Download unit 9: imaging

Reflection and Refraction of Light
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A laser pointer is aimed at the surface of a plane mirror. Use a protractor and straight-edge to construct the
laser beam after it reflects from the mirror.
Plane Mirror:
Normal:
Angle of incidence:
Angle of reflection:
Law of Reflection
Properties of Images formed by Plane Mirrors
Each object below is in front of a plane mirror (seen on edge). Sketch the image that you would see in each case if you
were looking into the mirror. Then, check your result by placing a plane mirror on top of this page at each location
and looking into it.
What are some properties of images formed by plane mirrors?
1.
4.
2.
3.
5.
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Locating Images using the Law of Reflection
Locate the image of this dot by means of two lines of sight.
1.
2.
How much of this 2.0 meter
tall mirror is actually needed
for the man to see the
reflection of his entire body?
Virtual Image:
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Spherical Mirrors
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Source of Parallel (paraxial) Rays:
Method of locating focal point:
Principal Axis:
Relationship between radius
of curvature and focal length
Center of Curvature (C):
Radius of Curvature (R):
Focal Point (F):
Focal Length (f):
Power (P):
Converging Mirror
Shape:
Focal Point:
Focal Length:
Images:
Examples:
Diverging Mirror
Shape:
Focal Point:
Focal Length:
Images:
Examples:
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Real and Virtual Images
Real image:
Properties:
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Virtual image:
1.
2.
Properties:
1.
2.
Ray Tracing to Locate Images
Ray #1:
Your Turn
Ray #2:
Your Turn
Ray #3:
Your Turn
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Image Properties
Case 1:
Your Turn
Image Properties
Case 2:
Image Properties
Case 3:
Image Properties
Case 4:
General Trend: As the object moves closer to the mirror . . .
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Image Properties Case 5:
Application:
Your Turn
Summary
Chart for
Converging
(Convex)
Mirror
Object
Position
Image
Position
Image
Properties
Convex Mirror
Image Properties:
Application:
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Calculating Locations and Sizes of Images
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p=
q=
ho =
hi =
Mirror
Equation
Linear
Magnification
Sign
Convention
1. A 2.0-cm-high object is placed 7.10 cm from a concave mirror whose radius of curvature is 10.20 cm.
a) Locate the image by means of a ray
diagram.
c) Calculate the magnification of the
mirror.
d) Calculate the size of the image.
b)
Calculate the location of the image.
e) Describe the image.
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2. An object is placed 6.00 cm in front of a concave mirror that has a 10.0-cm focal length.
a) Determine the location of the image.
b) The object is 1.2 cm high. Find the height of the image.
3. A convex mirror is used to reflect light from an object placed 66 centimeters in front of the mirror. The focal
point is 46 centimeters from the mirror. Find the location of the image.
In reality, why is a parabolic mirror preferred over a concave or convex mirror?
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Refraction of Light
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Refraction:
Complete the path of each light ray shown below.
Angle of refraction:
In which substance does light travel faster – air or water?
Snell’s Law for Refraction
Snell’s Law
Use a protractor and Snell’s law to construct
the refracted ray on the diagram at right.
air
flint glass
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Total Internal Reflection
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Total Internal Reflection:
Conditions for Total Internal Reflection:
1.
2.
Critical Angle (θc):
Formula:
1. What is the critical angle as light
exits from water into air?
2. What is the critical angle as light exits
from water into crown glass?
Applications of Total Internal Reflection
Fiber Optic Cables
How do fiber optic cables (optical fibres) work?
Cladding:
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Dispersion
Natural Dispersion: since the refractive index of a medium depends on the wavelength of the light
travelling through it, light of different wavelengths will travel through the glass core of an optical
fibre at different speeds
Result: a set of light rays of different wavelengths will reach the end of a fibre at different times,
even when following the same path
High-order-mode paths: rays that undergo many internal reflections over a given distance
Low-order-mode paths: rays that undergo few internal reflections over a given distance
Waveguide dispersion: the result of rays of the same wavelength following different paths (having
different order modes), thus reaching the end in different amounts of time
Multimode fibre vs Monomode fibre
Multimode fibres: Light of different wavelengths follow different paths, being subjected
to both material and waveguide dispersion; diameter of the core is about 100 µm
Monomode fibers: All light propagates along the same path; diameter of the core is very
small, about 10 µm, with considerably thicker cladding
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Step-index fibre vs Graded-index fibre
Step-index fibres: the refractive index of the core is constant, but is slightly higher than
the constant refractive index of the cladding.
