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Final Exam Review
Due
Dec 6 at 8am
Points
96.66
Questions
98
Available
Nov 23 at 12am - Dec 6 at 8am
Time Limit
None
Instructions
Attempt History
Attempt
Time
Score
LATEST
Attempt 1
4,984 minutes
54 out of 96.66
Submitted Dec 6 at 5:22am
Recap of Prelab Quiz Problems
1 / 1 pts
Question 1
What are photons?
Tiny quantized packets of light. Correct!
Correct!
neutrons name of a rock group protons electrons 1 / 1 pts
Question 2
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What happens in the photoelectric effect?
Light hits a metal and photons are emitted. Light hits a metal and protons are emitted. Light hits a metal and electrons are emitted. Correct!
Correct!
Electrons hit a metal and electrons are emitted. Electrons hit a metal and light is emitted. 1 / 1 pts
Question 3
READING 4
In the photometer, light strikes a metal and electrons are emitted. The
moving electrons constitute a current that flows through a resistor . The
digital multimeter (DMM) can be set to read the voltage drop across the
resistor . The DMM has a high input impedance (resistance) that is in
parallel with . Let's denote the input impedance . Note that .
will output a signal because of ambient electromagnetic waves, e.g., radio
waves in the air. These radio waves have an oscillating electric field that
pushes electrons in the cable connected to the DMM. These moving
electrons constitute a small current. Recall Ohm's law: .
For the DMM,
we can write So if is large, even a tiny current can produce a
finite voltage . This is why your DMM will read a voltage even if the input
Rp
Rp
Rp
RD
RD>>Rp
V=IR
V=IR
D
R
D
I
V
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cable is not connected to anything. The cable acts like an antenna.
What should the digital multimeter be set to read?
current voltage Correct!
Correct!
impedance resistance power 1 / 1 pts
Question 4
INTRODUCTION TO MEDICAL IMAGING:
To the human observer, the internal structures and functions of the human
body are generally invisible. However, by various technologies, images
can be created which allow a doctor to look into the body and diagnose
abnormal conditions and guide therapeutic procedures. The medical
image is a window to the body. No image window reveals everything.
Different medical imaging methods reveal different characteristics of the
human body. With each method, the range of image quality and structure
visibility can be considerable, depending on characteristics of the imaging
equipment, skill of the operator, and compromises associated with factors
such as patient radiation exposure and imaging time. [1]
READING 1:
Imaging methods create images that show the body from one of two
perspectives, through either projection (Lab Manual Fig. 1.2) or
tomographic (Lab Manual Fig. 1.3) imaging. There are advantages and
disadvantages to each.
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In projection imaging (radiography and fluoroscopy), images are formed
by projecting a beam through the patient's body and casting shadows onto
an appropriate receptor that converts the invisible beam image into a
visible light image. The primary advantage of this type of image is that a
large volume of the patient's body can be viewed with one image. A
disadvantage is that structures and objects are often superimposed so
that the image of a superfluous object interferes with the image of the
structure under study. In projection imaging, spatial distortion is generally
not a major problem in most clinical applications.
Tomographic imaging techniques such as conventional tomography,
computed tomography (CT), sonography, single photon emission
tomography (SPECT), positron emission tomography (PET), and
magnetic resonance imaging (MRI), produce images of selected planes or
slices of tissue of the patient's body. The general advantage of a
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tomographic image is the increased visibility of objects within the imaged
plane. One factor that contributes to this is the absence of overlying
objects. The major disadvantage is that only a small slice of a patient's
body can be visualized with one image. Therefore, most tomographic
procedures usually require many images to survey an entire organ system
or body cavity. [1]
[1] Perry Sprawls, PHYSICAL PRINCIPLES OF MEDICAL IMAGING,
(Rockville, Maryland: Aspen Publishers, Inc., 1987).
Which of the following DOES NOT describe projection imaging?
Produces images of selected planes or slices of tissue of the patient's body
Correct!
Correct!
A large volume of the patient's body can be viewed with one image Spatial distortion is generally not a major problem Images are formed by projecting a beam through the patient's body and
casting shadows onto an appropriate receptor
All of the above describe projection imaging. 1 / 1 pts
Question 5
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Which of the following DOES NOT describe tomographic imaging?
Produces images of selected planes or slices of tissue of the patient's body
Includes PET and MRI scans A large volume of the patient's body can be viewed with one image Correct!
Correct!
Increased visibility of objects within the imaged plane due to absence of
overlying objects
All of the above describes tomographic imaging 1 / 1 pts
Question 6
An ordinary dental X-ray uses which imaging technique(s)?
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Sonography Projection Correct!
Correct!
Tomography None of the above 1 / 1 pts
Question 7
READING 2
Depending on the application, a wide variety of particles and waves are
used to probe the body. The basic requirements for a suitable probe beam
are that it can penetrate the body (which excludes ordinary light), that is
interacts with the sample to produce useful information, and that harmful,
unwanted interactions with the body be minimal. One of the remarkable
results of modern physics is that all particles have wave properties and all
waves have particle properties. For example, "photons" and "light waves"
are both electromagnetic radiation. Conceptually, all imaging techniques
consist of some form of radiation interacting with matter. There are three
things that can happen when particles/waves strike a sample (Lab Manual
Fig. 1.4). Some of the particles (a fraction of the wave energy) are
reflected. If the wave is characterized by an initial intensity I
, the ratio of
the reflected intensity
I
to incident intensity is called the reflection
coefficient r. Some of the particles penetrate the sample. The ratio of
transmitted intensity
I
to initial intensity,
I / I
, is called the transmission
coefficient t. Finally, some energy can be absorbed into the sample.
Because energy is conserved, the sum of the reflected energy, the
transmitted energy, and the absorbed energy equals the energy in the
initial wave.
o
r
t
t
o
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In medical imaging, the reflections and absorption of various radiations by
structures in the body are exploited to infer the properties of internal body
structures. In the 3LC Lab, we study the interaction of some of these
radiations with matter.
Which of the following is NOT a requirement for a suitable probe beam?
It can penetrate the body Harmful, unwanted interactions with the body are minimal It interacts with the sample to produce useful information It interacts with a large number of cells Correct!
Correct!
All of the above are requirements for a suitable probe beam 1 / 1 pts
Question 8
Some wave energy can penetrate and pass through a sample. This is a
description of what type of interaction with matter?
Absorption Reflection Incidence Transmission Correct!
Correct!
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1 / 1 pts
Question 9
Which of the following is true for the reflection coefficient for a perfect
mirror?
I / I = 1
t o
The ratio of reflected intensity to initial intensity is zero I / I = 1
Correct!
Correct!
r
o
I / I = I / I r o
t
o
All of above 1 / 1 pts
Question 10
READING 3
The photometer is a device that converts light to voltage which is read out
by the digital multimeter (DMM). This is due to the photoelectric effect.
(Einstein won his Nobel Prize for his explanation of the photoelectric
effect.) In the photoelectric effect, light hits a piece of metal. Quantum
mechanics tells us that light can be thought of as tiny packets of energy
called photons. The photons are absorbed by electrons and increase the
energy of the electrons. Some of the electrons become so energetic that
they escape from the metal. In other words electrons are emitted from the
metal, producing a current that is amplified by the photometer. The current
goes through a resistor in the photometer. Ohm's law (V=IR) tells us that
current I going through a resistor R is associated with a voltage drop V.
The DMM measures the voltage drop across the resistor. Saturation
occurs when the metal emits the maximum flux of electrons that it can
produce. (Or it may be the maximum amplification of the photometer.)
What does a photometer do?
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It emits light. It measures the frequency of light. It converts light into electric current. Correct!
Correct!
It emits sound. 1 / 1 pts
Question 11
How does the flux of light from an isotropic source depend on the distance
r from the source?
r
1/2
r
1
r
-1
r
Correct!
Correct!
-2
r
2
None of the above
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1 / 1 pts
Question 12
Which statement is correct?
A desk lamp is a isotropic source. A laser is an isotropic source. A desk lamp is a collimated source. A laser is an isotropic source. A desk lamp is a collimated source. A laser is a collimated source. A desk lamp is an isotropic source. A laser is a collimated source. Correct!
Correct!
None of the above 1 / 1 pts
Question 13
INTRODUCTION: MEDICAL LASERS AND ACUITY
Three types of lasers used medically are: the argon laser, the carbon
dioxide CO2 laser and the yttrium-aluminum-garnet (YAG) laser. Each
laser, due to its physical characteristics has found its way into different
medical applications. The argon laser is a lower powered laser. It is
absorbed quickly by hemoglobin and is very effective in stopping bleeding
where many small blood vessels are clustered. It coagulates bleeding
ulcers, destroys vascular tumors of the bladder and cleans up lesions like
the portwine birthmark stain. The CO2 laser is high powered. When
teamed with an operating microscope, it is an excellent high-precision
surgical knife because it only cuts 0.1 mm for every pass made by the
surgeon. The CO2 laser is outstanding in neurosurgical work on
vulnerable areas such as the brain and spinal cord. While annihilating the
cell it touches, it leaves neighboring cells untouched - a plus when
vaporizing malignant brain tumors. It is also used to treat cervical cancer,
remove nodes from the larynx and open up blocked fallopian tubes. The
YAG laser can penetrate very deeply within tissue. Because the YAG
passes through clear liquid, it can be used in fluid-filled cavities such as
the eyeball. Since it can stop the bleeding in large vessels, it was first
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used on patients hemorrhaging from stomach ulcers. While it can vaporize
large tumors, it cannot be used for precision surgery. Recent reports have
this laser even destroying early bladder cancer [2].
READING 1
A cauterizing instrument such as a focused laser beam could have wide
application in general surgical procedures. The organ or tissue to be
excised can be cut free at the same time that the vascular input is sealed
so as to minimize subsequent bleeding. (Major organs have been
removed using laser surgery, and relatively few complications are reported
[3].) The laser light is absorbed in a small volume of tissue in which the
energy is quickly converted to heat. Direct contact of tissue by the laser is
not required, and the depth and confinement of the laser instrumentation
can be controlled. Because the laser beam is intense and has very low
divergence its energy can be focused onto a small spot. When the
temperature exceeds 100 degrees Celsius at the tissue, vaporization of
tissue water can occur. As the laser power and doses are increased,
greater amounts of steam are produced without much further increase in
temperature. If sufficient steam is generated rapidly within the tissue,
physical separation or cutting occurs. Tissues bordering the cut are
heated sufficiently by dissipated thermal energy to effect cautery [4]. Thus,
the effect is one of intense heating, with perhaps 1000 J being delivered to
the lesion [3].
[2] Paul Lavrakas, "Laser: The Healing Light of Medicine," CONSUMER'S
RESEARCH MAGAZINE, October 1985.
[3] Eugene Ackerman, BIOPHYSICAL SCIENCE, 2nd Ed. (Englewood
Cliffs, NJ: Prentice Hall Inc., 1979).
[4] Physics Survey Committee; National Research Council, PHYSICS
THROUGH THE 1990'S: SCIENTIFIC INTERFACES AND
TECHNOLOGICAL APPLICATIONS, (Washington D.C.: National
Academy Press, 1986).
To see some short QuickTime movie clips of laser surgery in action,
click here.
(http://www.lbah.com/laser.htm)
Why is damage to neighboring tissue minimized?
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Because the laser beam is intense and has very low divergence its energy
can be focused onto a small spot.
Correct!
Correct!
Because the laser is operated at an extremely low power level. Because direct contact of the tissue by the laser confines the damage to
those cells.
