004 - Option C Chapter 15 Imaging PDF
004 - Option C Chapter 15 Imaging PDF
004 - Option C Chapter 15 Imaging PDF
15
Imaging
ESSENTIAL IDEAS
The progress of a wave can be modelled using the ray or the wavefront. The change in
wave speed when moving between media changes the shape of the wave.
Optical microscopes and telescopes utilize similar physical properties of lenses and
mirrors. Analysis of the universe is performed both optically and by using radio
telescopes to investigate different regions of the electromagnetic spectrum.
Total internal reflection allows light or infrared radiation to travel along a transparent
fibre. However, the performance of a fibre can be degraded by dispersion and
attenuation effects.
The body can be imaged using radiation generated from both outside and inside.
Imaging has enabled medical practitioners to improve diagnosis with fewer invasive
procedures.
the progress
of a wave can be modelled via the ray or the wavefront; the change in wave speed
when moving between media changes the shape of the wave
cornea + lens
point image
light spreads in
all directions from
a point object
eye
(not to scale)
The eye uses refraction to bring light rays diverging from a point on an object back to a point on
the image. This process is called focusing the light to form an image.
Additional
Perspectives
15 Imaging
lens
vitreous humour
aqueous
humour
retina
cornea
fovea
pupil
blind spot
iris
ligaments
ciliary muscle
optic nerve
be received by the retina in order to see clearly. The aqueous humour is a watery liquid
between the cornea and the lens; the vitreous humour is a clear gel between the lens and
the retina.
The ciliary muscles can change the shape of the lens; this is how the eye is able to focus on
objects that are at different distances away.
1 If images are not formed on the surface of the retina, the eye will not be able see clearly.
Suggest possible reasons why this may happen.
2 Suggest the purpose of the optic nerve.
a
b
Figure 15.3 How curved
interfaces between transparent
media affect wavefronts and rays
We know from Chapter 4 that when wavefronts enter a different medium and
their speed changes, they can refract and change direction. If the interface
(boundary) between the two media has a curved surface, then the refracted
wavefronts will change shape. Figures 15.3a and 15.3b show plane wavefronts
(parallel rays) crossing an interface into a medium where they travel slower.
Rays showing the direction of travel of the wavefronts are also included (rays are
always drawn perpendicular to wavefronts). In 15.3a the incident waves arrive at
a convex surface and the transmitted wavefronts and rays converge. In 15.3b the
waves are incident on a concave surface and the transmitted wavefronts and rays
diverge.
The eye contains a lens that helps to focus the light. Manufactured lenses made of
transparent materials (such as glass or plastic) use the effect shown in Figure 15.3 to
focus light and form images. This usually involves light travelling from an object through air and
then through a transparent lens that has two smooth, curved surfaces. Refraction then occurs at
both surfaces as shown in Figure 15.4, which shows the effects of the two basic types of lens on
plane wavefronts. The wavefronts inside the lenses have not been included in these diagrams.
Light rays will refract and change direction at both surfaces of the lens, unless they are incident
along a normal. However, in the rest of this chapter we will usually simplify the diagrams in
order to show the rays changing direction only once in the centre of the lens.
converging rays
wavefronts
converging
(convex)
lens
image
formed at
focus
wavefronts
diverging
(concave)
lens
diverging
rays
In Figure 15.4a the wavefronts converge to a focusfor this reason this type of lens is often
called a converging lens. Because of the shape of its surface, this type of lens is also called a
convex lens. Despite their name, converging lenses do not always converge light (magnifying
glasses are the exception). Figure 15.4b shows the action of a diverging lens (concave surface).
Lenses are made in a wide variety of shapes and sizes, but all lenses can be described as either
converging/convex or diverging/concave.
Lenses have been in use for thousands of years in many societies around the world. The
oldest were crafted from naturally occurring translucent rock (see Figure 15.5). They may have
been used for magnification or for starting fires.
15 Imaging
Thin lenses
Although real lenses will not behave exactly as the idealized descriptions and equations
presented in this chapter, lens theory can be applied confidently to thin lenses (which have
surfaces with small curvatures) and for light incident approximately perpendicularly (normally)
close to the middle of such lenses.
Terminology
Figure 15.6 illustrates the basic terms used to describe lenses.
Figure 15.6
Defining the basic
terms used to
describe lenses
focal point, F
F'
principal
axis
focal length, f
b
focal length, f
principal
axis
principal
axis
F'
principal
axis
F'
focal
point
principal
axis
focal
length, f
Figure 15.6 shows ray diagrams, and for the rest of this chapter we will continue to use rays
because they are usually the easiest way of representing the behaviour of optical systems.
However, as an example, Figure 15.7 shows how the behaviour of the converging lenses shown in
Figure 15.6b could be represented using wavefronts.
Figure 15.7
Wavefronts being
focused by a
converging lens
wavefronts
focal point
1
f
15 Imaging
Worked example
1 What are the powers of lenses that have focal lengths of:
a +2.1m
b +15cm
c 50cm?
1
1
=
= +0.48D
f 2.1
1
1
b P = =
= +6.7D
f
0.15
1
1
= 2.0D
c P = =
f 0.5
a P =
Linear magnification, m
The linear magnification, m, of an image is defined as the ratio of the height of the image, hi, to
the height of the object, ho.
hi
ho
m=
Angular magnification, M
Sometimes the dimensions of an object and/or an image are not easily determined, or sometimes
quoting a value for a linear magnification may be unhelpful or misleading. For example, an
image of the Moon that had a diameter of 1m would be impressive, but its linear magnification
would be m = hi/ho = 1/(3.5 106) = 2.9 107. In such cases the concept of angular
magnification becomes useful. See Figure 15.8.
Figure 15.8
The concept of angular
magnification
top
rays from object
bottom
eye
top
bottom
eye
Angular magnification, M, is defined as the angle subtended at the eye by the image, i, divided
by the angle subtended at the eye by the object, o. Because it is a ratio, it has no unit.
M=
i
o
15 Imaging
6 When a magnifying glass was used to look at a small insect it appeared to have a length 3.7mm. If the
linear magnification of the lens was 4.6, what was the real length of the insect?
7 A picture of width 4.0cm and height 2.5cm is projected on to a screen so that it is 83cm wide.
a What is the linear magnification?
b How tall is the image?
c By what factor has the area of the image increased?
8 The angular magnification of a telescope was 12 when it was used to look at a tree 18m tall. If the tree
was 410m away, what angle was subtended by the image of the tree at the eye of the observer?
by converging lenses
The position and properties of an image can be predicted theoretically by using one of two methods:
scale drawing (ray diagrams)
the (thin lens) equation, which links the object and image positions to the focal length of the lens.