Graded-index fibres: the refractive index decreases smoothly from the center of the core
to the outer edge of the core, with a constant value in the cladding
Attenuation
Attenuation: A loss of power as a signal is traveling through a medium
Causes in an optical fibre: the scattering of light and the impurities in the glass core
Amount of attenuation depends on: the wavelength of the light being transmitted
Formula:
units:
General rule of thumb:
Specific Attenuation: The power loss in decibels per unit length travelled
Formula:
units:
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1. An amplifier amplifies an incoming signal of power 0.38 mW to 2.7 mW. Calculate the power
gain of the amplifier in decibels.
2. A source produces an initial power output of 12 mW. What is the final power output after it
experiences a power loss of 18 dB while traveling along a cable?
3. A signal of power 15 mW is input into a cable of specific attenuation 3.0 dB km-1. Calculate the
power of the signal after it has travelled 8.0 km
Advantages of Optical Fibres
Advantages of optical fibres over coaxial cables (and also over twisted copper wires):
1) Fast (travels at “c”) and cheap transfer of information in digital form
2) Low attenuation
3) No interference from stray EM signals
4) Greater capacity (bandwidth), making possible the transmission of many signals
5) Security against “tapping”, i.e. unauthorized extraction of information from the
signal
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Lenses
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Why do lenses converge or diverge light?
How can the focal length of a
converging lens be found?
Shape:
Converging Lens
Focal Point:
Focal Length:
Images:
Converging lenses are . . .
Diverging Lens
Shape:
Focal Point:
Focal Length:
Images:
Diverging lenses are . . .
Factors Affecting Focal Length
a) Thickness
c) Frequency
b) Refractive Index
n = 1.52
n = 1.68
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Image Formation by Converging Lenses
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Sample of a ray diagram used to locate and describe image
Three Principal Rays used in Ray Diagramming
1. Ray parallel to principal axis . . .
2. Ray through F . . .
3. Ray through the center of the lens . . .
Case 1
Properties of image:
Case 2
Properties of image:
Case 3
Properties of image:
Case 4
Properties of image:
As the object moves closer to the lens . . .
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Virtual Images
Case 5
Properties of image:
Case 6
Properties of image:
Applications
State the type of lens, locate the object and image, and describe the image for each device below.
a) Camera
b) Projector
c) Magnifying Glass
Lens:
Lens:
Lens:
Object:
Object:
Object:
Image:
Image:
Image:
d) Security “Peephole”
Lens:
Object:
Image:
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The Thin-Lens Equation and Linear Magnification
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f = focal length
p = object distance
q = image distance
ho = height of object
hi = height of image
Thin-Lens Equation
Linear Magnification
Sign Conventions
positive =
negative =
1. A 3.0 cm high object is placed 15 cm from a converging lens whose focal length is 6 cm. Determine the location
of the image and describe its properties. Determine the magnification of the lens and the height of the image.
2. A 20 cm high object is placed 10 cm in front of a convex lens whose focal length is 30 cm. Determine the
location and properties of the image.
3. A 10 cm high object is placed 20 cm in front of a diverging lens whose focal length is 60 cm. Determine
the location and properties of the image.
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Comparisons – Mirrors and Lenses
Converging
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Diverging
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Lens Combinations: Virtual Objects
By combining multiple lenses, the image produced by the first lens can serve as the object for the
second lens.
Overall Magnification:
4. An object lies on a table. A converging lens of focal length 8.0 cm is placed 6.0 cm above the
object.
a) Determine the image formed by this lens.
b) A second converging lens of focal length 5.0 cm is now placed 3.0 cm above the first lens.
Determine the image formed by this combination of lenses.
5. An object is placed 10.0 cm to the left of a converging lens of focal length 5.0 cm. A second
diverging lens of focal length 7.0 cm is placed 4.0 cm to the right of the converging lens. Determine
the image of the object in the two-lens system, AND verify your results with a scaled ray diagram.
