Version 001 – Test 6 Wave Phenomena – tubman – (IBII201516)
This print-out should have 25 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
1
In the Doppler effect for sound waves, which
factors affect the frequency that the observer
hears?
1. II and III only
AP B 1998 MC 49
001 10.0 points
A small vibrating object on the surface of a
ripple tank is the source of waves of frequency
20 Hz and speed 60 cm/s. The source S is
moving to the right with speed 20 cm/s, as
shown.
2. None is true.
3. All are true.
4. III only
5. II only
D
6. I and III only correct
A
S
C
7. I and II only
8. I only
B
At which of the labeled points will the frequency measured by a stationary observer be
greatest?
1. It will be the same for all four points.
2. A
3. D
4. C correct
5. B
Explanation:
The Doppler effect for stationary observers
is
vsound
f′ =
f0 .
vsound − vsource
Since the source is moving directly toward
point C, the measured frequency at point C
by a stationary observer will be greatest.
AP B 1993 MC 58
002 10.0 points
Consider the following:
I. the speed of the source;
II. the loudness of the sound;
III. the speed of the observer.
Explanation:
In the Doppler effect of sound waves, both
the speed of the source and the speed of the
observer affect the frequency that the observer
hears. (The directions of the movement are
factors, too.)
On the other hand, the loudness of the
sound (the intensity of the sound wave) is
irrelevant.
Falling Sky Diver
003 (part 1 of 2) 10.0 points
In order to be able to determine her speed,
a skydiver carries a tone generator. A friend
on the ground at the landing site has equipment for receiving and analyzing sound waves.
While the skydiver is falling at terminal speed,
her tone generator emits a steady tone of
1310 Hz.
If her friend on the ground (directly beneath the skydiver) receives waves of frequency 2330 Hz, what is the skydiver’s speed
of descent? Assume the air is calm and the velocity of sound in air is 343 m/s, independent
of altitude.
Correct answer: 150.155 m/s.
Explanation:
Version 001 – Test 6 Wave Phenomena – tubman – (IBII201516)
Let :
fe = 1310 Hz ,
fg = 2330 Hz ,
v = 343 m/s .
2
You will pick up an echo at what approximate frequency? The speed of sound is
350 m/s .
and
Since the diver (the source of the sound
waves) is moving toward her friend on the
ground (the receiver of the waves) the Doppler
formula takes the form
v
fg = fe
v − vdiver
fe
vdiver
=1−
fg
v
fe
vdiver = v 1 −
fg
1310 Hz
= (343 m/s) 1 −
2330 Hz
1. 1.1 kHz correct
2. 1.05 kHz
3. 0.9 kHz
4. 1.5 kHz
5. 0.95 kHz
Explanation:
Let :
= 150.155 m/s .
004 (part 2 of 2) 10.0 points
If the skydiver were also carrying soundreceiving equipment sensitive enough to detect waves reflected from the ground, what
frequency would she receive?
Correct answer: 3350 Hz.
Explanation:
When the waves are reflected, and the skydiver is moving toward them,
v + vdiver
frec = fg
v
= (2330 Hz)
343 m/s + 150.155 m/s
×
343 m/s
= 3350 Hz .
Doppler Shift of an Echo
005 10.0 points
Suppose you are at the bottom of a canyon.
You are driving toward the canyon wall at
35 m/s . A stationary siren behind you produces a sound also at the bottom of the canyon
at a frequency of 1.00 kHz.
v = 35 m/s ,
cs = 350 m/s ,
f = 1 Hz .
and
The Doppler shift in frequency according
to an observer (the driver) moving toward a
stationary source (the echo from the canyon
wall) is
v
f =f 1+
c
s
35 m/s
= 1.1 Hz .
= (1 Hz) 1 +
350 m/s
′
Serway CP 24 09
006 10.0 points
Waves from a radio station have a wavelength
of 68 m. They travel by two paths to a home
receiver 16.5 km from the transmitters. One
path is a direct path, and the second is by
reflection from a mountain directly behind
the home receiver.
What is the minimum distance from the
mountain to the receiver that produces destructive interference at the receiver? (Assume that no phase change occurs on reflection from the mountain.)
Correct answer: 17 m.