(not to scale)
object
F
2F'
F'
2F
image
(not to scale)
F
2F'
F'
2F
F
2F'
F'
2F
image
Object at 2F'
Image is real, same
size as object and inverted
object
2F
F'
2F'
image
2F
F'
2F'
image
Object at F'
Image is at infinity
object
2F'
F
F'
2F
10 15 Imaging
If an object is placed closer to the lens than the focal point, the emerging rays diverge and
cannot form a real image. Used in this way, a lens is acting as simple magnifying glass, and
a virtual, magnified image can be seen by an eye looking through the lens, as shown in
Figure15.12, which will be discussed later in this chapter.
9 a Draw a ray diagram to determine the position and size of the image formed when an object 10mm tall
is placed 8.0cm from a convex lens of focal length 5.0cm.
b What is the linear magnification of the image?
10 a Draw a ray diagram to determine the position and size of the image formed when an object 20cm tall is
placed 1.20m from a convex lens of power 2.0D.
b What is the linear magnification of the image?
11 Construct a ray diagram to determine where an object must be placed in order to project an image of
linear magnification 10 on to a screen that is 2.0m from the lens.
12 An image of an object 2.0cm in height is projected on to a screen that is 80cm away from the object.
Construct a ray diagram to determine the focal length of the lens if the linear magnification is 4.0.
13 a Describe the properties of images that are formed by cameras.
b Draw a sketch to show a camera forming an image of a distant object.
c How can a camera focus objects that are different distances away?
ho
F
F'
image
hi
object distance, u
image distance, v
ToK Link
Conventions
Could sign convention, using the symbols of positive and negative, emotionally influence scientists?
The real is positive convention is used in this course, but there is another widely used alternative (which
is not included). There are other situations in physics were we need to decide on a convention (for
example, current flowing from positive to negative). And the choice of positive charge for protons and
negative for electrons could easily have been the other way around. Provided that everyone understands
the convention that is being used, it is not of great significance which system is used, although through
cultural influences we may subjectively be inclined to wrongly believe that positive is more important
than negative.
Worked example
2 a Use the thin lens formula to calculate the position of the image formed by a converging lens of focal
length 15cm when the object is placed 20cm from the lens.
b What is the linear magnification?
c Is the image upright or inverted?
1 1 1
= +
v u
f
1
1
1
=
+
15 20 v
v = 60cm
60
v
b m = = = 3.0
20
u
c The negative sign confirms that the image is inverted.
a
12 15 Imaging
Additional
Perspectives
hi v
= .
ho u
Comparing the two equations, it is clear that:
But we have already that
v vf
=
u
f
vf = uv uf, or vf + uf = uv
Dividing by uvf, we get:
1 1 1
= +
f v u
The important simplifying assumptions made in this derivation are that:
the ray parallel to the principal axis changes direction in the middle of the lens
the ray passing through the middle of the lens does not deviate because it is incident
normally.
These assumptions are only valid for rays striking a thin lens close to the principal axis.
1 Draw a ray diagram showing the formation of a real image by the refraction of rays at both
surfaces of a converging lens.
Use the thin lens formula to answer the following questions about forming real images with convex lenses.
14 In an experiment investigating the properties of a converging lens, image distances were measured for a
range of different object distances.
a Sketch the shape of a graph that would directly represent the raw data.
b How would you process the data and draw a graph that would enable an accurate determination of the
focal length?
15 a Determine the position of the image when an object is placed 45cm from a converging lens of focal
length 15cm.
b Calculate the linear magnification.
16 a Where must an object be placed to project an image on to a screen 2.0m away from a lens of focal
length 20cm?
b What is the linear magnification?
17 An object is placed 10cm away from a converging lens and forms an image with a linear magnification of
3.5. What is the focal length of the lens?
18 What power lens is needed to produce an image on a screen 12cm away, so that the length of the image
is 10 per cent of the length of the object?
19 a Derive the equation m = f/(u f).
b What focal length of converging lens will produce a magnification of 2 when an object is placed 6.0cm
away from the lens?
hi
virtual image
at near point
object
ho
F'
i
i
u
v=D
Figure 15.12 A simple magnifying glass forming an image at the near point of the eye (not to scale)
The object must be placed closer to the lens than the focal point, so that the rays diverge into
the eye, which then sees an upright virtual image. The image distance v is equal to D, assuming
that the lens is close to the eye.
14 15 Imaging
Worked example
3 A converging lens of focal length 8.0cm is used to magnify an object 2.0mm tall.
a Where must the object be placed to form an image at the near point (v = 25cm)?
b What is the height of the image?
c Is the image upright or inverted?
This question could be answered by drawing a ray diagram, but we will use the thin lens formula.
1 1 1
= +
f v u
1
1
1
=
+ remembering that a virtual image must be given a negative image distance
8.0 25 u
u = 6.1cm
25
v
= 4.1
b m = =
6.1
u
so the height of the image is 4.1 2.0 = 8.2mm
a
But the height of the image cannot be measured directly, so we are usually more concerned
about the angular magnification, M, of a magnifying glass than its linear magnification, m.
M near point =
angle subtended at the eye by the image formed at the near point
angle subtended at the eye by the object placed at the near point
Mnear point = i =
= i
o ho/D ho
Note that this is numerically the same as the linear magnification, m (= v/u), but because
the height of a virtual image is not easily measurable we need to find an alternative method
of calculating the magnification, and it is also desirable to be able to calculate the possible
magnification directly from a knowledge of the focal length of the lens.
Looking at the similar triangles in Figure 15.12 containing the angle i, we see that:
h D
Mnear point = i =
ho u
But we want an equation that gives us the angular magnification in terms of f, not u.
Multiplying the lens equation ( 1 = 1 + 1 ) throughout by v gives us:
f v u
v v v
= +
f v u
Remember that in this situation v = D (the negative sign is included because the image is
virtual), so we get:
D
= 1 Mnear point
f
or
Mnear point =
D
+ 1
f
25
8
+ 1 = 4.1, as before.
Image at infinity
Forming an image at the near point provides the largest possible magnification, but the image
can also be formed at infinity and this allows the eye to be more relaxed. Figure 15.13 shows
that the object must be placed at the focal point.
virtual image
at infinity
ho
F'
f
Figure 15.13 A simple magnifying glass with the image at infinity
Minfinity = i =
o
ho
f
ho
D
D
Minfinity =
f
This equation is given in the Physics data booklet.
By adjusting the distance between the object and the lens, the angular magnification can
be adjusted from D to D + 1, but the lens aberrations (see later) of high-power (small f) lenses
f
f
limit the magnification possible with a single lens. A typical focal length for a magnifying
glass is about 10cm, which will produce an angular magnification between 2.5 and 3.5. More
magnification would require a lens of greater curvature and too many aberrations.