All of the above 1 / 1 pts
Question 14
READING 2
Pulsed ruby lasers have been used to repair retinal tears; the
photocoagulation process takes about 1.0 milliseconds. With argon lasers,
which are also used, photocoagulation takes about 100 milliseconds. After
the iris is dilated, a laser focal spot ~ 100 micrometers in diameter is
imaged on the loose tissue. The actual weld made by the laser occurs on
the pigmented epithelium (the layer furthest to the back of the retina)
because the first layers of the retina are transparent at optical
frequencies. The mechanism of photocoagulation is essentially thermal,
and a small scar is left at the laser focal spot. The green light of the argon
laser is preferred over the red light of pulsed ruby lasers because green
light is more heavily absorbed in the pigmented tissue and somewhat
sharper foci are possible, with 50 micrometers being the nominal limit.
Since the fovea of the retina is approximately 1000 micrometers across,
the decreased size of the scar left by the argon laser can be significant in
treatment of the patient's problem [3].
[3] Eugene Ackerman, BIOPHYSICAL SCIENCE, 2nd Ed. (Englewood
Cliffs, NJ: Prentice Hall Inc., 1979).
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Where does the actual 'weld' take place on the retina, based upon the
physiology of the eye?
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Pigmented Epithelium Correct!
Correct!
Sphincter or Constrictor Pupillae Stroma of Cornea Outer Epithelium None of the above 1 / 1 pts
Question 15
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READING 3
The light sensing retina consists of a collection of sensors called rods and
cones, which are packed closely together on the retinal surface. The
human eye contains about cones and rods. The rods and
cones are not distributed evenly over the retinal surface; the fovea is a
region of the retina that contains only cones, which are tightly packed.
These cones are ineffective at low levels of illumination, but are
responsible for all the visual information concerning color [5].
6 x 106
120 x 106
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Our ability to see is described in terms of how well we can distinguish two
closely spaced point sources, which is referred to as the acuity of the eye.
Maximum acuity is possible only for light that strikes the fovea, where the
density of receptors is exceptionally large. Highest acuity is therefore
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obtained when viewing objects on the visual axis within a very narrow
cone. If the image falls upon the retina outside the fovea, the acuity is
considerably decreased because of the lower density of receptors.
Secondly, acuity is influenced by the level of illumination; if light is intense,
the low-sensitivity cones are effective. If the level of light is low, the cones
fail, and since there are no rods in the fovea, the acuity in this region falls
[5].
Without this occurring, the two point sources would not be resolved - they
would appear as only one point source to the viewer.
[5] Jerry B. Marion, William F. Hornyak, GENERAL PHYSICS WITH
BIOSCIENCE ESSAYS, 2nd Ed. (New York: John Wiley & Sons, 1985).
What factors DO NOT affect visual acuity?
Region of retina image falls upon Density of receptors Level of illumination All of the factors above affect visual acuity Correct!
Correct!
1 / 1 pts
Question 16
To resolve two point sources, what distribution of cones must occur where
the image strikes the retina?
There must be a cone producing less than the full response between two
cones that produce the full response.
Correct!
Correct!
There must be an overexcited cone between two cones that produce the
full response.
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Two adjacent cones must both produce the full response. None of the above 0 / 1 pts
Question 17
Answer 1:
Unanswered
Unanswered
A common technique in analysis of scientific data is normalization. The
purpose of normalizing data is to eliminate irrelevant constants that can
obscure the salient features of the data. The goal of this experiment is to
test the hypothesis that the flux of light decreases as the square of the
distance from the source. In this case, the absolute value of the voltage
measured by the photometer is irrelevant; only the relative value conveys
useful information. Suppose that in Part 2.2.2 of the experiment, students
obtain a signal value of 185 mV at a distance of 4 cm and a value of 82
mV at a distance of 6.2 cm. Normalize the students' data to the value
obtained at 4 cm. (Divide the signal value by 185.) Then calculate the
theoretically expected (normalized) value at 6.2 cm.
___________________ Normalized experimental
value at 6.2 cm
____________________Theoretically expected
normalized value at 6.2 cm
(You left this blank)
ou Answered
ou Answered
0.443243243243243 orrect Answer
orrect Answer
0.4 orrect Answer
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0.44 orrect Answer
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0.443 orrect Answer
orrect Answer
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Answer 2:
0.4432 orrect Answer
orrect Answer
0.44324 orrect Answer
orrect Answer
(You left this blank)
ou Answered
ou Answered
0.416233090530697 orrect Answer
orrect Answer
0.4 orrect Answer
orrect Answer
0.42 orrect Answer
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0.416 orrect Answer
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0.4162 orrect Answer
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0.41623 orrect Answer
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0.416233 orrect Answer
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0 / 1 pts
Question 18
Unanswered
Unanswered
Suppose you want to take a chest X-ray with an X-ray source that has a
divergence of 5
. If the film is 1 meters from the (point) source, how big is
the spot size at the film in centimeters?
_______________________________cm
o
ou Answered
ou Answered
orrect Answers
orrect Answers
8.72664626044111 8.7 8.73 8.727 8.7266 8.72665
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0 / 1 pts
Question 19
Unanswered
Unanswered
A bothersome feature of many physical measurements is the presence of
a background signal (commonly called "noise"). In Part 2.2.4 of the
experiment, some light that reflects off the apparatus or from neighboring
stations strikes the photometer even when the direct beam is blocked. In
addition, due to electronic drifts, the photometer does not generally read
0.0 mV even in a dark room. It is necessary, therefore, to subtract off this
background level from the data to obtain a valid measurement. Suppose
the measured background level is 5.1 mV. A signal of 20.7 mV is
measured at a distance of 29 mm and 15.8 mV is measured at 32.5 mm.
Correct the data for background and normalize the data to the maximum
value. What is the normalized corrected value at 32.5 mm?
____________________
ou Answered
ou Answered
orrect Answers
orrect Answers
0.685897435897436 0.7 0.68 0.69 0.686 0.6859 1 / 1 pts
Question 20
In Part 3.2.2 of the experiment, the angles are defined differently than in
Lab Manual Fig. 3.1. In the experiment, the angles between the laser and
the optical bench and between the laser and the normal to the
mirror are measured (Lab Manual Fig. 3.7).
bench mirror
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What is the relationship between and the customary angles?
bench = r-i
bench = i-r
bench = r
bench = i+r
Correct!
Correct!
bench None of the above 1 / 1 pts
Question 21
What is the relationship between and the customary angles?
mirror = r-i
mirror
= r+i
mirror
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= i
mirror
= i
Correct!
Correct!
mirror
None of the above 1 / 1 pts
Question 22
INTRODUCTION: MEDICAL APPLICATIONS OF FIBER OPTICS
Fiberscopes are used for magnified, illuminated viewing of internal
structures when there is no direct access between the eyepiece and the
object. (The endoscope, another instrument which utilizes fiber optics, is
used when there is a direct line of access to the structure being viewed).
Fiberscopes can negotiate multiple curves with small radii [6]. The
angioscope, one type of fiberscope, is a slender medical instrument used
for direct intraoperative visualization of the arterial lumen (a lumen is a
cavity enclosed by cell walls). This instrument is introduced into the
arterial system by either a special plastic tube inserted directly into the
artery (catheterization) or directly through an incision into the artery
(arteriotomy). Direct visualization is accomplished by temporary
obstruction of the blood flow and liquid purging in order to avoid the
absorption of visual lumination energy by the blood [7].
There are two major components within the distal tip of the angioscope
(the end used for visualization of the arterial lumen): an objective lens
system focused onto a coherent image bundle, and optical incoherent
illumination fibers [7].
READING 1
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Because of the multiple reflections that occur within a single optical fiber,
an image cannot be passed through a single fiber. But an image can be
broken down into a series of fine dots of various shades of light and dark
and each portion is sent through a single fiber [8]. The basic requirement
to transfer an image is that the fibers must maintain identical relative
positions on the input and output faces of the bundle [9]. This is called a
coherent fiber bundle, and in this way, the variation of light intensity used
to create the image is maintained. Thus, it is necessary to use a coherent
bundle of fibers with aligned fiber ends in order to transmit the image
through the angioscope. Since light does not require such an organization
of fibers in order to illuminate the visualized area, incoherent bundles can
be used. Incoherent bundles do not maintain identical relative positions on
the input and output faces, and thus are not used for image transfer.
[6] D. A. Hill, FIBRE OPTICS, (London: Business Books Ltd., 1977).
[7] Frank D. D’Amelio, Steven T. DeLisi, Anthony Rega, ”Fiber Optic
Angioscopes,” PROCEEDINGS OF SPIE-NOVEL OPTICAL FIBER
TECHNIQUES FOR MEDICAL APPLICATIONS, Vol. 494, 21 August
1984.
[8] Leonard H. Greenburg, PHYSICS FOR BIOLOGY AND PRE-MED
STUDENTS, (Philiadelphia, W. B. Saunders Co., 1975).
[9] Jeff Hecht, UNDERSTANDING FIBER OPTICS, (Indianapolis: Howard
W. Sams Co., 1987).
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The History of Fiber Optics
(http://www.sff.net/people/Jeff.Hecht/history.html)
How Fiber Optics Works
(http://www.howstuffworks.com/fiber-
optic.htm)
Medical Applications of Fiber Optics
(http://www.us.schott.com/fiberoptics/english/products/healthcare/)
Which of the following describe the characteristics of a bundle of fiber
used for image viewing?
Coherent; no specific alignment of the fiber ends is necessary Incoherent; aligned fiber ends Incoherent; no specific alignment of the fiber ends is necessary Coherent; aligned fiber ends Correct!
Correct!
1 / 1 pts
Question 23
Which of the following describe the characteristics of a bundle of fiber
used for illumination?
Incoherent; aligned fiber ends Coherent; no specific alignment of the fiber ends is necessary Incoherent; no specific alignment of the fiber ends is necessary Correct!
Correct!
Coherent; aligned fiber ends 1 / 1 pts
Question 24
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READING 2
The length of the fiberscope is limited by the spectral capability (what
wavelengths can be conveyed or transmitted) and spectral capacity (how
much of those wavelengths are conveyed). Most optics can transmit
visible light and some infrared radiation, but will transmit ultraviolet rays
with only poor efficiency. The length of the fiber is limited mostly by
spectral capacity; as the length of the lightguide increases, the percentage
of input light transmitted decreases. This is due to absorption within the
fiber and coupling losses at the exit and entry points [6]. There is a limit to
the bending radius of the fiberscope; as the thickness increases, the
flexibility decreases because the minimum bending adius becomes larger.
However, the jacket that covers the fiber optics and protects it from
chemical attack prevents this limit from being exceeded. Magnification is
restricted to lower orders (x20) because the outline of the fibers can
become superimposed upon the image [6].
[6] D. A. Hill, FIBRE OPTICS, (London: Business Books Ltd., 1977).
The length of a fiberscope is mostly limited by what?
Spectral Capacity - how much of the transmitted wavelengths are
conveyed
Correct!
Correct!
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Jacket Thickness Bending Radius Spectral Capability - what wavelengths can be conveyed or transmitted 1 / 1 pts
Question 25
READING 3
Laser coronary angioplasty has received considerable interest recently as
a potential adjunct to coronary bypass surgery and transarterial treatment
by balloon angioplasty. Angioplasty is a technique that removes
interarterial plaque (buildup), thereby unclogging the artery for blood flow.
The mechanism by which the laser removes the arterial plaque is by
vaporization or thermal decomposition. Presently, this technique has been
used in unclogging a femoral artery; however, the eventual aim is to
perform laser angioplasty in the coronary artery; further development in
the ability to view, steer, and aim inside small arteries is required [10].