Explanation:
Version 001 – Test 6 Wave Phenomena – tubman – (IBII201516)
3
Explanation:
Given : λ = 68 m and
L = 16.5 km .
Let :
Note that we are neglecting any phase
changes which may occur upon reflection from
the mountain. If d is the distance to the
mountain, we have a path difference of δ = 2d.
For destructive interference :
1
δ = m+
λ
2
1
λ
2d = m +
2
y = 1.53871 m ,
L = 15 m , and
λu = 2.92 m .
y
d
L
The location of A is the central maximum
2dmin =
dmin =
=
=
λ
2
λ
4
68 m
4
17 m
ymax =
and the location of B is the first minimum
1
λu L m +
λu L
2
=
ymin =
d
2d
Serway CP 24 11
007 10.0 points
A riverside warehouse has two open doors
as shown. Its interior is lined with soundabsorbing material. A boat on the river
sounds its horn. To person A the sound is loud
and clear. To person B the sound is barely
audible. Persons A and B are 1.53871 m
apart and a distance 15 m from the doors.
The principal wavelength of the sound waves
is 2.92 m.
B
A
1.53871 m
15 m
Assuming person B is at the position of
the first minimum, determine the distance
between the doors, center to center.
Correct answer: 14.2327 m.
λu L m
=0
d
since m = 0. The distance between them is
λu L
2d
(2.92 m) (15 m)
λu L
=
d=
2y
2 (1.53871 m)
∆y =
= 14.2327 m .
The minimum path difference for destrucλ
tive interference must be .
2
p
λ
= L + y − L2 + y 2
2
= (15 m) + (1.53871 m)
q
− (15 m)2 + (1.53871 m)2
= (16.5387 m) − (1.53871 m)
λ
2.92 m
= , as expected .
= 1.46 m =
2
2
Serway CP 24 29
008 10.0 points
Helium-neon laser light of wavelength
Version 001 – Test 6 Wave Phenomena – tubman – (IBII201516)
y
825.7 nm is sent through a 0.286 mm wide
single slit.
What is the width of the central maximum
on a screen 0.94 m from the slit?
S1
S2
Explanation:
L
Given : λ = 825.7 nm = 8.257 × 10−7 m ,
L = 0.94 m , and
a = 0.286 mm = 0.000286 m .
What is the ratio
viewing
screen
θ
a
Correct answer: 5.42768 mm.
4
y2
?
y1
Correct answer: 1.25532.
Explanation:
The first minimum is at
y
θ
L
We use the fact that for small θ.
λ
y
, and sin θ =
L
a
For small θ, tan θ = sin θ, so
y
λ
= ,
L
a
and since the width x of the central maximum
is equal to 2y,
x = 2y
2λL
=
a
2 (8.257 × 10−7 m) (0.94 m) 1000 mm
=
·
0.000286 m
1m
= 5.42768 mm .
Pattern Distance Ratio
009 10.0 points
Consider the setup of a single slit experiment
shown in the figure.
The first minimum for 470 nm light is at y1 .
The first minimum for 590 nm light is at y2 .
λL
a
y1 =
λ1 L
a
For the first light,
For the second light,
y2 =
tan θ ≈ sin θ .
tan θ =
y=
λ2 L
a
Therefore
λ2
y2
=
y1
λ1
(590 nm)
=
(470 nm)
= 1.25532 .
First Order Dark Band 02
010 10.0 points
Monochromatic light with a wavelength of
355 nm passes through a single slit and falls
on a screen 83 cm away. If the distance of
the first-order dark band is 0.49 cm from the
center of the pattern, what is the width of the
slit?
Correct answer: 0.00601327 cm.
Explanation:
For single slit interference,
x
λ
=
L
w
Version 001 – Test 6 Wave Phenomena – tubman – (IBII201516)
5
so that
Lλ
= 0.00601327 cm
x
Dimensional analysis for w:
w=
cm · nm 1 m 102 cm
= cm
cm 109 nm 1 m
Taillights Blend Together
011 10.0 points
Suppose you are standing on a straight highway and watching a car moving away from
you at 26.3 m/s. The air is perfectly clear,
and after 2.24 min you see only one taillight.