20 a Draw an accurate ray diagram to show the formation of the image when an object is placed 5.0cm
away from a converging lens of focal length 8.0cm.
b Use the diagram to determine the linear magnification.
21 Use the thin lens formula to predict the nature, position and linear magnification of the image formed by a
converging lens of power +20D when it is used to look at an object 4.0cm from the lens.
22 What is the focal length of a converging lens that produces a virtual image of length 5.8cm when viewing
a spider of length 1.8cm placed at a distance of 6.9cm from the lens?
23 a Calculate the angular magnification produced by a converging lens of focal length 12cm when
observing an image at the near point.
b In what direction would the lens need to be moved in order for the image to be moved to infinity and
for the eye to be more relaxed?
c When the lens is adjusted in this way, what happens to the angular magnification?
24 What power lens will produce an angular magnification of 3.0 of an image at infinity?
25 Two small objects that are 0.10mm apart can just be distinguished as separate when they are placed at
the near point. What is the closest they can be together and still be distinguished when a normal human
eye views them using a simple magnifying glass that has a focal length of 8.0cm?
26 a Where must an object be placed for a virtual image to be seen at the near point when using a lens of
focal length 7.5cm?
b Calculate the angular magnification in this position.
by diverging lenses
Because they do not form real images, diverging lenses have fewer uses than converging lenses.
However, ray diagrams and the thin lens equation can be used for them in the same way as for
converging lenses.
16 15 Imaging
Worked example
4 A 2.0cm tall object is placed 6.0cm from a diverging lens of focal length 4.0cm. Determine
the properties of the image by:
a using a ray diagram
b using the lens equation.
a Figure 15.14 shows an image that is virtual, upright, 0.8cm tall and 2.4cm from the centre of the lens.
F
O
F'
Figure 15.14 Virtual upright image formed by a diverging lens (not to scale)
1 1 1
= +
f v u
1
1
1
= +
4.0 v 6.0
12
v =
= 2.4cm
5
The negative sign represents a virtual image.
2.4
v
= +0.40; so the image size = 0.40 2.0 = 0.80cm
=
m =
u
6.0
The positive sign represents an upright image.
b
Combining lenses
If two lenses are used in an optical system, the final image can be predicted by treating the image
formed by the first lens as the object for the second lens. Figure 15.15 shows an example in which
the virtual image formed by the diverging lens in Figure 15.14 is used to form a second, real image
by a converging lens of focal length 3.0cm with its centre 3.1cm from the centre of the diverging
lens. The blue lines are just construction lines used to locate the top of the final image.
Figure 15.15
Combining lenses
(not to scale)
F
O
I2
I1
From a scale drawing we can see that the final image is real and inverted. It is located 6.6cm
from the converging lens and its size is 1.0cm.
v 6.6
= 1.2
=
u
5.5
so:
final image size = 1.2 0.80 = 0.96cm
The negative sign represents an inverted image.
retina
+63 D
4 D
Figure 15.16 Correcting short-sight (the red
lines show the paths that the rays would follow
without the lens)
18 15 Imaging
Figure 15.17 represents spherical aberration. This is the
inability of a lens, which has surfaces that are spherically
shaped, to focus parallel rays that strike the lens at different
distances from the principal axis to the same point.
Spherical aberration results in unwanted blurring and
distortion of images (see Figure 15.18), but in good-quality
lenses the effect is reduced by adjusting the shape of the lens.
However, this cannot completely remove aberration for all
circumstances. The effects can also be reduced by only letting
light rays strike close to the centre of the lens. In photography
the size of the aperture (opening) through which light passes
Figure 15.17 Spherical aberration of monochromatic
before it strikes the lens can be decreased to reduce the effects
light (exaggerated)
of spherical aberration. This is commonly known as stopping
down the lens, but it has the disadvantage of reducing the
amount of light passing into the camera and may also produce
unwanted diffraction effects.
Figure 15.19 represents chromatic aberration. Chromatic aberration is the inability of a lens
to refract parallel rays of light of different colours (wavelengths) to the same focal point. Any
transparent medium has slightly different refractive indices for light of different frequencies, so
that white light may be dispersed into different colours when it is refracted. Typically, chromatic
aberration leads to the blurring of images and gives images red or blue/violet edges.
white
light
object
image
Chromatic aberration can be reduced by combining two or more lenses together. For example, a
converging lens can be combined with a diverging lens (of a different refractive index), so that
the second lens eliminates the chromatic aberration caused by the first (see Figure 15.20).
Figure 15.20
Combining lenses of
different refractive
indices to correct for
chromatic aberration
diverging lens
white
light
converging lens
In the modern world we are surrounded by optical equipment capable of capturing video and
still pictures and the quality of lenses has improved enormously in recent years. The quality of
the images produced by the best modern camera lenses is highly impressive (Figure 15.21) and
the improved detection of low light levels has meant that lenses (in mobile phones for example)
can be very small, so that aberrations are less significant.
Figure 15.21
This lens achieves
top-quality images
by having a large
number of lens
elements
27 a What is the focal length of a diverging lens that will produce an image 8.0cm from its centre when an
object is placed 10.0cm from the lens?
b List the properties of the image.
28 Suggest how the focal length of a diverging lens can be determined experimentally.
29 Two converging lenses with focal lengths of 10cm and 20cm are placed with their centres 30cm apart.
What is the linear magnification produced by this system when an object is placed 75cm from the
midpoint between the two lenses? Does this question have two different answers?
30 Make a copy of Figure 15.19 and show on it where a screen would have to be placed to obtain an image
with blue edges.
31 Draw a diagram(s) to illustrate the improved focusing achieved by stopping down a lens.
32 Suggest why lens aberrations tend to be worse for higher-power lenses.
33 In order to reduce chromatic aberration a converging lens of power +25D was combined with a diverging
lens of power 12D. What is the focal length of this combination?
converging
(concave) mirror
centre of
curvature
mirrors
normals
principal
axis
focal
length, f
radius of curvature
diverging
(convex) mirror
centre of
curvature
F
focal
length, f
radius of curvature
Figure 15.22 Reflection by spherical surfaces
principal
axis
20 15 Imaging
O
F
C
I
O
C
I
O
C
image at
infinity
I
O
C
Figure 15.23 shows us that as the object gets closer to the lens, the real inverted image gets
larger and further away from the lens. But if the object is closer than the focal point the image is
virtual, upright and magnified.