A key factor for the effectiveness of a laser in medical applications is the
laser delivery system. High transmission, light weight, small size, and
flexibility make fiber optics well suited to deliver the laser power. The best
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optical fibers with regard to transmission, flexibility, and nontoxicity are the
silica based fibers, used in conjunction with the Er:YAG laser.
[10] L. Esterowitz, C. A. Hoffman, D. C. Tran, ”Angioplasty with laser and
fiber optics at 2.94 micrometers,” PROCEEDINGS OF THE SPIE-
OPTICAL AND LASER TECHNOLOGY IN MEDICINE, Vol. 605, 23-24
January 1986.
What factors DO NOT make fiber optics well suited to deliver the laser
power?
Stiffness Correct!
Correct!
Light Weight Small Size High Transmission All of the above factors make fiber optics well suited to deliver laser power 1 / 1 pts
Question 26
READING 4
Another medical application of fiber optics is physical and chemical
measurements. A typical miniature fiber optic pH sensor has been applied
in opthamology. The operation of this sensor is based upon a color
change of a pH-sensitive agent, which is contained within a small
chamber with ion permeable walls. The subminiature chamber is attached
to a standard optical fiber 150 micrometers in diameter. The
measurements are performed by comparing colored light optical pulses to
white light optical pulses. This miniature, elastic medical pH sensor
measures pH changes in real time and in vivo. This technology has been
further applied to perform calcium and other ion concentration
measurements in inaccessible places [11].
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Oxygen saturation can also be determined utilizing optical fiber sensors.
Oxygen saturation refers to the amount of oxygen carried by the
hemoglobin in red blood cells relative to the maximum carrying capacity
(typical arterial blood is 95venous blood may be 75in the cardiovascular or
cardiopulmonary system. Optical measurement of oxygen saturation
depends on the difference in the absorption spectra of oxyhemoglobin and
hemoglobin. The absorption coefficients of hemoglobin and
oxyhemoglobin (absorption coefficient is a characteristic of the absorbant;
the larger the coefficient, the better the absorption capacity of the
medium) are sufficiently low at wavelengths above 620 nm that sufficient
light transmission through whole blood can occur over distances that are
compatible with the geometric requirements of practical fiber-optic
catheters. At 650 nm, the signal is measured where there is a useful
difference in absorbance between oxyhemoglobin and hemoglobin [12].
Optical sensors fall into two categories: those that sense changes in the
specific optical properties of the material of interest (direct sensors) and
those that use a reagent that changes its optical properties in response to
changes in the material of interest (indirect sensors) [13]. The pH sensor
is an indirect sensor because it utilizes a pH sensitive agent, which
changes its optical properties in response to the pH. On the other hand,
the oxygen saturation sensor is direct; it measures the difference in optical
properties between the two materials of interest. Other fiber optic sensors
include those which detect changes in flow, pressure, temperature, and
other chemicals within the body.
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[11] Ryszard S. Romaniuk, ”Directions of applications of optical fiber
technology in Medicine and Health protection systems,” PROCEEDINGS
OF THE SPIE-OPTICAL FIBERS IN MEDICINE II, Vol. 713, 17-19
September 1986.
[12] J. I. Peterson, G. G. Vurek, ”Fiber Optic Sensors for Biomedical
Applications,” SCIENCE, Vol. 224, 13 April 1984.
[13] Gerald G. Vurek, ”In Vivo Optical Chemical Sensors,”
PROCEEDINGS OF THE SPIE-NOVEL OPTICAL FIBER TECHNIQUES
FOR MEDICAL APPLICATIONS, Vol. 494, 21 August 1984.
Which type of sensor is the pH sensor? Why?
Direct; It senses changes in the specific optical properties of the material of
interest
Indirect; It senses changes the specific optical properties of the material of
interest
Direct; It uses a reagent that changes its optical properties in response to
changes in the material of interest
Indirect; It uses a reagent that changes its optical properties in response to
changes in the material of interest
Correct!
Correct!
0 / 1 pts
Question 27
Unanswered
Unanswered
The index of refraction of the core of a typical fiber optic is n
= 1.46; the
cladding has n
= 1.4. Calculate the critical angles for the total internal
reflection i
and .
core clad crit crit
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Answer 1:
Answer 2:
i
= _________________degrees,
= ___________________ degrees
crit crit (You left this blank)
ou Answered
ou Answered
73.5170298580243 orrect Answer
orrect Answer
73.5 orrect Answer
orrect Answer
73.52 orrect Answer
orrect Answer
73.517 orrect Answer
orrect Answer
73.5170 orrect Answer
orrect Answer
73.51703 orrect Answer
orrect Answer
(You left this blank)
ou Answered
ou Answered
24.4718615917547 orrect Answer
orrect Answer
24.5 orrect Answer
orrect Answer
24.47 orrect Answer
orrect Answer
24.472 orrect Answer
orrect Answer
24.4719 orrect Answer
orrect Answer
24.47186 orrect Answer
orrect Answer
0 / 1 pts
Question 28
Unanswered
Unanswered
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The speed of light in a transparent medium is 1.3 x 10
m/s. What is the
index of refraction of this medium?
8 ou Answered
ou Answered
orrect Answers
orrect Answers
2.30769230769231 2.3 2.31 2.308 2.3077 2.30769 0 / 1 pts
Question 29
Unanswered
Unanswered
In Part 3.2.2 of the experiment, a pair of students measure = 18
degrees. To find , they use the theoretical relationship = 1/2 . Then they measure = 47 degrees. What is the discrepancy
between theory and experiment for ?
mirror
bench
mirror bench
bench
bench
ou Answered
ou Answered
orrect Answers
orrect Answers
30.5555555555556 31 30.5 30.6 30.56 30.556 30.5556
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1 / 1 pts
Question 30
Which of the following statements is correct?
A convex lens has both sides rounded outward causing the light rays to
converge towards one another. A concave lens has both sides rounded
inward, causing the light rays to diverge away from one another.
Correct!
Correct!
A convex lens has both sides rounded outward causing the light rays to
converge towards one another. A concave lens has both sides rounded
inward, causing the light rays to go parallel to one another.
A convex lens has both sides rounded outward causing the light rays to
diverge away from one another. A concave lens has both sides rounded
inward, causing the light rays to converge towards one another.
A convex lens has both sides rounded inward causing the light rays to
converge towards one another. A concave lens has both sides rounded
outward, causing the light rays to diverge away from one another.
None of the above 1 / 1 pts
Question 31
THE EYE
READING 1
Consider the indices of refraction of the various materials of the eye.
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The eye accomplishes focusing by bending light which is also known as
refraction. Refraction of light is greatest at the interface of materials with
widely differing indices of refraction. At the aqueous humor-lens or lens-
vitreous humor interface, the indices of refraction differ by 0.07. However
at the air-cornea interface, the indices differ by 0.38. Therefore, the lens is
only of secondary importance for image formation for animals that live in
air [14].
Under water, the human eye cannot focus sharply because the air-cornea
interface is replaced by a water-cornea interface. For creatures that live
underwater, the eye cannot rely on the corneal surface for focusing.
Without the cornea, all of the focusing must be done by the lens, which
has to be of much higher power than the human eye. There are several
modifications that take place in the fish eye. First, since there is no
focusing done by the cornea, its shape doesn’t matter; the eye can be
bulbous or flat. The lens takes on a different shape - in humans, it is an
ellipse (seen from the side), but in the fish it is spherical. Furthermore, the
index of refraction for the fish lens is very high, typically 1.65 [8].
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[8] Leonard H. Greenburg, PHYSICS FOR BIOLOGY AND PRE-MED
STUDENTS, (Philadelphia: W. B. Saunders Co., 1975).
[14] F. R. Hallett, P. A. Speight, R. H. Stinson, INTRODUCTORY
BIOPHYSICS (New York: John Wiley Sons, 1977).
Find your blind spot! Are you color blind? This site has some fun
eye tests.
(http://www.csun.edu/~vceed002/biology/senses/vision_labs/vision.htm)
Where does the majority of the focusing in the human eye occur?
Cornea - aqueous humor interface Lens - vitreous humor interface Aqueous humor - lens interface Air - cornea interface Correct!
Correct!
1 / 1 pts
Question 32
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Where does the focusing in the eye of a fish occur?
Cornea - aqueous humor interface Water - cornea interface the lens Correct!
Correct!
1 / 1 pts
Question 33
READING 3
One type of aberration that occurs in the eye is SPHERICAL
ABERRATION (Lab Manual Figure 4.6). This is found on all lenses bound
by spherical surfaces (the cornea). The marginal portions of the lens bring
rays to a shorter focus than the central region. The image of a point is
therefore not a point, but a small, blurred circle. The cornea is flatter at its
margin than at its center, and the lens is denser in the center and hence
refracts light more strongly at its core than at its outer layers [15].
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For an image to come into focus, the light received by the eye must be
bent so that the rays converge at the fovea. The closer the object being
viewed, the more the light must be bent if the object is to be seen clearly.
The cornea, vitreous humor, and aqueous humor have a fixed index of
refraction, but the lens has the ability to adapt. The lens can sharpen the
curvature of its front and back surfaces, increasing its focusing power.
When the eye is focused upon infinity (for human beings, this begins
about 20 feet away), ciliary muscles that encircle the equator of the lens
relax, and thus expand. This causes the zonules (which connect the ciliary
muscles to the lens) to be pulled taut, flattening out the front and back of
the lens and increasing the diameter of the equator. In this state, the
ability of the lens to bend light is at a minimum; at this point the combined
refractivities of the components of the eye allow it to focus on a distant
object. When the eye focuses on a closer object, ciliary muscles contract,
relieving stress on the zoncules and the lens undergoes an elastic
recovery. The lens becomes thicker, front to back, its surfaces become
more sharply curved and the diameter of the equator shrinks. This gives
the lens the added refractive power needed for focusing upon closer
objects [16].
[15]George Duncan, PHYSICS FOR BIOLOGISTS (New York: John Wiley
Sons, 1975).
[16] Jane F. Koretz, George H. Handleman, ”How the Human Eye
Focuses,” SCIENTIFIC AMERICA, July 1988.
What types of tests do eye doctors perform?
(http://www.allaboutvision.com/eye-exam/expect.htm)
How has the eye accomplished correction for spherical aberrations? Fill in
the blanks. The____________
is flatter at its___________________ than at its_____________ ,and
the_________________ is denser in the center and hence refracts more
strongly at its core than at its outer layers.
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lens; center; margin; cornea cornea; margin; center; lens Correct!
Correct!
lens; margin; center; cornea cornea; center; margin; lens 0 / 1 pts
Question 34
How is the lens able to change its refractive power?
The lens can INCREASE its index of refraction, thus increasing its focusing
power.
orrect Answer
orrect Answer
The lens can sharpen the curvature of its front and back surfaces,
increasing its focusing power.
ou Answered
ou Answered
The lens can DECREASE its index of refraction, thus increasing its
focusing power.
None of the above 0 / 1 pts
Question 35
Unanswered
Unanswered
Assume you have a convex lens with f = 10 cm. If the object is placed 18
cm from the lens, how far is the image from the lens? What is the lateral
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Answer 1:
Answer 2:
magnification?
cm, times
Book Companion Site: Glossary of Terms for Mirrors and Lenses
(http://www.brookscole.com/cgi-brookscole/course_products_bc.pl?
fid=M20b&product_isbn_issn=0030351146&discipline_number=13)
Note: Select the chapter on Mirrors and Lenses and then select the
glossary of terms.
(You left this blank)
ou Answered
ou Answered
22.5 orrect Answer
orrect Answer
(You left this blank)
ou Answered
ou Answered
-1.25 orrect Answer
orrect Answer
-1.3 orrect Answer
orrect Answer
1 / 1 pts
Question 36
Referring to last problem, is the image upright or Inverted?