If the diameter of your pupil is 5.34 mm
and the index of refraction of your eye is 1.53,
estimate the width of the car. (Taillights are
red, so use a wavelength of 650 nm.)
Correct answer: 0.34308 m.
Explanation:
The wavelength λi of the light inside your
eye will be
λ
λi =
n
6.5 × 10−7 m
=
1.53
= 4.24837 × 10−7 m .
The Rayleigh’s criterion in this case is
w
λi
= ,
d
L
where d is the pupil diameter, w is the distance between the light sources (i.e. the width
of the car), and L is the distance from the car
to you. The distance L is equal to the product
of the velocity and the elapsed time:
θmin = 1.22
Binary Star System
012 10.0 points
A binary star system in the constellation
Orion has an angular separation between the
two stars of 1.52 × 10−5 rad.
If the wavelength is 232 nm, what is the
smallest diameter a telescope can have and
just resolve the two stars?
Correct answer: 1.86211 cm.
Explanation:
Let : θ = 1.52 × 10−5 rad
λ = 232 nm .
λ
,
D
λ
D = 1.22
θ
θ = 1.22
102 cm
232 nm
·
1.52 × 10−5 rad 109 nm
= 1.22
= 1.86211 cm .
Holt SF 13A 01
013 (part 1 of 3) 10.0 points
An electric guitar’s amplifier is at a distance
of 4.8 m.
Find the intensity of its sound waves when
its power output is 0.33 W.
Correct answer: 0.00113978 W/m2 .
Explanation:
L = vt.
Therefore
1.22 λ v t
w=
nd
(1.22)(650 nm)(26.3 m/s)(2.24 min)
=
(1.53)(5.34 mm)
(1.22)(6.5 × 10−7 m)(26.3 m/s)(134.4 s)
=
(1.53)(0.00534 m)
= 0.34308 m .
and
Let :
I=
r = 4.8 m and
P = 0.33 W .
P
0.33 W
=
4 π r2
4 π (4.8 m)2
= 0.00113978 W/m2 .
Version 001 – Test 6 Wave Phenomena – tubman – (IBII201516)
014 (part 2 of 3) 10.0 points
Find the intensity of the sound waves when
the amplifier’s power output is 0.51 W.
Correct answer: 0.00176148 W/m2 .
Explanation:
Let :
I=
P = 0.51 W .
0.51 W
= 0.00176148 W/m2 .
4 π (4.8 m)2
015 (part 3 of 3) 10.0 points
Find the intensity of the sound waves when
the amplifier’s power output is 2.1 W.
Correct answer: 0.00725315 W/m2 .
7λ
L
2a
λ
L
6. y3 =
2a
2λ
7. y3 =
L
a
5λ
8. y3 =
L
a
9λ
9. y3 =
L
2a
λ
10. y3 = L
a
Explanation:
The third minimum occurs at β = 6 π,
which corresponds to a path difference between two end rays:
5. y3 =
Explanation:
b3 =
Let : P = 2.1 W .
2.1 W
I=
= 0.00176148 W/m2 .
4 π (4.8 m)2
×10
θ
S2
L
viewing
screen
a
S1
y3
Exam Single Aperture
016 10.0 points
Consider the setup of a single slit experiment.
Hint: Use a small angle approximation; e.g.,
sin θ = tan θ .
Determine the height y3 , where the third
minimum occurs.
3λ
L
2a
5λ
2. y3 =
L
2a
4λ
3. y3 =
L
a
3λ
L correct
4. y3 =
a
1. y3 =
6
β
k
6π
2π
λ
= 3λ
b3
θ=
a
y3
=
L
b3
y3 =
L
a
3λ
L.
=
a
=
Eye Resolution
017 10.0 points
Assume a lens can act like a one-dimensional
single slit, with the diameter of the lens equivalent to the slit width. The resolution of the
lens is then equivalent to the distance from
the middle of the central bright band to the
first-order dark band. Suppose the image
formed on the retina of the eye shows the effect of diffraction. The diameter of the iris
opening in bright light is 3.048 mm and the
distance from the iris to the retina is 2.42 cm.
Find the resolution of the eye for light of
wavelength 539.2 nm. Assume the index of
refraction of the interior of the eye is 1.33 .