Linear magnification, m =
hi
v
=
(as with lenses)
u
ho
i
(as with lenses)
o
Angular magnification, M =
Worked example
5 When a 3.2cm tall object is placed 5.1cm from a converging mirror, a magnified virtual image is formed
9.7cm from the mirror.
a What is the linear magnification?
b How tall is the image?
v 9.7
=
= 1.9
u 5.1
hi
b m =
ho
h
1.9 = i
3.2
hi = 3.2 1.9 = 6.1cm
a m =
Diverging mirrors
Figure 15.24 show the formation of a diminished, upright, virtual image by a diverging mirror.
This can be very useful when we need to see a wide field of vision, Figure 15.25 shows one
application a cars wing mirror.
small, upright
virtual image
Mirror combinations
FA
FA
mirror B
FB
IA
IB
FB
CB
22 15 Imaging
IA then provides the object that produces the (still) inverted, virtual image, IB, when the rays
reflect off the diverging mirror, B. (The blue line is a construction line used to determine the
position of the top of the image.)
The same idea can be used in reverse. If a point source of light (or other radiation) is placed
at, or near, the focus of a parabolic reflector, the emerging beam will be parallel, or have a low
divergence. The beams from a torch, car headlight or spotlight (Figure 15.29) are good examples
of beams with only small divergence, which are produced by parabolic reflecting surfaces.
Figure 15.29 The low
divergence beam from
a spotlight
34 Draw a ray diagram to determine the properties of the image formed when an object 1.5cm tall is placed
7.0cm from a converging mirror of focal length 5cm.
35 a Draw a ray diagram of a diverging lens forming an image of an object placed between the mirror and its
focal point.
b Describe the properties of the image.
36 a A make-up/shaving mirror uses a curved mirror. Describe the image seen.
b What kind of mirror is used, and typically how far away would a face be when using such a mirror?
c Suggest a suitable focal length for such a mirror.
37 Draw a ray diagram to locate the final image formed by the following optical arrangement. An object is
placed 20cm away from a large converging mirror of focal length 8cm; the image formed is located 4cm
in front of a small converging mirror of focal length 5cm. The two mirrors face each other.
38 Draw ray diagrams to represent:
a spherical aberration in a diverging mirror
b the production of the light beam from a car headlight.
optical
microscopes and telescopes utilize similar physical properties of lenses and mirrors;
analysis of the universe is performed both optically and by using radio telescopes to
investigate different regions of the electromagnetic spectrum
In this section we will look at how lenses and/or mirrors can be combined to produce optical
images better than can be seen by the eye or by the use of a single lens. Similar ideas can then
be applied to the use of other parts of the electromagnetic spectrum for imaging, in particular
the use of radio waves in astronomy. Extending the range of human senses in these ways has
contributed enormously to our expanding knowledge of both the microscopic world and the rest
of the universe.
eyepiece
lens
real, magnified
image produced
by the objective
object
Fo
Fe
Fo
final inverted,
magnified,
virtual image
D
Figure 15.30 Compound microscope with the final image at the near point (normal adjustment)
construction line
Fe
24 15 Imaging
The object to be viewed under the microscope is placed just beyond the focal
point of the objective lens, so that a real image is formed between the two lenses
with a high magnification. The eyepiece lens is then used as a magnifying glass,
and its position is adjusted to give as large an image as possible with the final
virtual image usually at, or very close to, the near point of the eye. This is called
normal adjustment.
To locate the image by drawing you need to find the point where the
construction line through the centre of the eyepiece from the top of the first
image meets the extension of the ray from the first image that passes through
the focal point.
A model of a simple compound microscope can be investigated in a darkened
room as shown in Figure 15.31. To begin with, a converging lens with a focal
length of about 5cm is used to form an inverted image of a brightly illuminated
object (e.g. graph paper) on a translucent screen. Then, the position of a second,
less-powerful lens is adjusted until a second (virtual) image of the first image is
seen when looking through this eyepiece. The screen can then be removed and
the two lenses used together to observe the scale, so that the magnification of
the image can be estimated. As with many optical experiments, keeping the eye
and all the components aligned is important for success.
eyepiece lens
translucent
screen
bright
light
objective lens
Angular magnification
The angular magnification produced by a compound microscope is equal to the product of
the linear magnification of the objective lens multiplied by the angular magnification of the
eyepiece lens. For an image at the near point:
Moverall = mobjective Meyepiece =
D
v
+1
f
u objective
eyepiece
This equation is not given in the Physics data booklet. If the final image is at infinity (for less eye
strain) the +1 term can be omitted.
Worked example
6 A compound microscope contains an objective lens of focal length 0.80cm and an eyepiece lens of focal
length 5.4cm. The microscope is adjusted to form a final image at the near point of the eye for an object
placed 0.92cm in front of the objective.
a Determine the distance between the two lenses.
b What is the angular magnification of the image?
a First find the image distance for the objective:
1
1
1
=
+
f
v
u
1
1
=
+ 1
0.80
v
0.92
v = 6.1cm
b M =
( uv )
objective
( ) (
M = 6.1 25
0.92
5.4
D
f
+1
eyepiece
) = 37
+1
The exact angular magnification of a microscope clearly depends on where the object and
final image are located, but an approximate guide to the angular magnification of a compound
microscope can be obtained from the focal lengths and the distance between the lenses, L:
M DL . (This equation predicts M 45 for Worked example 6.) This confirms that shorter
ff
oe
focal lengths will produce higher magnifications; but, as with magnifying glasses, the higher
curvatures associated with higher of shorter focal lengths will introduce lens aberrations and
reduce the quality of the images produced.
Resolution
Although the magnification produced by a microscope is obviously important, resolution is
usually of more significance in optical instruments. In general, high resolution can be described
as the ability to see detail in an image. Two images that can be seen as separate are said to be
resolved. To understand the difference, consider a picture on a phone or computer screen it can
be very easy to magnify, but that often results in a poorer-quality image.
A high-quality microscope, or telescope, will produce images with good magnification and
resolution; a poorer-quality instrument may produce high magnification but the resolution will
be disappointing.
The magnification of an optical system is largely dependent on the focal lengths of the lenses,
but the resolution of a system depends on the quality of the lenses, the diameter of the objective
and the wavelength of the radiation being detected. High overall resolution also assumes that the
properties of the surface detecting the waves, for example the separation of
pixels in a camera or the separation of cells on the retina of the eye, will not
angular resolution
have an adverse effect.
two objects
A normal human eye can see two similar objects placed at the near point
that can
as separate if they are approximately at least 0.1mm apart. Resolution is best
just be
0.1
resolved
represented by the angle subtended by these points, = 250 4 104rad
Figure 15.32 Angular resolution
(see Figure 15.32). Better resolution is represented by a smaller angle between
of the eye
two points that can just be seen as separate.