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Optical Instruments
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PURPOSE: To increase the
angular size of an image in
order to aid the eye to see it
more clearly
Linear Size (ho or hi): height of object or image – measured in meters
Angular Size (θo or θi): angle subtended at the eye by the object or image – measured in radians
An object has one linear size but different angular sizes depending on how far it is from the observer.
Far Point: distance between the eye and the furthest
object that can be comfortably brought into focus
1. Taken to be infinity – approx. > few meters in practice
2. Most comfortable viewing arrangement
Near Point: distance between the eye and the nearest
object that can be comfortably brought into focus
1. “least distance of distinct vision”
2. D = 25 cm for normal eye– depends on age
Standard Case – When the
object is at the near point,
the it occupies the greatest
angular size. The
magnification of any optical
instrument is measured
against this case.
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The purpose of an optical instrument is to magnify the angular size of an image.
Without an optical instrument
to magnify image
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With an optical instrument to
magnify image
Angular Magnification (M): ratio of the angle
subtended at the eye by the image to the angle
subtended at the eye by the object
Formula:
Possible Angular Magnifications of a Magnifying Glass
Extreme Case 1: Image is formed “at infinity”
1. object placed at focal point (u = f)
2. image formed at infinity (v →∞) – at far point
3. smallest possible angular magnification for
mag. glass
Angular Magnification Derivation
1. A stamp collector wishes to look at a 1.20 cm square stamp for
awhile comfortably with a magnifying glass whose focal
length is 10.0 cm.
a) Where should she place the stamp for comfortable
prolonged viewing?
b) Where will the image appear?
c) What is the angular magnification of the lens for this
situation?
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Extreme Case 2: Image is formed at the near point
1. object placed in front of focal point
2. image formed at near point (v = -25)
3. largest possible angular magnification
for mag. glass
Angular Magnification Derivation
2. The stamp collector now wishes to examine a
particular detail on the stamp more closely so
she wishes to have the maximum possible
angular magnification.
a) Where will the image be formed?
b) Where should the stamp be placed?
c) What is the angular magnification of the
lens for this situation?
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The Compound Microscope
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Purpose:
Objective Lens:
Eyepiece Lens:
lens close to object,
lens very near eye
Short focal length (f < 1 cm)
Longer focal length (f ≈ few cm)
Forms real image to be used for
second lens
Forms virtual larger image
Focal Lengths:
1.
2.
Principles of Operation
1. The object is placed just outside the objective lens’ focal point in order to form a real image for the eyepiece
lens.
2. A real, inverted, larger image is formed more than twice the objective lens’ focal length away on the other side
of this lens.
3. The eyepiece lens is placed so that it acts as a simple magnifying glass, that is, so that the real image falls
within its focal length. This lens then forms the final image – a virtual, non-inverted, larger image somewhere
behind the real image.
4. The eye is placed very near to the eyepiece lens and the position of this lens is adjusted so that the final image
is located at the near point for maximum angular size. This is called “normal adjustment.”
Linear Magnifications: (pg 21 book packet)
Overall Angular Magnification at normal adjustment of the microscope: (pg 20/21 of book packet)
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1. a) Locate the final image in the diagram below by means of ray tracing.
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\
b) The objective and eyepiece of the compound microscope above have focal lengths of fo =
15.0 mm and fe = 25.5 mm. A distance of 61.0 mm separates the lenses. The microscope is
being used to examine an object placed 24.1 mm in front of the objective. Find the final
image distance and linear magnification.
c) Where would the eyepiece need to be placed for the microscope to yield
the maximum possible angular magnification?
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The Astronomical Telescope
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Purpose:
Objective Lens:
large diameter – to collect more light, reduce diffraction,
increase resolution
Long focal length
Object approximately at infinity
Forms real image in focal plane of lens
Eyepiece Lens:
Focal Lengths:
short focal length
1.
Acts as magnifying glass with object at focal
point
2.
Forms large virtual image at infinity
Principles of Operation
1. Object (star, planet, etc.) is very far away – approximately at infinity. As a result, wavefronts arriving
from object are approx. parallel as are the rays.