Upright Inverted Correct!
Correct!
Impossible to determine
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0 / 1 pts
Question 37
Referring to the last two problems, is the image real or imaginary?
Real orrect Answer
orrect Answer
Imaginary ou Answered
ou Answered
Impossible to determine 0 / 1 pts
Question 38
Answer 1:
Unanswered
Unanswered
In Part 4.2.2, you will determine the focal length of a convex lens by
focusing on an object across the room. If the object is 10 m away and the
image is 9.8 cm, what is the focal length? (Hint: use Lab Manual Equation
4.2: 1 / o+ 1 / i = 1 / f, convert m into cm) Suppose one estimated the
focal length by assuming f = i. What is the discrepancy between this
approximate value and the true value? (Hint: When the difference
between 2 numbers is much smaller than the original numbers, round-off
error becomes important. So you may need to keep more digits than usual
in calculating the discrepancy.)
_____________________________ cm (keep 1 digit
after decimal point), _____________________ %
(round answer to integer number)
Book Companion Site: Glossary of Terms for Mirrors and Lenses
Note: Select the chapter on Mirrors and Lenses and then select the
glossary of terms.
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Answer 2:
(You left this blank)
ou Answered
ou Answered
9.70489205783323 orrect Answer
orrect Answer
9.7 orrect Answer
orrect Answer
9.70 orrect Answer
orrect Answer
9.705 orrect Answer
orrect Answer
9.7049 orrect Answer
orrect Answer
9.70489 orrect Answer
orrect Answer
(You left this blank)
ou Answered
ou Answered
0.980000000000008 orrect Answer
orrect Answer
1 orrect Answer
orrect Answer
0.98 orrect Answer
orrect Answer
0.980 orrect Answer
orrect Answer
0.9800 orrect Answer
orrect Answer
0.98000 orrect Answer
orrect Answer
0 / 1 pts
Question 39
Unanswered
Unanswered
In Part 4.2.5 of the experiment, the expected magnification of the
microscope is given by Lab Manual Equation 4.3: m = i
L / O
f
. Refer
also to Fig. 4.4 for a definition of the components and distances used in
Eq. 4.3. Suppose you obtain the following data. The distance between the
object and the objective lens is 15 cm. The distance between the objective
lens and the real, inverted image is 38 cm. The focal length of the
eyepiece is 10 cm. When viewing the ruled screen (as described in Part
4.2.5), you observe 2 magnified, millimeter divisions filling the 78 mm
1
1 2
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width of the screen. What eye-to-object distance is consistent with this
data? Round up answer to integer.
____________cm
ou Answered
ou Answered
orrect Answers
orrect Answers
153.947368421053 154 153.9 153.95 153.947 153.9474 153 1 / 1 pts
Question 40
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Blue light is bent more by a prism than orange light. Does blue light or
orange light have a larger index of refraction in glass?
Demonstration of relationship between wavelength and index of
refraction.
(http://www.rpi.edu/dept/phys/Dept2/APPhys1/optics/optics/node7.html)
Note: you need to click on the picture of the two circles labeled 1 and 2 to
see the video.
Additional Information on the Index of Refraction
(http://scienceworld.wolfram.com/physics/IndexofRefraction.html)
Orange Blue Correct!
Correct!
1 / 1 pts
Question 41
READING 2
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An x-ray beam is made up of photons with different energies. As the beam
passes through many materials, photons of certain energies penetrate
better than others. This selective attenuation of photons, according to their
energy, is referred to as FILTRA TION. Low energy photons are filtered
out by diagnostic x-ray equipment, usually by placing several millimeters
of aluminum in front of the beam before it enters the patient. These pho
tons are filtered out because they do not contribute to the image formation
(they are quickly attenuated by tissue and therefore do not penetrate the
body well to form the image); they only contribute to the patient’s
exposure to the harmful radiation [1].
[1] Perry Sprawls, PHYSICAL PRINCIPLES OF MEDICAL IMAGING,
(Rockville, Maryland: Aspen Publishers, Inc., 1987).
What photon energies are selectively filtered (higher or lower energies) in
medical imaging techniques?
Lower Correct!
Correct!
Higher 1 / 1 pts
Question 42
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Why is it necessary to collimate the light source before using the prism to
disperse the light?
So that the light rays entering the prism are focused to a small spot So that the light rays entering the prism are filtered for a specific
wavelength
So that the light rays entering the prism are parallel Correct!
Correct!
All of the above 1 / 1 pts
Question 43
Filter manufacturers sell ’interference filters’ that only transmit light within
a narrow range of wavelengths. For example, an interference filter exists
for use with a HeNe laser that strongly attenuates light at wavelengths
other than the red laser wavelength (
= 632nm). Suppose that in Part
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5.2.4 of the experiment you held one of these filters between the prism
and the telescope. What would you see?
Additional information on Interference Filters
(http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/intfilt.html)
An blue line A spectrum of colors A red line Correct!
Correct!
A yellow line None of the above 1 / 1 pts
Question 44
Why does a gas discharge tube (e.g., a neon light) have a certain color?
How do neon lights work?
(http://www.sciam.com/askexpert_question.cfm?articleID=00045ACB-372B-
1C71-84A9809EC588EF21)
The tungsten filament in the tube glows.
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The gas radiates a continuous spectrum of colors. The gas radiates only a few discrete wavelengths. Correct!
Correct!
The discharge tube is surrounded by a filter. None of the above 1 / 1 pts
Question 45
CHROMATIC ABERRATION, X-RAY FILTRATION, CT, AND MRI
READING 1
All lenses made of a single material refract rays of shorter wavelength
more strongly than those of longer wavelength, and so bring blue light to a
shorter focus than red, a phenomenon called CHROMATIC ABERATION
(Lab Manual Figure 5.4). The result is that the image of a point of white
light is not a white point but a blue circle fringed with color. The error is
actually moderate between the red end of the spectrum and the blue-
green, but it increases rapidly at shorter wavelengths - the blue-violet and
ultraviolet. The first device which helps the eye compensate for this type
of aberration is the lens, which acts as a color filter. The lens passes the
visible spectrum, but cuts off sharply at the violet end, where chromatic
aberration is worst. Secondly, there is a difference in spectral sensitivity
between the two types of photoreceptors. Rods are maximally sensitive in
the blue-green, whereas cones have their maximum in the green. Thus,
as one moves from dim light to bright light, the pattern of maximum
sensitivity of the eyes moves away from the end of the spectrum where
aberration is at its worst. The third device is found in the fovea. A region
around it, called the macula lutea, is colored with a yellow pigment,
xanthophyll. The pigment absorbs maximally in the violet and blue regions
of the spectrum, just where absorption by the lens falls to low values, and
hence remove further parts of the spectrum where chromatic aberration is
worst [15].
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[15]George Duncan, PHYSICS FOR BIOLOGISTS (New York: John Wiley
Sons, 1975).
Another explanation of Chromatic Aberration
(http://www.yorku.ca/eye/chroaber.htm)
Correcting for Chromatic Aberrations
(http://hyperphysics.phy-
astr.gsu.edu/hbase/geoopt/aber2.html)
Which of the following does NOT help the eye compensate for chromatic
aberrations?
The lens acts as a color filter. The fovea reflects radiation in the yellow region of the spectrum. Correct!
Correct!
The macula lutea absorbs maximally in the violet and blue regions. There is a difference in spectral sensitivity between the two types of
photoreceptors.
All of the above help the eye compensate for chromatic aberrations. 1 / 1 pts
Question 46
Why are lower energy photons filtered out in medical x-ray imaging?
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They only contribute to the patient exposure to the harmful radiation. They do not contribute to the image formation. They are quickly attenuated by tissue and therefore do not penetrate the
body well.
All of the above are reasons why these photons are filtered out. Correct!
Correct!
1 / 1 pts
Question 47
READING 3
CT (computed tomography), like conventional projection x-ray imaging,
uses x-rays. However, CT produces a clearer picture of soft tissue. The
CT generates its image from multiple views. A thin, fan shaped x-ray beam
is projected through the edges of the body slice being imaged. The
detectors do not see a complete image of the body slice, only a profile
from one direction. The profile data are measurements of the x-ray
penetration along each ray. In order to produce enough information to
create a full image, the x-ray beam is rotated, or scanned, around the
body section to produce views from many angles. Then, tissue density
values for each view are fed into a computer, where they are organized
into a two-dimensional matrix. The matrix is displayed on the system
monitor, and appears as a gray-shaded image. The image represents a
cross-sectional slice of human anatomy, typically a thickness of 1 cm [17].
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[17] Douglas E. Bahniuk, ”Engineering Paces Medical Advances,”
MACHINE DESIGN, 12 November 1987.
How can the CT image soft tissues with x-rays?
It measures the x-ray penetration along each ray. Tissue densities are fed into a computer and viewed as a gray-shaded
image.
The beam is rotated around the body section to produce views from many
angles.
All of the above contribute to created the CT image of soft tissues with x-
rays.
Correct!
Correct!
1 / 1 pts
Question 48
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READING 4
MRI (Lab Manual Figure 5.5) stands for Magnetic Resonance Imaging, a
medical imaging technique that gives very clear pictures. MRI exploits the
fact that certain materials, when placed in a magnetic field, absorb and
reradiate electromagnetic radiation that has a specific frequency. Like the
radiation from a gas lamp, it is an example of line radiation. For MRI, the
radiation is associated with the nucleus rather than the atom; this is why
its original name is Nuclear Magnetic Resonance (NMR). The radiation is
typically in the form of radio signals. Since the frequency of the emission
is proportional to the magnetic field, spatial information is obtained by
placing the subject in a spatially varying magnetic field; the frequency of
the radio emission then becomes proportional to position in the subject.
The intensity of the emission is proportional to the density of resonant
nuclei. The magnetic resonance image is actually an array of pixels (bins
or picture elements) showing the intensity of the radio signals originating
from each tissue volume element within the body section [18, 19]. Paul
Lauterbur and Peter Mansfield won the 2003 Nobel Prize in Physiology
and Medicine for their development of MRI.
For more information on the Nobel Prizes
[18] Craig M. Coulam, Jon J. Erickson, F. David Rollo, THE PHYSICAL
BASIS OF MEDICAL IMAGING, (New York: Appleton-Century- Crafts,
1981).
[19] J. Valk, C. MacLean, P. R. Algra, BASIC PRINCIPLES OF NUCLEAR
MAGNETIC RESONANCE IMAGING, (New York: Elsevier, 1985), Ch. 1-2.
What is the similarity between MRI and radiation from a gas lamp?
The radiation from both is created by a static magnetic field.
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They are both examples of line radiation. Correct!
Correct!
They are both examples of continuous spectrum radiation. The radiation from both is found in the visible portion of the
electromagnetic spectrum.
0 / 1 pts
Question 49
Unanswered
Unanswered
In Part 5.2.3 of the experiment, you will measure the index of refraction of
yellow light using Lab Manual Equation 5.2. Suppose the minimum angle
of deviation is 29 degrees. What is the index of refraction?
ou Answered
ou Answered
orrect Answers
orrect Answers
1.4018185286593 1.4 1.40 1.402 1.4018 1.40182 1 / 1 pts
Question 50
READING
When light passes through a slit and hits a screen, a pattern of bright and
dark spots appear on the screen if the wavelength of the light is
comparable to the size of the slit. To understand this pattern, we can use
Huygens’ principle and imagine the wavefront that hits the slit as
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consisting of a series of point sources of light. Each point source is like a
tiny light bulb radiating light in isotropically every direction. Suppose the
slit has a right edge and a left edge. We can think of the piece of the
wavefront that hits the right edge as a point source radiating light that hits
the screen. Similarly the left edge also can be associated with a point
source. The light from each point source is a wave that has maxima
(crests) and minima (troughs). When the waves from the two edges of the
slit hit a point on the screen, they may interfere either constructively or
destructively. If they interfere constructively, then the maxima arrive at the
same time and the waves are in phase. So a bright spot results. If the two
waves interfere destructively, then the maximum from one wave arrives at
the same time as the minimum from the other wave, and they cancel each
other out. In this case the waves are out of phase and a dark spot results.