Version 001 – Test 6 Wave Phenomena – tubman – (IBII201516)
Explanation:
The angle of resolution for the Paul Revere’s pupils is
Correct answer: 0.000321883 cm.
Explanation:
θmin = 1.22
Let : a = 3.048 mm = 0.003048 m ,
L = 2.42 cm ,
λ = 539.2 nm = 5.392 × 10−7 m ,
neye = 1.33 .
λ
d
= .
D
L
Therefore
and
The wavelength of the light in the eye is
λeye =
7
λ
5.392 × 10−7 m
=
neye
1.33
= 4.05414 × 10−7 m .
x
w
L
For single slit interference,
λeye
y
=
L
a
L λeye
y=
a
(2.42 cm) (4.05414 × 10−7 m)
=
0.003048 m
= 0.000321883 cm .
λL
D
(570 nm) (2.05 mi)
= 1.22
4.26 mm
(5.7 × 10−7 m) (3298.45 m)
= 1.22
0.00426 m
= 0.538437 m .
d = 1.22
An Optical Coating
019 (part 1 of 2) 10.0 points
A light ray is traveling in a medium with an
index of refraction n1 and it is reflected at the
boundary of a second medium with an index
of refraction n2 .
Considering the change of the relative
phases ∆φ due to their reflections, which of
the following conditions is correct?
1. If n1 > n2 , then ∆φ = 0 and
π
if n1 < n2 , then ∆φ = .
2
π
2. If n1 > n2 , then ∆φ = and
2
if n1 < n2 , then ∆φ = π.
π
and
2
π
if n1 < n2 , then ∆φ = .
2
3. If n1 > n2 , then ∆φ =
Paul Revere and Resolution
018 10.0 points
On the night of April 18, 1775, a signal was
to be sent from the Old North Church steeple
to Paul Revere, who was 2.05 mi away: “One
if by land, two if by sea.” Assume that Paul
Revere’s pupils had a diameter of 4.26 mm
at night, and that the lantern light had a
predominant wavelength of 570 nm.
At what minimum separation did the sexton have to set the lanterns so that Revere
could receive the correct message? One mile
is approximately equal to 1.609 km.
Correct answer: 0.538437 m.
4. If n1 > n2 , then ∆φ = 0 and
if n1 < n2 , then ∆φ = π. correct
5. If n1 > n2 , then ∆φ = 0 and
if n1 < n2 , then ∆φ = 0.
6. If n1 > n2 , then ∆φ = π and
if n1 < n2 , then ∆φ = π.
Explanation:
If a light ray is traveling in a medium with
an index of refraction n1 and it is reflected
at the boundary of a second medium with an
Version 001 – Test 6 Wave Phenomena – tubman – (IBII201516)
index of refraction n2 , the phase change is
∆φ = 0 when n1 > n2 and is π when n1 < n2 .
020 (part 2 of 2) 10.0 points
Consider the optical coating on a glass lens
where the index of refraction of the coating
is n, where n is greater than the index of
refraction of the air.
1 2
θ ≈ 0◦ θ
air
n
t
lens
Assume: The index of refraction of the lens
is greater than that of the coating.
To minimize the reflection of a ray with
a wavelength λ = 500 nm and incident angle
θ ≈ 0, what is the minimum nonzero thickness
t of the coating?
1. t =
2. t =
3. t =
4. t =
5. t =
6. t =
7. t =
8. t =
nλ
8
nλ
4
3λ
4n
λ
correct
4n
λ
8n
nλ
2
3nλ
4
λ
n
9. t = n λ
10. t =
λ
2n
Explanation:
Destructive interference occurs when the
difference between the phase angle of the incident ray reflected from the outer surface of the
8
coating (ray 1) and the phase angle of the ray
reflected from the inner surface of the coating
(ray 2) are at π, 3 π, 5 π etc. The phase differences are due to the path difference of the two
rays and the change of their relative phases
due to reflections. Putting them together, it
gives
π, 3 π, 5 π... = (2 t) kn + |π − π| = 2 t kn .
2π
With kn =
, it implies that the minimum
λn
λ
thickness is given by t =
.