Using a good microscope with an angular magnification of, say, 50 could improve the
resolution by the same factor, so that it would then be possible to separate two points subtending
4
an angle of (4 10 ) = 8 106rad, which corresponds to a linear separation of 2 103mm at
50
the near point.
Rayleighs criterion
The diffraction of waves as they pass into the eye, or an optical instrument, is the main factor
limiting resolution, and the amount of diffraction depends on the wavelength, , and the width
of the aperture, b (Chapter 4). Rayleighs criterion is a guide to resolution for waves passing
through circular apertures (the theory was discussed in Chapter 9, but is not needed here):
Two objects are considered to be resolvable if the angle, , that they subtend at the eye or
optical instrument is bigger than 1.22/b.
Worked example
7 Use Rayleighs criterion to estimate the angular resolution of the human eye.
Assuming the average wavelength of light is 6 10 7m and the diameter of the pupil (in bright light) is
2mm:
6 10 7
1.22
=
4 10 4 rad
= 1.22
b
2 10 3
Which is in reasonable agreement with actual observations.
Applying Rayleighs criterion to the resolution achieved by optical instruments, we can see that
resolution would be improved by using shorter wavelengths and wider apertures. Using wider
apertures has the added advantage of admitting more light and producing brighter images,
although larger lenses may have aberration problems. Using light of smaller wavelengths (the
26 15 Imaging
blue/violet end of the spectrum) can improve resolution but, of course, any coloured effects
would be lost.
Rayleighs criterion can be used as a guide to resolution, but there are other factors
involved for example placing an oil with a high refractive index between the specimen and
the eyepiece can improve resolution.
Utilizations
Electron microscopes
The resolution of a microscope will be improved if radiation with a shorter wavelength than
light can be used to examine the object. Waves of the electromagnetic spectrum with shorter
wavelengths than light are ultraviolet, X-rays and gamma rays, but none of these are as easily
produced, controlled and detected as a beam of electrons.
Like all particles, electrons have wave properties but, because their mass is so small, electron wave
properties are relatively easily observed; electrons in a beam have typical wavelengths of about
1010m. This is about 5000 times smaller than the average wavelength
of visible light, so that resolution can be improved by the same factor
by using a beam of electrons rather than a beam of light photons.
Beams of electrons can be produced by accelerating potential
differences of a few thousand volts and their wavelength can be
adjusted by changing the p.d. Because electrons are charged, they can
be focused by electric or magnetic fields.
Of course, electrons cannot be seen using our eyes, so their
energy needs to be converted to light to form a visible image (see
Figure15.33).
Figure 15.33
1 F igure 15.33 shows a living bed-bug. Use the internet to find out if
placing such an organism in an electron beam is harmful.
39 An object is placed 5.0mm from the objective lens of a two-lens compound microscope. The eyepiece of
the microscope has a focal length of 4.0cm.
a If the linear magnification produced by the objective lens is 5.0, what is its focal length?
b What is the overall angular magnification of the microscope when observing an image of this object at
infinity?
40 a If the diameter of the objective lens in a microscope is 1.2cm, estimate its angular resolution assuming
an average wavelength of light of 5.5 10 7m.
b If the microscope produces an angular magnification of 80, estimate the minimum separation of two
points that can just be resolved by a normal human eye.
c What assumption did you make in answering (b)?
41 Suggest why placing a transparent oil between a specimen and an objective lens can improve the
resolution of a microscope.
scale
bright
light
objective lens
translucent
screen
eyepiece
lens
objective
lens
eyepiece
lens
fo
parallel rays
all from top of
distant object
fe
construction line
Fe
Fo
h1
i
i
virtual image
at infinity
Figure 15.35 Simple refracting telescope with the final image at infinity (normal adjustment)
The angular magnification (in this adjustment) can be determined by examining the two
triangles involving h1:
h1
i
fe
M=
=
o
h1
fo
M=
fo
fe
28
15 Imaging
The quality and diameter of the objective lens are the most important factors when considering
the quality of the image in any kind of optical instrument. A larger objective has two
advantages:
Most importantly, it collects more light to produce a brighter image (to see dimmer and
more distant or smaller objects).
There will be less diffraction, which improves the resolution of images.
However, larger objectives also have aberration problems, which inevitably reduce the quality
of the images.
42 A student hopes to construct an astronomical refracting telescope that produces an angular magnification
of at least 100.
a If she chooses an objective lens of focal length 68 cm, what is the minimum power needed for the
eyepiece?
b Explain why this telescope may not produce high-quality images.
43 Venus has a diameter of about 12 000 km.
a What angle does it subtend at the eye when it is a distance of 2.0 108 km from Earth? (Assume that all
of Venus is visible.)
b A refracting telescope with an objective lens of focal length 120 cm is used to observe Venus. What is
the diameter of the image formed by this lens?
c An eyepiece lens of focal length 1.5 cm is used to form a final virtual image of Venus at infinity. What is
the angle subtended by the image at the eyepiece?
d Use your answers to (a) and (c) to confirm that the overall angular magnification of the telescope is
given by M = fo/fe.
44 A refracting telescope consists of lenses of focal length 86 cm and 2.1 cm.
a Which lens is the eyepiece?
b Calculate the angular magnification in normal adjustment.
c If the objective lens was replaced with another having twice the diameter but the same focal length,
how would the image change?
45 Suggest how the use of a third lens in a refracting telescope can result in an upright image.
ToK Link
Can we trust our own senses?
However advanced the technology, microscopes and telescopes always involve sense perception.Can technology
be used effectively to extend or correct our senses?
Our eyes collect and focus light, and then electrical signals are sent along the optic nerve to our brains.
The brain processes the information and the result is what we call an image, and we would describe this as
real (seeing is believing). But our ability to observe is well known to be fallible, and simple optical illusions
demonstrate how easily we can be fooled. In Figure 15.36 our eyes tell us that squares A and B are different
shades, but scientific measurement would correctly inform us that they are the same.