When light passes through a slit whose size is comparable to the
wavelength of the light, the screen shows which of the following
bright spots due to constructive interference of light waves, and dark spots
due to destructive interference of light waves.
Correct!
Correct!
bright and dark spots due to destructive interference of light waves. bright spots due to destructive interference of light waves, and dark spots
due to constructive interference of light waves.
nothing. bright and dark spots due to constructive interference of light waves. None of the above. 1 / 1 pts
Question 51
What is a diffraction grating?
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bright and dark spots that appear on the screen. a large single slit. a grid of perpendicular lines. a series of bright and dark rings. a large number of equally spaced parallel adjacent slits. Correct!
Correct!
None of the above. 1 / 1 pts
Question 52
READING 2
The resolving power of the eye depends on the distribution of light
sensing elements on the retina as well as diffraction effects. The average
diameter of a foveal cone is about 2.0 micrometers. Assuming cone-cone
contact, let’s determine the angular separation of the cones from the lens
(given that distance between the cones and the lens is 23 mm), and thus
the limit of resolution of the eye due to cone density (at its maximum).
Clearly, if two point sources are to be resolved, the images cannot evoke
equal responses from adjacent cones. Instead, there must be an
unexcited cone between the two cones that produce the full response.
Therefore, the separation of two excited cones with one unexcited cone
between them is 4 micrometers. The angular separation of these cones,
measured from the eye lens is : = 4.0x10
/ 23x10
= 1.8x10
rad.
Since the angle of resolution for diffraction-limited resolution governed by
the diameter of the pupil is a larger angle, the diameter of the pupil under
optimal condition limits eye resolution. However, the two values still
coincide very closely [5].
[5] Jerry B. Marion,William F. Hornyak, GENERAL PHYSICS WITH
BIOSCIENCE ESSAYS, 2nd Ed. (New York: John Wiley Sons, 1985).
What limits the resolution of the eye under optimal conditions?
-6 -3 -4
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Diameter of the pupil Correct!
Correct!
Separation between adjacent cones Cornea None of the above 0 / 1 pts
Question 53
Unanswered
Unanswered
In Part 6.2.2, you will determine the wavelength of the laser by shining the
laser beam on a ”diffraction grating”, a set of regularly spaced lines.
Suppose the pattern is displayed on a screen a distance L from the
grating and the spots are separated by s. If the screen is 9 m away, the
spots are 3.5 cm apart, and the lines of the grating are separated by 0.3
mm, what is ?
___________ nm
Information on diffraction gratings
(http://hyperphysics.phy-
astr.gsu.edu/hbase/phyopt/grating.html)
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ou Answered
ou Answered
orrect Answers
orrect Answers
1166.66666666667 1200 1170 1166 1167 1166.7 1166.67 1166.667 0 / 1 pts
Question 54
Unanswered
Unanswered
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Suppose you have measured the diffraction pattern of a grating with d =
0.05 mm and have found that the spots were separated by s = 1.1 cm.
Now you want to determine d for an unknown grating. With the unknown
grating, the spots are separated by s = 4.3 cm. What is d?
_________ mm
Check out this website!
(http://www.physics.uq.edu.au/people/mcintyre/applets/grating/grating.html)
You can simulate any diffraction grating with any wavelength and see the
results.
More information on diffraction gratings
(http://hyperphysics.phy-
astr.gsu.edu/hbase/phyopt/grating.html)
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ou Answered
ou Answered
orrect Answers
orrect Answers
0.0127906976744186 0.01 0.013 0.0128 0.01279 0 / 1 pts
Question 55
Unanswered
Unanswered
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A beam of microwaves with = 0.9 mm is incident upon a 13 cm slit. At a
distance of 2 m from the slit, what is the approximate width of the slit’s
image?
_____________ cm
ou Answered
ou Answered
orrect Answers
orrect Answers
13
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0 / 1 pts
Question 56
Unanswered
Unanswered
Babinet’s principle of complementary screens. A surprising result of wave
theory is that the diffraction pattern of a dark screen with a hole cut in it (of
arbitrary shape) is identical to the pattern of an object with the same
shape as the hole. (Lab Manual Fig. 6.3).
An application of this principle is that a line mounted on transparent slide
casts the same diffraction pattern as a dark film with a slot of equal size
cut in it. In Part 6.2.5 of the experiment, you will exploit this principle to
measure the width of a hair. If the distance between the first spot and the
central minimum is s = 2.4 cm, L = 9 m, and = 4 x 10
m, what is the
width of the hair?
_________ mm
-7 ou Answered
ou Answered
orrect Answers
orrect Answers
0.15 0 / 1 pts
Question 57
Unanswered
Unanswered
DIFFRACTION IN EYES, MICROSCOPES, AND ULTRASOUND
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READING 1
Two point sources can be resolved by an optical system if the
corresponding diffraction patterns are sufficiently small or sufficiently
separated. By definition, the ”minimum resolvable separation” is when the
maximum of the diffraction pattern of one source falls on the first minimum
of the diffraction pattern of the other. For a circular aperture, this distance
in angular measure (radians) is given by:
= 1. 22
/ d
where is the wavelength of light visualized, d is the diameter of the
aperture, and is the angular separation. The wavelength of light in a
material is smaller than the wavelength in vacuum by the ratio =
/ n, where n is the index of refraction. So the formula can be
generalized to:
= 1. 22
/ ( n * d)
This formula determines the diffraction-limited resolving power of an
aperture with diameter d. This angular separation can be converted to an
approximate linear separation, D, at the retina using:
= D / 0.025
where 0.025 is the corneal-retinal distance in meters. (Note: The lab
manual has a typo and should have D in this last equation. D is not the
same as d.)
Limitations of the human eye
(http://csep10.phys.utk.edu/astr162/lect/light/limitations.html)
What is the linear separation D at the retina? The minimum diameter of
the pupil is about 2 mm, the eye is most sensitive to wavelengths of about
500 nm (air), the index of refraction of the aqueous humor is n = 1.33, and
the distance between the cornea and retina is about 2.5 cm.
____________ micrometers
(This is a very small distance - approximately 1/100 the size of a period.)
m
m m
o
m
o
o ou Answered
ou Answered
orrect Answers
orrect Answers
5.733
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5.7 5.73 0 / 1 pts
Question 58
Unanswered
Unanswered
READING 3
Ultrasonic waves, like all other waves, exhibit diffraction. Recall that the
minimum half angle of a beam is given by:
= 1.22
/d
The wavelengths used in ultrasound imaging techniques are generally in
the range from 0.3 mm to 0.75 mm [longer waves (lower frequencies)
penetrate more deeply but provide less resolution of detail than the short
waves]. In order for an ultrasonic beam to spread by less than a
centimeter after it has traveled across a human torso (20 cm), the source
of the waves must be about 10 or 20 times larger than the wavelength
[20]. The source of the waves is called a transducer.
m
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[20] Harold J. Metcalf, TOPICS IN CLASSICAL BIOPHYSICS,
(Englewood Cliffs, NJ: Prentice Hall, Inc., 1980).
Obstetric Ultrasounds
(http://www.ob-ultrasound.net/)
A transducer is 19.5 mm in diameter and emits a wave of wavelength 0.50
mm. The beam travels a total distance of 37 cm through the patient and
back to the transducer. How much has the beam spread after returning to
the transducer? (Hint: Start with the formula, = 1.22
/d, and recall
that this is the minimum HALF angle of the beam.)
____________ mm
m ou Answered
ou Answered
orrect Answers
orrect Answers
23.1562717605895 23 23.2 23.1 23.16 23.156 23.1563 0 / 1 pts
Question 59
Unanswered
Unanswered
READING 4
The resolution of a microscope is diffraction limited, just as the eye, but
gains a significant advantage because an object can be placed very close
to its objective lens. The smallest resolvable linear separation xm for an
ordinary light microscope is:
X
= /( 2nsin
)
where is the wavelength of light used, and is the half viewing angle
of the objective. The quantity ’n sin ’ is called the numerical aperture of
the objective, where n is the index of refraction. The minimum resolvable
separation of two objects can be reduced by placing a liquid with a large
m
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index of refraction between the object and the objective, as in an oil
immersion microscope [14].
Generated by ©
WeBWorK, http://webwork.maa.org, Mathematical
Association of America
[14] F. R. Hallett, P. A. Speight, R. H. Stinson, INTRODUCTORY
BIOPHYSICS (New York: John Wiley Sons, 1977).
With light of wavelength 490 nm and a lens with a numerical aperture of
0.47, what is the minimum resolvable separation of two objects?
____________ micrometers
ou Answered
ou Answered
orrect Answers
orrect Answers
0.521276595744681 0.5 0.52 0.521 0.5213
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0.52128 0 / 1 pts
Question 60
Unanswered
Unanswered
In Part 6.2.2, you will determine the wavelength of the laser by shining the
laser beam on a ”diffraction grating”, a set of regularly spaced lines.
Suppose the pattern is displayed on a screen a distance L from the
grating and the spots are separated by s. If the screen is 9 m away, the
spots are 3.5 cm apart, and the lines of the grating are separated by 0.3
mm, what is ?
___________ nm
Information on diffraction gratings
(http://hyperphysics.phy-
astr.gsu.edu/hbase/phyopt/grating.html)
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ou Answered
ou Answered
orrect Answers
orrect Answers
1,166.66 (with margin: 12)
0 / 1 pts
Question 61
Unanswered
Unanswered
Suppose you have measured the diffraction pattern of a grating with d =
0.05 mm and have found that the spots were separated by s = 1.1 cm.
Now you want to determine d for an unknown grating. With the unknown
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grating, the spots are separated by s = 4.3 cm. What is the spacing
between lines d
in the unknown grating?
_________ mm
Check out this website!
(http://www.physics.uq.edu.au/people/mcintyre/applets/grating/grating.html)
You can simulate any diffraction grating with any wavelength and see the
results.
More information on diffraction gratings
(http://hyperphysics.phy-
astr.gsu.edu/hbase/phyopt/grating.html)
ou Answered
ou Answered
orrect Answers
orrect Answers
0.0128 (with margin: 0.0003)
0 / 1 pts
Question 62
Unanswered
Unanswered
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A beam 14 cm in diameter (a) consisting of microwaves with wavelength = 0.9 mm is incident upon a d=13 cm slit. At a distance of 2 m from the
slit, what is the approximate width of the slit’s image?
_____________ cm
ou Answered
ou Answered
orrect Answers
orrect Answers
13 (with margin: 0)
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0 / 1 pts
Question 63
Unanswered
Unanswered
Babinet’s principle of complementary screens. A surprising result of wave
theory is that the diffraction pattern of a dark screen with a hole cut in it (of
arbitrary shape) is identical to the pattern of an object with the same
shape as the hole. (Lab Manual Fig. 6.3).
An application of this principle is that a line mounted on transparent slide
casts the same diffraction pattern as a dark film with a slot of equal size
cut in it. In Part 6.3.4 of the experiment, you will exploit this principle to
measure the width of a hair. If the distance between the first spot and the
central minimum is s = 2.4 cm, L = 9 m, and = 4 x 10
m, what is the
width of the hair?