4n
Coating on Glass
021 (part 1 of 2) 10.0 points
A material with an index of refraction of 1.23
is used to coat glass. The index of refraction
of glass is 1.5.
What is the minimum thickness of the coating that will minimize the reflection of light
with a wavelength of 6810 Å?
Correct answer: 0.138415 µm.
Explanation:
Phase Changes for Reflecting Waves.
Since the coating has a refraction that is less
than that for glass (but greater than that for
air), we know that the reflected light from
both the glass and the coating will undergo
a 180◦ phase shift. This means that the total trip inside the coating must be exactly
one half of the wavelength of light inside the
coating. (Hence the waves reflected from the
glass and from the coating will interfere destructively.) The wavelength of light inside
λ
where n is the index of rethe coating is
n
fraction of the coating. This implies that the
thickness of the coating must be one fourth
the wavelength of light inside this medium.
Hence (calling the thickness of the material t,
the wavelength of light in air λ, and the index
of refraction of the medium n)
6.81 × 10−7 m
λ
=
t=
4n
(4) (1.23)
t = 1.38415 × 10−7 m
= 0.138415 µm .
Version 001 – Test 6 Wave Phenomena – tubman – (IBII201516)
022 (part 2 of 2) 10.0 points
Now assume that the coating’s index of refraction is 1.67. Assume that the rest of the
system (from the previous question) remains
the same.
What is the minimum thickness of the coating needed to minimize the reflection of this
light now?
The spacing between adjacent slits is d =
1
, so
n
d sin θ = m λ
mλ
= arcsin(λ n)
θ = arcsin
d
= arcsin[(5.225 × 10−7 m)
×(3.67 × 105 lines/m)]
Correct answer: 0.203892 µm.
= 11.0554◦ .
Explanation:
This part is really solved in the same way
as Part 1 except now the index of refraction of the material is greater than that for
glass. Hence the light reflected from the material surface undergoes a 180◦ phase transition,
but the light reflected from the glass within
the material goes through a 0◦ phase transition. This means that the light must travel
a full wavelength within this material in order to interfere destructively. Using the same
notation as before, we then say:
t=
λ
6.81 × 10−7 m
=
2n
2 (1.67)
Holt SF 16B 02
024 (part 1 of 2) 10.0 points
A diffraction grating with 4545 lines/cm is
illuminated by direct sunlight. The first-order
solar spectrum is spread out on a white screen
hanging on a wall opposite the grating.
a) At what angle does the first-order maximum for blue light with a wavelength of 490
nm appear?
Correct answer: 12.868◦ .
Explanation:
Basic Concepts:
d(sin θ) = mλ, m = 0, ±1, ±2, · · ·
t = 2.03892 × 10−7 m
= 0.203892 µm .
d=
1
n
Given:
Angle of First Order Maximum
023 10.0 points
Monochromatic 522.5 nm light is incident on
a diffraction grating containing 3670 lines per
centimeter. Find the angle of the first-order
maximum.
Correct answer: 11.0554◦.
Explanation:
Let :
λ = 522.5 nm
= 5.225 × 10−7 m ,
n = 3670 lines/cm
= 3.67 × 105 lines/m ,
m = 1.
9
100 cm
1m
5
= 4.545 × 10 lines/m
1m
λblue = 490 nm · 9
10 nm
= 4.90 × 10−7 m
m=1
n = 4545 lines/cm ·
Solution:
sin θ =
=
and
mλ
d
mλ
1
n
= mλn
θ = sin−1 (m λ n)
Version 001 – Test 6 Wave Phenomena – tubman – (IBII201516)
θ = sin−1 (m λblue n)
h
= sin−1 (1) 4.9 × 10−7 m
i
· (4.545 × 105 lines/m)
= 12.868◦
025 (part 2 of 2) 10.0 points
b) At what angle does the first-order maximum for red light with a wavelength of 778
nm appear?
Correct answer: 20.7077◦.
Explanation:
Given:
1m
109 nm
= 7.78 × 10−7 m
λred = 778 nm ·
Solution:
θ = sin−1 (m λred n)
h
−1
(1) 7.78 × 10−7 m
= sin
i
· (4.545 × 105 lines/m)
= 20.7077◦
10