Figure 15.36 Squares A and B are exactly the same shade of grey!
incident
light
plane mirror
converging
objective mirror
Worked example
8 Two stars separated by a distance, so, subtend an angle, o, of 5.3 10 5rad when viewed from Earth. The
stars are observed through a Newtonian telescope with a converging mirror of focal length 3.4m.
a Calculate the separation, si, of these two stars in the image formed by the converging mirror. (Assume
they are the same distance from Earth.)
b If the eyepiece has a focal length, fe, of 4.5cm and is used to form a final image at infinity, what is the
overall angular magnification produced by the telescope?
a
si
v
=
so u
The incident light rays are (almost) parallel, so the image is formed at the focal point, and therefore v = f.
s
si = o f
u
s
But o = o , so:
u
si = o f = 5.3 10 5 3.4
= 1.8 10 4m
m=
b The image from the mirror must be formed at the focal point of the eyepiece lens for the final
image to be at infinity.
s
1.8 10 4
angle subtended by image, i = i =
fe
0.045
= 4.0 103rad
4.0 10 3
= 75
M= i =
o 5.3 10 5
f
This could be found directly from M = o .
fe
Figure 15.38 A
reflecting telescope
with a Cassegrain
mounting
eyepiece
diverging mirror
The Cassegrain mounting (designed by Laurent Cassegrain in 1672) is an alternative design that
reflects the rays off a second (diverging) mirror back through a hole in the objective mirror see Figure
15.38. This arrangement enables the user to look in the same direction as that from which the light is
coming. Using a diverging mirror in this way produces extra magnification in a compact design.
30 15 Imaging
Worked example
9 Consider Figure 15.38. The rays converging to the small diverging mirror would otherwise have formed
an image 24cm behind the mirror, but have been reflected to a focus 1.36m away near the hole in the
converging objective mirror. What extra magnification has this introduced compared to the use of a
plane mirror in a Newtonian mounting with an objective mirror and eyepiece of similar focal lengths?
v 1.36
m= =
= 5.6
u 0.24
Reflecting telescopes are still popular today and they have some important advantages over
refracting telescopes, including:
The light does not have to pass through a refracting medium, so there is no chromatic aberration.
The light does not have to pass through a refracting medium, so there is no absorption or
scattering.
The objective has only one active surface so high-quality, larger-diameter objectives of the
right shape are easier and cheaper to produce.
For these reasons the majority of optical astronomical telescopes used for research are reflectors.
They are also popular with amateur astronomers. Figure 15.39 show the supporting structure of
the worlds largest reflecting telescope, Gran Telescopio Canarias. Figure 15.40 shows a smaller
reflecting telescope for individual use.
46 Determine the angular magnification of a Newtonian reflecting telescope that has a converging mirror of
focal length 6.7m and an eyepiece lens of focal length 1.8cm. Assume that the final image is at infinity.
47 Increasing resolution and light-gathering ability is achieved by using larger mirrors. Explain why telescopes
cannot be improved by simply making them bigger and bigger.
48 Use the internet to find out some of the reasons why an amateur astronomer would choose one of the
two basic types of reflecting telescope described in this section (rather than the other).
Satellite-borne telescopes
Terrestrial telescopes (those on the Earths surface) receive waves that have passed through the
Earths atmosphere. However, the atmosphere reflects, scatters, refracts and absorbs some of the
incoming radiation, and these effects can significantly affect the quality of images formed by
terrestrial telescopes. Examples of the effects of the atmosphere on radiation include stars viewed
from the Earths surface twinkling because of the constantly changing effects of refraction, and
clear skies appearing blue because shorter wavelengths of light scatter more from the atmosphere
than longer wavelengths.
Optical astronomical telescopes also have particular problems they can only be used at
night, they can only be used if there are no clouds, and they are affected by light from the Earth
being scattered by the atmosphere after dark (light pollution).
The effects of the atmosphere are very dependent on the wavelengths of the radiation
involved, as shown in Figure 15.41.
100
50
0
1010
108
106
104
102
102
Wavelength, /m
An obvious way of reducing the effects of the atmosphere is to place telescopes in locations that
are at high altitudes, with good weather conditions and which are far from towns and cities. See
Figure 15.42.
With the considerable improvements in satellite technology in recent years, it has become
possible to place significant numbers of space telescopes on satellites (satellite-borne) in orbit
around the Earth. This has produced an enormous amount of data (collected and analysed using
high-power computers), new discoveries of less intense or more distant sources, or those emitting
different kinds of radiation, and impressive high-resolution images.
The Hubble telescope is probably the most well-known orbiting
telescope, with many of its spectacular images well known and
freely available around the world. The telescope was launched in
1990 and was named after the famous American astronomer Edwin
Hubble. It has a mass of about 11tonnes and orbits approximately
560km above the Earths surface, taking 96minutes for one
complete orbit. One of the greatest achievements of astronomers
using the Hubble telescope has been accurately determining the
distances to very distant stars, enabling a much improved estimate
for the age of the universe.
Because such projects are expensive and the data obtained of
interest to astronomers worldwide, they are often joint ventures between
countries.
Utilizations
32 15 Imaging
The purpose of the structures was to measure time and the apparent motions of the planets
and stars, but also to be impressive structures in themselves and to stimulate interest in the
newly developing science of astronomy. In India at that time, astronomy and astrology were
closely connected, as they had been throughout the world in nearly all civilizations (and even
today for many people).
1 Many people believe that the positions of the Moon, stars and planets can influence our
individual lives and our futures. Do you think that this is possible? Explain your answer.
Non-optical telescopes
Nature of Science Technological advances in astronomy
The invention of the optical telescope happened over 500 years ago; there is no general
agreement about who was responsible although a German spectacle maker, Hans Lippershay,
is often credited. Certainly Galileo adapted and improved early designs and his observations
of moons orbiting Jupiter are well known. This was presented as evidence that the Earth may
not be the centre of the universe and is an early example of the dramatic advances in human
knowledge that can achieved by using instruments to extend our observations.
For most of the subsequent 500 years, astronomy has relied on the detection of visible light
to provide information, but radiation from all other parts of the electromagnetic spectrum
also arrive at the top of the Earths atmosphere from space. Telescopes and sensors capable
of detecting infrared, ultraviolet and X-rays have now been placed on orbiting satellites, and
the data obtained leads to knowledge about the universe that can be very different from that
obtained from light alone. For example, new sources of radiation have been discovered (for
example gamma ray bursts and X-ray binary stars) and our knowledge of how the universe began
has improved considerably.
Radio astronomy, in particular, is a highly advanced technology.
Radio telescopes
Radio waves are emitted by a wide variety of sources in space and they are mostly unaffected by
passing through the Earths atmosphere (see Figure 15.41), so radio telescopes can be terrestrial
and, unlike visible light telescopes, they can be used 24 hours a day.
Some sources have been discovered from their radio wave emissions because they do not
emit significant visible light, but radio waves are also emitted as part of the spectrum of elements
(for example hydrogen, the most common element in the universe, emits a significant radio
wavelength of 21cm). In this way radio astronomy has helped to map the universe.
We know that a guide to the angular resolution of a telescope may be determined from 1.22
,
b
where b is the width of the aperture/dish. The wavelengths to be investigated are predetermined
by the nature of the investigation, and because radio wavelengths are much longer than light
wavelengths, good resolution is much more difficult to achieve.