_________ mm
-7 ou Answered
ou Answered
orrect Answers
orrect Answers
0.15 (with margin: 0.002)
0 / 1 pts
Question 64
Unanswered
Unanswered
DIFFRACTION IN EYES, MICROSCOPES, AND ULTRASOUND
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READING 1
Two point sources can be resolved by an optical system if the
corresponding diffraction patterns are sufficiently small or sufficiently
separated. By definition, the ”minimum resolvable separation” is when the
maximum of the diffraction pattern of one source falls on the first minimum
of the diffraction pattern of the other. For a circular aperture, this distance
in angular measure (radians) is given by:
= 1. 22
/ d
where is the wavelength of light visualized, d is the diameter of the
aperture, and is the angular separation. The wavelength of light in a
material is smaller than the wavelength in vacuum by the ratio =
/ n, where n is the index of refraction. So the formula can be
generalized to:
= 1. 22
/ ( n * d)
This formula determines the diffraction-limited resolving power of an
aperture with diameter d. This angular separation can be converted to an
approximate linear separation, D, at the retina using:
= D / 0.025
where 0.025 is the corneal-retinal distance in meters. (Note: The lab
manual has a typo and should have D in this last equation. D is not the
same as d.)
Limitations of the human eye
(http://csep10.phys.utk.edu/astr162/lect/light/limitations.html)
What is the linear separation D at the retina? The minimum diameter of
the pupil is about 2 mm, the eye is most sensitive to wavelengths of about
500 nm (air), the index of refraction of the aqueous humor is n = 1.33, and
the distance between the cornea and retina is about 2.5 cm.
____________ micrometers
(This is a very small distance - approximately 1/100 the size of a period.)
m
m m
o
m
o
o ou Answered
ou Answered
orrect Answers
orrect Answers
5.733 (with margin: 0.057)
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0 / 1 pts
Question 65
Unanswered
Unanswered
READING 3
Ultrasonic waves, like all other waves, exhibit diffraction. Recall that the
minimum half angle of a beam is given by:
= 1.22
/d
The wavelengths used in ultrasound imaging techniques are generally in
the range from 0.3 mm to 0.75 mm [longer waves (lower frequencies)
penetrate more deeply but provide less resolution of detail than the short
waves]. In order for an ultrasonic beam to spread by less than a
centimeter after it has traveled across a human torso (20 cm), the source
of the waves must be about 10 or 20 times larger than the wavelength
[20]. The source of the waves is called a transducer.
[20] Harold J. Metcalf, TOPICS IN CLASSICAL BIOPHYSICS,
(Englewood Cliffs, NJ: Prentice Hall, Inc., 1980).
m
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Obstetric Ultrasounds
(http://www.ob-ultrasound.net/)
A transducer is 19.5 mm in diameter and emits a wave of wavelength 0.50
mm. The beam travels a total distance of 37 cm through the patient and
back to the transducer. How much has the beam spread after returning to
the transducer? (Hint: Start with the formula, = 1.22
/d, and recall
that this is the minimum HALF angle of the beam.)
____________ mm
m ou Answered
ou Answered
orrect Answers
orrect Answers
23.17 (with margin: 0.232)
0 / 1 pts
Question 66
Unanswered
Unanswered
READING 4
The resolution of a microscope is diffraction limited, just as the eye, but
gains a significant advantage because an object can be placed very close
to its objective lens. The smallest resolvable linear separation xm for an
ordinary light microscope is:
X
= /( 2nsin
)
where is the wavelength of light used, and is the half viewing angle
of the objective. The quantity ’n sin ’ is called the numerical aperture of
the objective, where n is the index of refraction. The minimum resolvable
separation of two objects can be reduced by placing a liquid with a large
index of refraction between the object and the objective, as in an oil
immersion microscope [14].
m
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Generated by ©
WeBWorK, http://webwork.maa.org, Mathematical
Association of America
[14] F. R. Hallett, P. A. Speight, R. H. Stinson, INTRODUCTORY
BIOPHYSICS (New York: John Wiley Sons, 1977).
With light of wavelength 490 nm and a lens with a numerical aperture of
0.47, what is the minimum resolvable separation of two objects?
____________ micrometers
ou Answered
ou Answered
orrect Answers
orrect Answers
0.5213 (with margin: 0.005)
1 / 1 pts
Question 67
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Which of the following safety rules should you follow when working with
radioactive substances?
When counting samples, do not stand beside the source. Wash your hands following the laboratory Mouth pipetting, eating, drinking, and smoking are prohibited. Return all sources to the closed, labeled container handled by your T.A. All of the above Correct!
Correct!
1 / 1 pts
Question 68
In Part 7.2.2 of the experiment, a pair of students measure the signal from
a cobalt source for 10 sec and find C = 25 counts. They decide to repeat
the measurement, but this time read only C = 20 counts. What is wrong?
Nothing is wrong; cobalt has a really short half-life Nothing is wrong; there is a certain amount of statistical uncertainty in the
measurement
Correct!
Correct!
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They did not count for long enough time intervals The Geiger counter is probably broken 1 / 1 pts
Question 69
BIOLOGICAL EFFECTS OF RADIOACTIVITY
Reading 1
As ionizing radiation penetrates living tissue, it gives up its energy through
a series of random collisions and interactions with molecules along its
path. An injury to a cell results from the molecular changes caused by
radiation-induced ions and free radicals. From a chemical point of view,
living protoplasm is a complex system consisting of water and a variety of
small and large molecules with diverse biological functions. Some of them
are found in high concentrations, so that very high radiation doses are
needed to destroy or inactivate a significant percentage of them. Others
are fewer in number and play such an important role in the life of
protoplasm that even the inactivation of a few of them results in serious
damage to certain vital activities of the cell (e.g., nucleic acids, ATP,
myosin, and hemoglobin) [21].
Since water is the most abundant molecule in the system, the interaction
of water and radiation is important. It has been firmly established that
every water molecule acted directly upon by an ionizing particle or photon
yields one hydrogen radical and one hydroxyl radical - two extremely
chemically reactive species. This in itself is not significant, but these free
radicals go on to interact with molecules of biological importance,
damaging or destroying them [21].
Among the many types of molecules that are affected by ionizing
radiation, the most critical is DNA because of the limited redundancy of
the genetic information encoded within it. The damage to DNA along a
track of radiation with a low LET (linear energy transfer) is likely to consist
of simple lesions such as single-strand breaks in the double helix of DNA.
Multiple lesions are likely with radiations of higher LET. Since simple types
of lesions can be repaired to a considerable extent by enzymes in the
cells, the damage to DNA that ultimately results may depend as much on
the effectiveness of the repair process as it does on the nature of the
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initial lesion. The effects on an unrepaired or misrepaired lesion in a DNA
molecule may be amplified many times as the DNA is transcribed and
translated; ultimately it can be transmitted to numerous daughter cells. If
only for this reason, damage to chromosomes and genes is more harmful
than other types of damage induced by low-level irradiations [22].
Differential sensitivity studies have indicated that certain cellular types are
relatively more sensitive to ionizing radiation. (Cellular sensitivity here
means the likelihood of cellular attrition in a given time after irradiation).
Cells having any of the following properties would be unusually
susceptible to damage by ionizing particles: high rate of mitosis, cellular
division throughout a major portion of the organism’s lifetime, and lack of
specialization in a cellular developmental sequence. Cell types especially
sensitive to radiation are: lymph tissue (especially lymphocytes), white
blood cells and immature red blood cells produced in the bone marrow,
cells lining the gastrointestinal tract, gonadic cells and the skin (especially
the proliferating layer) [23].
[21] Victor Arena, IONIZING RADIATION AND LIFE, (Saint Louis: C.V.
Mosby Company, 1971).
[22] Arthur C. Upton, ”The Biological Effects of Low-Level Ionizing
Radiation,” SCIENTIFIC AMERCIAN, Februrary 1982.
[23] Daniel S. Grosch, Larry E. Hopwood, BIOLOGICAL EFFECTS OF
RADIATIONS, 2nd Ed. (New York: Academic Press, 1979).
The Biological Effects of Nuclear Radation
(http://www.ratical.org/radiation/NRBE/NRBE5.html)
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Which of the following is NOT an example of a type of molecular change
within the cell typically leading to cellular damage?
A hydrogen radical and a hydroxyl radical combining to form a water
molecule.
Correct!
Correct!
Ionization of a water molecule into a hydrogen radical and a hydroxyl
radical.
A single strand break in double stranded DNA. All of the above are examples of molecular change leading to cellular
damage
1 / 1 pts
Question 70
Which of the following cellular properties would make the cell more
susceptible to damage by ionizing particles?
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Cellular division throughout a major portion of the organisms lifetime High rate of mitosis Lack of specialization in a cellular developmental sequence All of the above would make the cell more susceptible to damage Correct!
Correct!
1 / 1 pts
Question 71
READING 2
Two of the radioisotopes that are found in the fallout from nuclear
weapons tests and in the waste products of nuclear power plants are Sr
(strontium) and Cs
(cesium). Cesium is chemically similar to potassium
and strontium is chemically similar to calcium [5]. Strontium is a bone
seeker and is readily assimilated into the bones. It is eliminated very
slowly and has a long half-life. Thus, it is extremely damaging to the
bones and bone marrow over a long period of time, increasing the amount
of radiation damage done by strontium. Cesium, like potassium, becomes
concentrated in the muscle, liver and brain. The nature of these
radionuclides causes concern over the long-term hazards from their
absorption.
[5] Jerry B. Marion,William F. Hornyak, GENERAL PHYSICS WITH
BIOSCIENCE ESSAYS, 2nd Ed. (New York: John Wiley Sons, 1985).
Why are the similarities of cesium to potassium and strontium to calcium
important?
90
137
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They are readily assimilated into the body, INCREASING the amount of
radiation damage.
Correct!
Correct!
They are readily assimilated into the body, DECREASING the amount of
radiation damage.
Cesium and strontium are easily converted into potassium and calcium,
respectively, INCREASING their half-life.
Cesium and strontium are easily converted into potassium and calcium,
respectively, DECREASING their half-life.
1 / 1 pts
Question 72
READING 3
In medical applications, attention must be directed to the radioactive
isotope administered to the patient. The amount of radioactivity given for a
scan of an organ is greater than the amount necessary for a study of an
organ’s functioning. In selecting a nuclide several factors should be
considered. One of these considerations is that the EFFECTIVE half-life
within the patient should be short. Rapid decay or rapid turnover of the
nuclide limits the accumulation of radiation exposure. A short physical
half-life is desirable as it automatically limits patient exposure. In some
cases a material of longer physical half-life may be used if it has a shorter
biologic half-life (the time for the total body to eliminate one-half of the
dose of any substance by the regular process of elimination) [24]. Various
factors may affect the removal time, the pattern of distribution, the
metabolic rate and, therefore, the degree and speed of elimination of the
radioisotope. Factors that affect the biological half-life are physical (size of
particles), chemical (water solubility of material, metabolic affinity of the
element), ecological (balance of calcium, iodine, etc.) and physiological
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(mode of intake, metabolic condition of the organism). Thus, both physical
and biological factors affect the biological half-life. A radionuclide can have
a long half-life, but if it is retained in the body for a short time it will have a
short biological half-life and a correspondingly short effective half-life. The
effective half-life is given by the formula [1]:
T
= ( T
xT
)/( T
+ T
)
where T
is the biological half-life and T
is the physical halflife.
[1] Perry Sprawls, PHYSICAL PRINCIPLES OF MEDICAL IMAGING,
(Rockville, Maryland: Aspen Publishers, Inc., 1987).
