1
I3 = I 2
4
34
(208)
(209)
(210)
31
84
31
3
I2
=
=
F raction =
I0
84
32
I3 = I 0
(211)
(212)
(d) How much light would get through if the second and third polarizers
were exchanged?
Nothing!!
(e) Which way are the polarizers in your sunglasses oriented. Explain
why.
Glare comes mostly from reflections off of horizontal surfaces. This
glare is highly polarized parallel to the surface (i.e. horizontally).
To cut the glare, you orient the axis of your polarizing singlasses
vertically.
(f) Explain how polarizing glasses are used to view high quality 3D
movies.
3D movies operate on the principle of sending different images into
each eye. The images are selected so that they view the same scene
from different angles. To make a high quality 3D movie, you use two
psojectors sending light to the same screne. The two projectors send
polarizations that are perpendicular to each other. By using “special
glasses” (i.e. with polarizing filters over the two eyes) you can see the
depth of the image. All one does is match the polarization direction
of projector 1 to the left eye, and projector 2 to the right eye. Bottom
line is different images to different eyes and a 3D movie results.
35