The most significant factor controlling the resolution of a single-dish radio telescope is the width
of the dish. Larger dishes will produce higher resolution but, unfortunately, larger dishes are also
much more expensive; it is also more difficult to maintain their precise shape and more difficult
to steer them to the desired direction. It should also be stressed that larger dishes will collect more
energy, so that more distant and dimmer sources can be detected. The largest single-dish radio
telescope in the world is at the Arecibo Observatory in Puerto Rico. It has a diameter of 305m
this is only possible because the surrounding landscape has been used to help support the structure.
Worked example
10 a Determine the resolution of the Parkes telescope (Figure 15.44) if it was used to detect the 21cm
wavelength of hydrogen.
b Would this telescope be able to resolve two stars emitting radio waves that were 2.3 1013km apart
and (both) 6.7 1015km from Earth?
a
1.22
0.21
= 1.22
b
64
= 4.0 10 3
34 15 Imaging
There will be a path difference between a wavefront emitted from an astronomical source
arriving at different telescopes (Figure 15.45). When the signals are combined, an interference
pattern (Chapters 4 and 9) will be produced, the spacing and centre of which can be used to
accurately determine the direction to the source of radiation.
Figure 15.45
Radio interferometry
telescopes
baseline
signals
combined
Resolution is improved by using more telescopes in a regular pattern (compare with the use of
diffraction gratings, discussed in Chapter 9) and there are a number of different ways in which
a pattern of telescopes might be arranged, however the details are not needed for this course.
Receiving enough energy from distant and dim sources is always an important issue, so the total
combined collecting area of the telescopes must still be as large as possible (which is another
reason why more dishes are preferable).
The individual telescopes may be arranged in an array relatively close together, see Figure
15.46, or they may be separated by a long distance (a long baseline), which can be as much as
thousands of kilometres long and in different countries, although longer distances introduce
technological problems.
Figure 15.46 An
interferometer array
Worked example
11 Compare the theoretical resolution of two radio telescopes of dish diameter 50m separated by
a distance of 1km with one of the telescopes used individually.
The effective aperture has been increased by a factor of 1000/50 = 20. So the angular resolution of the
two together is 20 times smaller (better). The combination will also have double the receiving area and
will therefore be able to detect less-intense sources.
49 Suggest some reasons why the weightless and airless environment of a satellite orbit has advantages for
reflecting telescopes.
50 Make a list of the advantages of placing telescopes on orbiting satellites.
51 When detecting radio waves of frequency 1420MHz, what is the minimum diameter of a single-dish
reflecting telescope needed to resolve two stars that are 4.7 1016m apart and 1.5 1019m from Earth?
52 What angular resolution would be obtained from an interferometer using two radio telescopes 540m
apart using radio waves of wavelength 0.18m?
53 An angular resolution of 1 arcsecond is obtained using interferometry techniques. What separation of two
telescopes would be needed to achieve this value when receiving radio waves with a frequency of 6GHz?
54 Use the internet to research the latest developments in the SKA project.
55 Suggest what problems may arise when using interferometry techniques between telescopes that are
thousands of kilometres apart.
15.3 (C3: Core) Fibre optics total internal reflection allows light
In Chapter 4 we introduced the ideas of critical angle and total internal reflection, and briefly
discussed how these could be used in, for example, optic fibre endoscopes to obtain images from
inside the human body. In this section we will discuss optic fibres in more detail, with respect to
their advantages in the transmission of data.
36 15 Imaging
Figure 15.47 Fourcore twisted pair
cabling
An alternative for reducing electromagnetic noise is the use of co-axial cable, which contains
a central copper wire surrounded by an insulator and then an outer copper mesh see
Figure15.48.
Figure 15.48
Co-axial cable
Figure 15.50 The
critical angle (v1 < v 2;
n1 > n2)
Medium 1:
optically denser
wave speed slower, v1
larger refractive index, n1
1 = c
2 = 90
Medium 2:
optically less dense
wave speed faster, v2
smaller refractive index, n2
1
sin c
1
sinc
1
= 0.65
sin
c=
1.54
c = 40
n1
1
, with n2 = 1.47
c n =
sinc
2
n1 =
1.47
= 0.95
1.54
c = 73
sin
c=
Nature of Science
Digital communication
Modern electronic communication is digital in nature; this means that rather than
communicating using voltages or light/infrared intensities, which vary continuously (analogue
signals), the data are transmitted as a very large number of pulses, each of which is intended to
have only one of two possible levels (commonly called 0 and 1, or low and high). Figure 15.51
38 15 Imaging
0
Time
represents a signal of only eight binary digits (bits), commonly called one byte. (The term
binary describes a number in which each digit can only have one of two possible values.)
This kind of digital signal can be sent along a cable as a series of voltage pulses, or pulses of
light/infrared.
56 Discuss the factors that affect the choice between using electromagnetic waves in cables or
electromagnetic waves in air for communicating data.
57 Suggest why electromagnetic noise (interference) is often a bigger problem at higher frequencies.
58 Explain, with the help of diagrams, why angles of incidence inside optic fibres will always be large if the
fibres are thin. Assume that the original signal is transmitted parallel to the axis of the optic fibre.
59 A typical optic fibre has a refractive index of 1.62.
a What is the critical angle for such a fibre if it is surrounded by air?
b What is the critical angle for such a fibre if it is surrounded by glass of refractive index 1.51?
c Signals travel slower in glass of higher refractive index. Discuss whether or not this is a significant factor
in choosing the type of glass used in an optic fibre.
60 Suggest why digital signals are used in preference to analogue signals for transferring data.
61 Suppose the digital signals shown in Figure 15.51 were transferred over a long distance and, as a result,
the powers of the pulses were halved and the pulse times were doubled. Assuming that the pulses remain
rectangular:
a How would their energy have changed?
b Sketch a graph of the received eight-bit signal.
Figure 15.53
Dispersion causes
pulses to overlap
long cable
signal
received
Intensity
Intensity
signal
transmitted
Time
Time
Waveguide dispersion
If all the rays of light or infrared radiation that is transmitted along an optic fibre are parallel to
begin with, they will follow parallel paths and take exactly the same time to reach any particular
point along the cable. But this is an idealized situation and is not possible in practice. Rays
representing a particular pulse can take slightly different paths and, therefore, different times to
reach their destination. Figure 15.54 illustrates this problem (but it is exaggerated for clarity). This
causes the spreading of pulses and the kind of dispersion
known as waveguide dispersion, which is sometimes also
cladding
known as modal dispersion, although this term will not be
core
rays
rays
exit at
enter
used in IB examinations. (An optic fibre is an example of a
different
at same
waveguide, which is a term used for any structure designed to
times
time
transfer waves along a particular route.)