[24] Sheldon Baum, Roland Bramlet, BASIC NUCLEAR MEDICINE (New
York: Appleton-Century-Crofts, 1975).
More Information on Decay and Half-Life
(http://www.iem-
inc.com/prhlfr.html)
What factors affect the biological half-life?
eff b
p
b
p
b p Mode of intake Water solubility of material Size of particle All of the above affect the biological half-life Correct!
Correct!
0 / 1 pts
Question 73
Unanswered
Unanswered
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At a distance of 8 cm from a presumably isotropic, radioactive source, a
pair of students measure 57 cps (cps = counts per second). On average,
how many counts per second do you expect at a distance of 29 cm? (Note
that the average number of counts per second need not be an integer.)
___________ cps
ou Answered
ou Answered
orrect Answers
orrect Answers
4.3377 (with margin: 0.1)
0 / 1 pts
Question 74
Unanswered
Unanswered
For an unknown sample of the experiment, students measure 1603
counts when they first receive their sample and 1382 counts 4.5 minutes
later. Calculate the half-life t
of their sample.
___________ min
1/2
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ou Answered
ou Answered
orrect Answers
orrect Answers
21.0264 (with margin: 0.21)
0 / 1 pts
Question 75
Unanswered
Unanswered
As an example, mercury-203 is a radioisotope that is important for kidney
function studies and renal imaging. The physical half-life, T
, is 46.6 days
and biological half-life, T
, of Hg
is 9.2 days. What is the effective half
life of mercury-203?
____________ days
p
b
203 ou Answered
ou Answered
orrect Answers
orrect Answers
7.6832 (with margin: 0.1)
0 / 1 pts
Question 76
Unanswered
Unanswered
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Semilog paper is a convenient tool for analyzing data that changes
exponentially. The scale on the y axis of the paper is logarithmic. Since
the logarithm of an exponential function is a linear function, exponentially
varying data appear as a straight line on semilog paper. Suppose that in
Part 7.2.3 of the experiment, students created the above graph based on
their data.
What is the half-life in minutes?
__________ minutes
More about Semilog Graph Paper
(http://www.science-
projects.com/SemiLogUse.htm)
ou Answered
ou Answered
orrect Answers
orrect Answers
1 (with margin: 0.1)
1 / 1 pts
Question 77
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An alpha cannot penetrate your skin. Once inside the body, however, an
alpha source is the most hazardous radiation since it causes the most
ionizing events per unit length. There is really only one way a sealed
alpha source can hurt you. What is it?
If it comes from a uranium source By direct contact with skin By ingestion Correct!
Correct!
None of the above 1 / 1 pts
Question 78
Suppose that in Part 8.2.4 a pair of students obtain data measuring the
range of gammas in lead using thicknesses of 0, 3, 6, and 9 cm. For 0 cm
thickness, they measure 2742 counts; for 3 cm thickness, they measure
638 counts; for the 6 cm thickness, they record 155 counts; and for 9 cm
thickness, they record 26 counts. Next, they plot their results on semilog
paper. Along the x-axis they plot r x, where x is the thickness. Along the y-
axis they plot the number of counts. Because they use semilog paper, this
is the same as if they did a linear plot with the y-axis being the logarithm
of the number of counts. Then they fit a line through the data, and use the
graph to determine the mass absorption coefficient . On the graph, the
mass absorption coefficient is given by
More about Semilog Graph Paper
(http://www.science-
projects.com/SemiLogUse.htm)
m
The y-intercept of the line (where the line hits the y-axis) The inverse of the slope of the line The slope of the line
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The negative of the slope of the line Correct!
Correct!
The x-intercept of the line (where the extrapolated line hits the x-axis) none of the above 1 / 1 pts
Question 79
READING 1
The three types of radiation (
, , ) have different effects upon the
biological system. This relative effect is called the QUALITY FACTOR (or
relative biological damage caused by the radiation) [21].
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Alpha particles are doubly charged, slow moving particles, which allows
them to interact effectively with atomic electrons and produce a high
degree of ionization. This is because the electric field of a slowly moving
particle can act on an atomic electron for a greater length of time, thus
increasing the probability of driving an electron out and causing ionization.
The ray, on the other hand, has no electric change (it consists of a
photon) and is rapidly moving. Furthermore, all the ionization
accomplished by rays is produced by secondary electrons. Thus, the
effect of rays is not as severe as particles. Note that this has an
inverse relationship to the penetrating power of the radiations: g rays can
penetrate the body, b particles penetrate a few millimeters of tissue, and
a particles will penetrate only to a depth of about 40 micrometers.
Neutrons are relatively damaging because they penetrate deep into tissue
and produce high energy protons, which behave like alphas [5].
[5] Jerry B. Marion,William F. Hornyak, GENERAL PHYSICS WITH
BIOSCIENCE ESSAYS, 2nd Ed. (New York: John Wiley Sons, 1985).
[21] Victor Arena, IONIZING RADIATION AND LIFE, (Saint Louis: C.V.
Mosby Company, 1971).
Fill in the blanks. Lower energy particles can act on an atomic electron for
a greater length of time, thus increasing the probability of driving an
electron out and causing ionization. Thus___________ ______________
are more damaging than __________________ ____________________.
Gamma rays; Alpha particles Gamma rays; Beta particles Alpha particles; Gamma rays Correct!
Correct!
Beta particles; Alpha particles 1 / 1 pts
Question 80
READING 2
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In medical imaging techniques, many different radioisotopes are utilized:
I
for thyroid studies, Fe
for blood studies, etc. However, in recent
years, the desirability of Tc
(m stands for metastable) and other
metastable isotopes for use in nuclear medicine has increased.
Metastable states are of particular interest in nuclear medicine because
they make possible the separation of electron and photon radiation. In a
diagnostic procedure it is undesirable to have electron radiation in the
body because it contributes to radiation dosage but not to image
formation. By using a nuclide that has already undergone an electron-
emitting transition and is in a metastable state, it is possible to have a
radioactive material that emits only gamma radiation [1].
[1] Perry Sprawls, PHYSICAL PRINCIPLES OF MEDICAL IMAGING,
(Rockville, Maryland: Aspen Publishers, Inc., 1987).
Which metastable radionuclides are preferable to other radionuclides for
diagnostic imaging in nuclear medicine?
131 59 99m Metastable radionuclides that have already undergone an electron-emitting
transition.
Correct!
Correct!
Metastable radionuclides that have a low quality factor. Metastable radionuclides that have very low gamma radiation. All of the above are desirable metastable radionuclides.
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0 / 1 pts
Question 81
READING 3
The thyroid needs iodine to function properly. (This is why iodine is added
to table salt.) The radioisotope most commonly used in thyroid studies is
I. Tracer doses of Na
I in the range of 5 to 50 microcuries are given
orally to the patient, either in a solution or in a capsule. An identical dose
is set aside to be used as a standard. An external radioassay of the
thyroid is done 24 hours later. Scintiscans of an abnormal thyroid may
reveal areas of greater or lesser radioactivity than the rest of the gland.
Some of these areas are called nodules, and they can be felt by
palpitation. The scintiscan will reveal whether a nodule is ”hot” (very
active, accumulates radioiodine) or ”cold” (inactive). Cold nodules may be
either centers of neoplastic malignancy or adenomas, cysts or
hemorrhagic areas. Hot nodules are almost always benign [21].
Radioiodine experiments can also determine the activity of the thyroid,
which is used to diagnose hyperthyroidism or hypothyroidism.
The diagnosis of hyperthyroidism/hypothyroidism is a similar procedure
(administration of compounds of I, followed by external measures of -radiation intensity in the thyroid). A hyperactive thyroid may absorb up to
80% of the tagged iodine; a hypoactive thyroid may absorb as little as
15in the body reduces concentration via excretion. Mapping of the thyroid
by I scintography is common practice. Both the outline of the organ
and its turnover rate can be obtained from maps made at different time
intervals after administration. The maximum activity of the emission is a
direct measure of the uptake of iodine by the thyroid [26].
131 131
131 131
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[21] Victor Arena, IONIZING RADIATION AND LIFE, (Saint Louis: C.V.
Mosby Company, 1971).
[26] E. J. Casey, BIOPHYSICS (New York: Reingold Publishing Co.,
1965).
What is palpitation?
(http://cancerweb.ncl.ac.uk/cgi-bin/omd?
query=palpitation&action=Search+OMD)
What is a scintiscan?
(http://www.nlm.nih.gov/medlineplus/ency/imagepages/1206.htm)
What is neoplastic?
(http://cancerweb.ncl.ac.uk/cgi-bin/omd?
query=neoplastic)
Information on thyroid nodules
(http://www.endocrineweb.com/nodule.html)
How do radioiodine experiments diagnose hyperthyroidism/
hypothyroidism?
Measure of the normal biological turnover rate of the radioactive source
via excretion.
ou Answered
ou Answered
A hyperactive thyroid may absorb up to 80a hypoactive thyroid may absorb
as little as 15% of I.
orrect Answer
orrect Answer
131
A hypoactive thyroid may absorb up to 80% of I, while a hyperactive
thyroid may absorb as little as 15% of I.
131
131
Administer radioiodine compound and then measure the relative
temperature of the thyroid with respect to a standard.
Administer Na, followed by external measures of -radiation intensity.
131
None of the above
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1 / 1 pts
Question 82
READING 4
In positron emission tomography (PET), the distribution of a positron
source within the body is reconstructed from the profile of gamma rays,
the product of positron annihilation. Short lived radioisotopes such as C,
N, O, and F (as a substitute for hydrogen) replace natural isotopes
of C, N and O in various organic compounds. These isotopes all decay by
the emission of positrons (antimatter electrons) that combine with
electrons to annihilate and produce two 511 keV rays that are emitted
180 apart. These gamma rays easily penetrate body tissues and can be
externally detected. The only radioisotopes of these elements that can be
detected outside the body are positron emitters. The use of these specific
radioisotopes leads to an attractive use of the PET technique. Positron
emission tomography is an analytical imaging technique that provides a
way of making in vivo measurements of the anatomical distribution and
rates of specific biochemical reactions. C, N, and O are used to
label biologically active substrates without disrupting their biochemical
properties. Radiation detectors record the emission of rays from the
tissue distribution of positron activity. Data collected are used to form a
tomographic image of the cross-sectional distribution of tissue
concentration of radioactivity according to the principles of computed
tomography. This provides quantitative, non-invasive measurements in
humans of biochemical activity [27].
11 13
15
18
11 13
15
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[27] Michael Phelps, John C. Mazziotta, ”Positron Emission Tomography:
Human Brain Function and Biochemistry,” SCIENCE, Vol. 228, No. 4701
(17 May 1985).
What is an anti-particle?
(http://www2.slac.stanford.edu/vvc/theory/antiquarks.html)
Fill in the blanks. In the PET technique, detectors record the emission
of_____________________ from active substrates to form a tomographic
image of the cross-sectional distribution of tissue concentration.
Alpha particles Beta particles Gamma rays Correct!
Correct!
Positrons
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0 / 1 pts
Question 83
Unanswered
Unanswered
Suppose that placing 0.3 inch of lead in front of a gamma source reduces
the count rate from 996 cps to 613 cps. What is in g / cm
? The
density of lead is 11.4 g / cm
.