2. Real, inverted, smaller image is formed in the focal plane of the objective lens.
3. The eyepiece lens is placed so that it acts as a simple magnifying glass - the real image is at the focal
point of the eyepiece lens and acts as an object for this lens. The eyepiece lens’ focal point then
coincides with the focal point of the objective lens. The real image now falls in the mutual focal plane of
the two lenses.
4. The eye is placed very near to the eyepiece lens. The final image is a virtual, non-inverted, larger image
at infinity.
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Your Turn
Magnification of the Astronomical Telescope
Angular Magnification:
Length of telescope tube:
1. A telescope has the following specifications: fo = 985 mm and fe = 5.00 mm. From these data, find
(a) the angular magnification of the telescope
(b) the approximate length of the telescope.
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Reflecting Telescopes
Reflecting telescopes use mirrors rather than the lenses of refracting telescopes. The largest
telescopes used are reflecting.
Advantages:
1) To see distant faint objects requires large lenses (to collect more light); but large lenses are
hard to make and can only be supported along their rim; large lenses may collapse under
their own weight
By contrast, large mirrors can be supported along the rim and at the back.
2) Mirrors do not suffer from chromatic aberration
3) Only 1 side has to be ground, as opposed to 2 for lenses
Two General types of reflecting telescopes:
1)
2)
Single-dish Radio Telescopes
A radio telescope receives and detects EM waves in the radiofrequency region.
What types of objects are known to radiate in this region?
Properties of single-dish radio telescopes:
1) Since frequencies are small, their wavelengths are large, so the diameter of the radio telescope
has to be large as well in order to achieve reasonable resolution (Arecibo radio telescope =300
m diameter , while Hubble Space Telescope = 2.4 m diameter)
2) Large radio telescopes are very heavy steel structures
3) Can be difficult to steer (if it can be steered at all)
4) Parabolic shape with a detector placed at the focus of the mirror
Note: Diffraction places limits on resolution, i.e. that ability of an instrument to see two nearby
objects as distinct.
Resolution:
Where: θA =
λ=
b=
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Radio Interferometry Telescopes
Radio interferometry:
How does it work?
Use a very large array of radio telescopes very far apart and appropriately combine the signals from
the individual dishes. This can achieve the same resolution as a single dish with a diameter equal to
the length of the entire array.
Satellite-borne Telescopes
Earthbound telescopes are limited for a number of reasons:
1) Light pollution (excess light in the atmosphere)
2) Atmospheric turbulence (mainly due to convection currents and temperature
differences)
3) Absorption of various wavelengths (especially X-rays and UV) by the atmosphere.
This makes observation at these wavelengths impossible.
These problems do not exist for satellite-based telescopes in orbit around the Earth.
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Lens Aberrations
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Aberration:
A single large converging lens is unable to form a perfectly sharp image for two reasons.
Spherical Aberration:
Result:
Method to reduce or eliminate aberration:
use a “stop” to reduce the size of the aperture– an aperture with
a disk with an adjustable hole to block some rays far from the
principal axis
Tradeoff: also reduces amount of light so image is sharper but
fainter
Chromatic Aberration:
Result:
Method to reduce or eliminate aberration:
“chromatic doublet” – a compound lens formed
by adding a second lens that is diverging,
usually of a different material, so that the
dispersion is reduced
Make a sketch to show how the
image of this object would be
distorted due to spherical aberration.
aberration
object
fix
Make a sketch to show how the
image of this object would be
distorted due to chromatic aberration.
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Medical Imaging and Diagnostic Techniques
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Various imaging techniques are used for diagnostic purposes:
1) X-rays: use high energy radiation
2) Computed Tomography (CT): 3D x-ray (also known as a CAT scan)
3) Ultrasound: uses high frequency sound waves
4) Nuclear Magnetic Resonance (NMR): uses magnetic fields (also known as MRI)
5) Lasers
Production of X-rays:
a) Electrons are boiled off a filament due to heating.
b) They are accelerated through a high potential difference.
c) They strike a metal target, often tungsten.
d) The inner shell electrons of the target (tungsten) jump to
high energy level and emit x-ray photons when they relax.
Imaging technique:
a) X-rays pass through person and fall on a photographic film
which darkens when x-rays hit it.
b) Bone absorbs more of x-rays than soft tissue so on the film
tissue appears darker and bone appears lighter.