_____________g / cm
-1
m 2
3
2
ou Answered
ou Answered
orrect Answers
orrect Answers
17.8968199140404 18 17.9 17.90 17.897 17.8968 17.89682 0 / 1 pts
Question 84
Unanswered
Unanswered
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Answer 1:
Answer 2:
The densities of cardboard, aluminum, and lead are 0.6 g= cm
, 2.7 g /
cm3 , and 11.4 g /cm
, respectively. Suppose that you are studying the
range of a (nonexistent) elementary particle, the Heidbrinkion, and that it
takes 53 cm of cardboard, or 43 cm of aluminum, or 12 cm of lead to stop
half of the Heibrinkions emitted from a source. Calculate the absorber
thickness for each material.
g / cm
for cardboard
g / cm
for aluminum
g / cm
for lead
3
3
2 2
2
(You left this blank)
ou Answered
ou Answered
31.8 orrect Answer
orrect Answer
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Answer 3:
(You left this blank)
ou Answered
ou Answered
116.1 orrect Answer
orrect Answer
(You left this blank)
ou Answered
ou Answered
136.8 orrect Answer
orrect Answer
0 / 1 pts
Question 85
Unanswered
Unanswered
The densities of cardboard, aluminum, and lead are 0.6 g= cm
, 2.7 g /
cm3 , and 11.4 g /cm
, respectively. Suppose that you are studying the
range of a (nonexistent) elementary particle, the Heidbrinkion, and that it
takes 53 cm of cardboard, or 43 cm of aluminum, or 12 cm of lead to stop
half of the Heibrinkions emitted from a source. Calculate the absorber
thickness for each material.
g / cm
for cardboard
g / cm
for aluminum
3
3
2 2
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Answer 1:
Answer 2:
Answer 3:
g / cm
for lead
2
(You left this blank)
ou Answered
ou Answered
31.8 orrect Answer
orrect Answer
(You left this blank)
ou Answered
ou Answered
116.1 orrect Answer
orrect Answer
(You left this blank)
ou Answered
ou Answered
136.8 orrect Answer
orrect Answer
0 / 1 pts
Question 86
Unanswered
Unanswered
A typical sheet of weighing paper is 4.5 inches by 4.3 inches and weighs
0.43 g. What is its absorber thickness in g / cm
?
______________g / cm
2
2
ou Answered
ou Answered
orrect Answers
orrect Answers
0.0034 (with margin: 0.0001)
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0 / 0.33 pts
Question 87
Unanswered
Unanswered
The densities of cardboard, aluminum, and lead are 0.6 g / cm
, 2.7 g /
cm3 , and 11.4 g /cm
, respectively. Suppose that you are studying the
range of a (nonexistent) elementary particle, the Heidbrinkion, and that it
takes 53 cm of cardboard, or 43 cm of aluminum, or 12 cm of lead to stop
half of the Heidbrinkions emitted from a source.
Calculate the absorber thickness for aluminum.
_____________g / cm
for aluminum
3
3
2
ou Answered
ou Answered
orrect Answers
orrect Answers
116.1 (with margin: 1.16)
0 / 0.33 pts
Question 88
Unanswered
Unanswered
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The densities of cardboard, aluminum, and lead are 0.6 g / cm
, 2.7 g /
cm3 , and 11.4 g /cm
, respectively. Suppose that you are studying the
range of a (nonexistent) elementary particle, the Heidbrinkion, and that it
takes 53 cm of cardboard, or 43 cm of aluminum, or 12 cm of lead to stop
half of the Heidbrinkions emitted from a source.
Calculate the absorber thickness for lead.
_____________g / cm
for lead
3
3
2
ou Answered
ou Answered
orrect Answers
orrect Answers
136.8 (with margin: 1.37)
1 / 1 pts
Question 89
When a wave strikes a new medium, three things can happen: reflection,
transmission, and absorption. Consider a wave propagates in air and is
transmitted through a piece of paper. Does increasing the frequency
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increase the amplitude of a sound wave transmitted through the piece of
paper?
No Correct!
Correct!
Yes 1 / 1 pts
Question 90
When a wave strikes a new medium, three things can happen: reflection,
transmission, and absorption. Consider an echo. Do you think reflection is
an important process when sound waves produce an echo?
Learn more about the behavior of sound waves.
(http://www.physicsclassroom.com/Class/sound/u11l3c.html)
Yes Correct!
Correct!
No
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1 / 1 pts
Question 91
This refers to the previous question in which a wave propagates in air and
is transmitted through a piece of paper. Does increasing the frequency
increase the amplitude of a sound wave transmitted through the piece of
paper?
No Correct!
Correct!
Yes 0 / 1 pts
Question 92
Unanswered
Unanswered
In an arrangement such as in Lab Manual Figure 10.1, the attenuators are
8 inches by 11 inches. Referring to the previous problem, compare the
wavelength with the size of the attenuator. Do you expect diffraction of
sound waves by the attenuators to be important? (Hint: Use the diffraction
equation (Equation (6.2)) from the lab manual to compare the wavelength
with the size of the attenuator. Use the wavelength that you found in the
previous question. Let d=8 inches in Eq. (6.2) and find the angle . If the
angle is large, then diffraction is important. It the angle is small (close to
zero), then diffraction is not important.)
Diffraction will be very important and will dominate. Diffraction will be somewhat important orrect Answer
orrect Answer
Diffraction will be negligible.
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Question 93
Unanswered
Unanswered
THE OSSICLES AND ULTRASOUND
READING 1
Of all the materials in the body, water produces by far the least
attenuation. This means that water is a very good conductor of ultrasound.
Water within the body, such as that within cysts and the bladder, forms
”windows” through which underlying images can be easily imaged.
However, the lung and bones act as a shield against easy access by
ultrasound to some parts of the body. Lung has a much higher attenuation
rate than either air or soft tissue. This is because the small pockets of air
in the alveoli are very effective in scattering ultrasound energy. (Any
interface involving air as one of the components will be an impenetrable
barrier for the ultrasound. For this reason, the transducer is carefully
coupled to the skin via mineral oil or gel in a manner that will eliminate all
air bubbles). Because of this, the normal lung structure is extremely
difficult to penetrate with ultrasound. Compared to the soft tissues of the
body, bone has a relatively high attenuation rate. Thus, bone shields some
parts of the body from access by ultrasound [1].
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[1] Perry Sprawls, PHYSICAL PRINCIPLES OF MEDICAL IMAGING,
(Rockville, Maryland: Aspen Publishers, Inc., 1987).
More about how a fetal ultrasound works
(http://my.webmd.com/content/healthwise/95/23646)
An interface between air and some other substance, like tissue or water,
strongly scatters ultrasound waves. Which of the following is NOT a
consequence of this?
Lungs have high ultrasound attenuation rates. Images are sharper with higher ultrasound frequencies. orrect Answer
orrect Answer
Sound transducers should be carefully coupled to the skin via gel. All of the above are consequences. 0 / 1 pts
Question 94
Unanswered
Unanswered
READING 2
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The reflection of ultrasound pulses by structures within the body is the
interaction that creates the ultrasound image. The reflection of an
ultrasound pulse occurs at the interface, or boundary, between two
materials. In order to form a reflection interface, the two materials must
differ in terms of a physical characteristic called acoustic impedance. This
is a characteristic of a material related to the density and elastic properties
of the material. For example, the acoustic impedance of water is 1.5x10
gm / cm
- sec, while the impedance of air is 41 gm / cm
- sec. At most
interfaces, only a portion of the ultrasound pulse is reflected. The
amplitude of the pulse that is reflected back to the transducer is smaller
than the amplitude of the incident pulse. This reduction is characteristic of
the type of interface the pulse was reflected from (blood-tissue, fat-
muscle, etc.) [1]. In most imaging situations, the ultrasound pulse passes
through several different materials on its way to a reflection site (interface)
and back to the transducer. The velocity of the pulse depends on the
characteristics of the material through which it is passing. Therefore, the
velocity of the pulse actually changes as it passes from one material to
another. However, distance errors because of variations in ultrasound
velocity are relatively small, except in solid materials such as bone.
When an ultrasound pulse returns to the transducer, it brings back two
kinds of information. The relative amplitude provides information as to the
type of interface that produced the reflection. The time interval between
when a pulse leaves and when it returns to the transducer is related to the
distance between the transducer and the reflecting interface. These two
pulse characteristics are used to form the ultrasound image [1].
In the ear, three small bones called the ossicles connect the eardrum and
the oval window. These ossicles transmit the pressure wave from the
eardrum to the oval window. In the outer and middle ear, sound waves
travel through the air but, in the cochlea, the sound wave propagates
through a fluid. If the waves acted directly on the cochlea, only a small
fraction of the wave energy would be transmitted (most of the energy
would be reflected). The ossicles convert the large amplitude, low
pressure of the eardrum to a low amplitude, high pressure vibration at the
window of the cochlea. In physics jargon, the ossicles match the
impedance of air to the impedance of the liquid in the cochlea [8].
5
2 2
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[1] Perry Sprawls, PHYSICAL PRINCIPLES OF MEDICAL IMAGING,
(Rockville, Maryland: Aspen Publishers, Inc., 1987).
[8] Leonard H. Greenburg, PHYSICS FOR BIOLOGY AND PRE-MED
STUDENTS, (Philiadelphia, W. B. Saunders Co., 1975).
What is acoustic impedance and why is it important?
(http://www.phys.unsw.edu.au/~jw/z.html)
How hearing works
(http://entertainment.howstuffworks.com/hearing.htm)
In order to form a reflection interface for sound waves, what must be true
of the materials on either side of the interface?
They must differ in their indices of refraction. They must be at different temperatures. They must be separated by a certain distance. They must differ in their acoustic impedance. orrect Answer
orrect Answer
None of the above
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0 / 1 pts
Question 95
Unanswered
Unanswered
Fill in the blanks
. A reflected ultrasound pulse brings back two kinds of
information. The ___________ of the reflected pulse relative to that of the
incident pulse gives information on the type of interface that produced the
reflection. The _______________ gives information about the distance
between the transducer and the reflecting interface.
travel time; frequency of the reflected pulse; amplitude; travel time between when a pulse leaves and when it returns to
the transducer;
orrect Answer
orrect Answer
frequency; amplitude of the reflected pulse; None of the above 0 / 1 pts
Question 96
Unanswered
Unanswered
The velocity of sound in air at STP (Standard Temperature and Pressure)
is approximately 344 m/s. If the frequency of the sound wave is = 5.3
kHz, what is the wavelength in cm?
__________cm
What is STP?
(http://www.scottecatalog.com/glossary.nsf/0/b6cc55c065e7d9f0852568a9007
OpenDocument)
Standard Wave Relationship
(http://hyperphysics.phy-
astr.gsu.edu/hbase/wavrel.html)
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ou Answered
ou Answered
orrect Answers
orrect Answers
6.49056603773585 6.5 6.49 6.491 6.4906 6.49057 6.490566 0 / 1 pts
Question 97
Unanswered
Unanswered
In Eq. (10.1) in the lab manual, the constant is 41.1 gmcm
sec
in
air. A piece of paper has a mass per unit area of approximately 7x10
gm/cm
. If the frequency is 4.6 kHz, what does theory predict for the ratio
of the transmitted amplitude to the incident amplitude of the sound wave?
(A sound wave is a pressure wave.)
_____________
-2
-1 -
3
2
ou Answered
ou Answered
orrect Answers
orrect Answers
0.376408789721231 0.38
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0.376 0.3764 0.37641 0.376409 0.3764088 0.37640879 0 / 1 pts
Question 98
Unanswered
Unanswered
What is the mass per unit area of an 7 inch by 10 inch lead sheet that
weighs 192 gm.
____________g / cm
2
ou Answered
ou Answered
orrect Answers
orrect Answers
0.425143707430272 0.4
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0.42 0.43 0.425 0.4251 0.42514
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A cylindrical specimen of Aluminum having a diameter of 12.8 mm and a gauge length of
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Load
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50.800
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50.902
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AN[2]
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