Attenuation:
a)
b)
Therefore,
Contrast:
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1. X-rays are incident on the muscle of a
patient as shown at right. The intensity
of the transmitted x-rays is measured
and plotted as a function of the
thickness of the muscle.
I0 =
I=
x=
a) What is the relationship between the transmitted intensity and the thickness?
Symbol:
Half-value Thickness:
Units:
b) What is the half-value thickness of the muscle shown above?
Intensity
Equation:
where μ =
Relationship between Half-value thickness and attenuation coefficient
Attenuation coefficient:
Symbol:
Units:
Mass absorption coefficient:
Symbol:
Units:
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1. The graph at right shows the transmitted intensity of a parallel
beam of X-rays versus the thickness of a certain type of tissue.
Determine the:
a) half-value thickness of this tissue
b) attenuation coefficient of this tissue
2. The half-value thickness for soft tissue (muscle) is about 20 cm and for bone is 150
times less than this. Determine the attenuation coefficient for bone and for soft tissue.
3. Compare the half-value thickness and attenuation coefficients for bone and muscle.
4. A parallel beam of X-rays passes through muscle and reduces to one-eighth of its initial intensity.
Determine the fraction of the intensity of this beam when it is transmitted through the same thickness
of fatty tissue. The half-value thickness of muscle is 4.0 mm and the half-value thickness of fatty
tissue is 6.0 mm.
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5. The half-value thickness of a 30 keV X-ray photon in aluminum is 2.4 mm. The initial intensity
of the X-ray beam is 4.0 x 102 kW/m2.
a) What is the beam intensity after passing through 9.6 mm of aluminum?
b) What is the beam intensity after passing through 1.5 mm of aluminum?
6. Analyze the intensity graph at right by straightening it.
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Computed Tomography (CT)
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CT or CAT scan: Computer-Assisted Tomography or Computerized Axial Tomography
Outline the basis of CT scanning
– X-ray image of target taken at different angles (many
different directions)
- computer produces detailed image of slice (these images
are combined using computers to form a two-dimensional
image of section)
– images of many sections/slices can be obtained
– combined to build up a 3D image so image can be rotated
for viewing from any angle
Compare X-ray imaging and CT scans
X-ray imaging
1) uses ionizing radiation (X-rays)
2) short duration of exposure
3) smaller dose of radiation
4) two-dimensional shadow image
CT scanning
1) uses ionizing radiation (X-rays)
2) long duration of exposure (hard for kids)
3) larger dose of radiation
4) three-dimensional image
Why/how are barium meals used to assist X-ray imaging of stomach or intestinal tract?
a) All tissues in the abdominal cavity have approximately the same attenuation coefficient so
there is little to no contrast on photographic film.
b) The attenuation coefficient for barium is greater than for the tissues in the abdominal cavity.
c) Barium meal lines the stomach.
d) Outline of stomach becomes clearer (greater contrast).
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Ultrasound Imaging
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Type of energy used:
Advantage:
What is ultrasound?
Typical operating frequencies of ultrasound waves:
Operating principles: High frequency sound waves are transmitted
from a transducer (a probe) into the patient’s body and are reflected at
each boundary between different types of tissue and bone. The same
probe both transmits and receives the ultrasound waves. By measuring
the time between transmission and reception, the distance to each
boundary can be calculated using the speed of sound and thus the
location and surface of each organ can be mapped.
Production and detection of ultrasound waves
Piezoelectric crystal: a quartz crystal that changes shape when a
potential difference is applied across it
Production: apply an AC voltage to generate a vibration at
desired frequency
Detection: received sound wave causes it to vibrate and generate
an AC voltage that can be measured
Factors affecting choice of diagnostic frequency:
a) Resolution: size of smallest object that can be imaged. Ultrasound is a wave so diffraction
effects must be minimized. Use smallest wavelength possible.
Favors:
b) Attenuation: absorption of signal. Attenuation increases as frequency increases. High
frequency ultrasound will be absorbed and not reflected back to receiver. High
frequency can’t scan organs deep in body – poor depth penetration
Favors:
Compromise:
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1. The time delay for an ultrasound pulse going through body fat to reach and be reflected from the
liver is 0.133 ms. The speed of the ultrasound through fat at the chosen frequency is 1450 m/s.
a) Calculate how far from the probe the liver is.
b) What are some assumptions made in this calculation?
Acoustic Impedance:
Formula:
Where
Units:
Z=
ρ=
c=
Density
(kg m-3)
Speed of sound
(m s-1)
Air (200 C)
1.21
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Muscle
1075
1580
Bone
1900
3780
Fat
900
1480
Soft Tissue (skin)
1060
1540
Material
Acoustic Impedance
(kg m-2 s-1)
2. Complete the table with the acoustic impedance of each of the above substances.
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Reflection at a boundary:
3. From what boundary will most of the energy of the
ultrasound waves be reflected?
4. What is the purpose of putting gel on probe and the patient’s skin?
Ultrasound of intensity Io is traveling in a medium of impedance Z1 and is incident on a medium
of impedance Z2. The intensity of the reflected ultrasound is IR and is given by the formula:
5. Calculate the fraction of ultrasound that would be reflected from a patient’s skin if gel were
not applied to the tip of the ultrasound wand.
Types of ultrasound scans
I. B-scan:
a)
b)
Presentation: Uses the signal strength to affect the brightness of a dot on the screen which can be
displayed as a real-time video. Many B-scans are combined to give an image of the
internal organs or baby.
Use:
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II. A-scan:
a)
b)
Presentation:
Uses:
The diagram above shows an ultrasound transducer (probe) in contact with the skin in an effort to
determine the depth and width of the organ shown. An A-scan of the results is shown also. The
average speed of the ultrasound used in tissue and muscle is 2.0 x 103 m/s.
a) Identify the source of each reflection labeled A, B, C, and D.
b) Calculate the depth of the organ.
c) Calculate the width of the organ.
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NMR and MRI
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NMR:
MRI:
Type of energy used:
1)
2)
Uses:
1) imaging blood flow and soft tissue in the body
2) Preferred for the brain and central nervous system
3) Detecting tumors, strokes, infections in brain, spine, joints
Operating Principles:
a) Large constant uniform magnetic field causes hydrogen atoms to line up (align their spin
axes) – act like tiny magnets
b) Small non-uniform magnetic field is superimposed on top of larger field – localized
magnetic field – weak oscillating field in the form of pulses of radio waves
c) If frequency of radio waves matches that of the hydrogen atoms (resonance) then the
smaller field makes some hydrogen atoms realign
d) When small non-uniform field is removed, atoms relax back to original alignment
e) As they relax they emit radio-waves
f) Time it takes to relax is measured
g) Frequency of emitted radio waves and relaxation times are processed to produce the NMR
image
Advantages:
1) Any situation where detailed tomography/slicing/imaging is required
2) Large scale investigations where dose of ionizing radiation would be too great
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Comparisons of Diagnostic Imaging Techniques
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1. What are the advantages and disadvantages of using
ultrasound instead of X-rays?
Advantage: not as harmful since no ionizing radiation
And so can be used for pregnant women
Disadvantages:
Small depth of penetration
Limit to size of objects that can be imaged
Blurring of images due to reflection at boundaries
2. Why are X-rays preferred over ultrasound for bone fractures?
Nearly all ultrasound is reflected by bone (at bone/tissue
boundary) but x-rays can penetrate bone therefore X-rays
show up internal structures.
3. What are the main advantages of each of the following imaging
techniques?
X-rays:
To detect broken bones because bone and tissue show different
attenuation/good distinction between bones and flesh
Ultrasound:
Any soft tissue analysis – takes advantage of reflections off
organ boundaries
Pre-natal scans because there is no risk from ionizing radiation
NMR:
Any situation where detailed tomography/slicing/imaging is
required
Large scale investigations where dose of ionizing radiation
would be too great
4. Apart from health hazards, why are different means of diagnosis needed?
a)
b)
c)
d)
e)
f)
g)
Different types of tissues and bone have different absorption/attenuation properties
Some are better at distinguishing boundaries of organs
Some provide two dimensional slice imaging – some provide complete three dimensional images
Some are better at monitoring static or dynamic conditions
Some are better to investigate at large or small scales
Some can be used to study concentrations of specific types of tissue or pharmaceuticals
Some are better at monitoring specific organ functions
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