To reduce the effects of waveguide dispersion it is
cladding
better to use very thin fibres and to try to ensure that the
Figure 15.54 Rays taking different paths and causing
light rays are parallel, and also to use graded-index fibres, as
waveguide dispersion
described next.
cladding
core
a
refractive
index
40 15 Imaging
Figure 15.55b represents a graded-index fibre in which the refractive index of the core optic
fibre increases gradually towards a maximum at the centre. The effect of this is to relatively
decrease the speed of the rays passing along the most direct (central) routes and relatively
increase the speed of rays that pass closer to the outer surfaces of the core. The overall effect of
having gradual changes in refractive index is to produce more central, curved paths, with less
time differences between them, as shown in Figure 15.56. This reduces waveguide dispersion.
Figure 15.56
Typical paths of rays
in a graded-index
fibre
ray paths
core
cladding
Material dispersion
Material dispersion is the name given to dispersion caused by the use of different wavelengths.
In this respect it is a problem similar to chromatic aberration in lenses.
We know from Chapter 4 that the refractive index of a medium depends on the wavelength
of the radiation. This effect is deliberately used in a prism to disperse light into a spectrum. For
example, red light travels faster in glass (than the other visible colours), so that it has a slightly
higher refractive index and it is the colour least deviated by passing through a prism. If different
wavelengths (representing the same pulse) travel along the same path through an optic fibre, they
will take slightly different times to reach their destination and will therefore produce dispersion.
The obvious solution to material dispersion is to use monochromatic radiation. Infrared LEDs
are the most common source.
62 a Explain why infrared radiation, which is normally internally reflected, could pass between optic fibres if
they came in contact with each other.
b Explain how the use of cladding will prevent this problem.
63 If the refractive indices of the cladding and the core in a stepped-index fibre are 1.60 and 1.55,
respectively, what is the critical angle in the core?
64 Explain why dispersion limits the rate at which digital data can be communicated over longer distances.
65 Summarize, without reference to a diagram, how the use of graded-index fibres reduces waveguide
dispersion.
66 Different data can be sent along the same optic fibre using different wavelengths of radiation. Discuss
whether or not material dispersion affects this process.
Utilizations
Figure 15.57
An underwater optic
fibre cable
Attenuation
Attenuation is the gradual loss of intensity of a signal as it passes through a material.
intensity,
I0
The high quality of the glass in optic fibres means that absorption and scattering should not be
too significant over short distances, but they become important whenever long distances are
involved. Scattering is the main cause of attenuation in optic fibres, but dispersion also affects
attenuation. Consider again Figure 15.53. Even in the idealized example of no absorption or no
scattering (so that the total power received is the same as transmitted, which is shown by equal
areas) the intensity has still been reduced.
It may be assumed that signal intensity is reduced by equal factors in
intensity,
I
equal distances, which means that intensity varies exponentially with
distance. This means that we can quote a value for the attenuation per
unit length of a system.
attenuation = log I
I0
Attenuation calculated in this way is given the unit bel, but a smaller unit, the decibel (dB) is
usually preferred:
attenuation (dB) = 10log
I
I0
This equation is given in the Physics data booklet. It can also be applied to powers:
Attenuation (dB) = 10log
P
P0
Because I < I0, the attenuation in dB will be a negative number. We will not be using a symbol
to represent attenuation in this topic.
42 15 Imaging
Worked example
13 The attenuation in an optic fibre of length 100km is 53dB. If the input power is 0.0028W, what power
would be received:
a 100km away?
b 200km away?
a attenuation (dB) = 10 log
53 = 10log
P
(0.0028
)
P
P0
P
0.0028
P = 1.4 10 8W
P
b 5.01 10 6 =
1.4 10 8
P = 7.0 10 14W
5.01 10 6 =
It is common practice to quote attenuation per unit length, for example in dBkm1. Table15.1
gives some examples, but remember that this is just a rough guide because the values are also
frequency dependent. (These numbers are sometimes called attenuation coefficients.)
Table 15.1 Typical
attenuations per unit
length
Type of cable
twisted pairs (1MHz)
co-axial (200MHz)
optic fibre (1014Hz)
Attenuation (dBkm1)
50
100
1
Whatever kind of cable is used to communicate over long distances, attenuation will result in
the signal intensity falling below an acceptable level unless the pulses can be amplified (and
reshaped). The devices that perform this task are called repeaters or regenerators.
67 Sketch a graph to show how the intensity of a signal varies with distance as it travels along an optic fibre.
68 a What is the overall attenuation of a cable if the intensity of a signal is halved by passing through it?
b If a cable has an attenuation loss of 10dB, by what percentage is the intensity of the output lower than
the input?
69 The attenuation in a cable is rated at 0.36dB for each 100m. If the input power is 6.8mW, what length
of cable will reduce the power to 5.0 10 10W?
70 The input power to a very long optic cable is 0.15W. When the power falls below a certain value (P) the
signal needs to be amplified/regenerated. Determine the value of P if the minimum distance between
repeaters is 80km and the cable has an attenuation loss of 1.8dBkm1.
71 a Use the internet to determine the infrared frequency most commonly used in optical fibre
communication.
b Why is this particular frequency chosen?
72 Suggest why attenuation in an optic fibre is frequency dependent.
73 The advantages of transferring data using glass rather than copper seem considerable. Use the internet to
research possible reasons why copper wiring is still in widespread use.
the
body can be imaged using radiation generated from both outside and inside; imaging
has enabled medical practitioners to improve diagnosis with fewer invasive procedures
In recent years the rapid growth of computing power and technological advances have resulted
in a dramatic increase in the number of utilizations of physics in medicine around the world.
Nuclear medicine was mentioned in Chapter 12 and there are many applications of lasers, but
in this section we will discuss the various physics principles that can be used to obtain images of
bones, organs and tissues located inside the human body.
We begin with the use of X-rays and ultrasound, both of which involve sending penetrating
waves into the body from outside, then we will discuss nuclear magnetic resonance.
X-rays
X-rays are useful in medical imaging because they are penetrating and some can pass deep
into the body and out of the other side. However, some of the X-ray photons are absorbed, so
when the transmitted X-rays are detected on the other side of the body, a shadow or negative
picture can be produced. Figures 15.59a and 15.59b show how the bones have absorbed
more of the X-rays than the rest of the hand. In this simple example, the X-rays are detected
photographically (like light) by transferring the photons energy to produce chemical changes
in the film.
This technique is still widely used around the
world. It is relatively inexpensive but the image
must be chemically developed, which means
there is a delay before the image is available to
be viewed. Detection by CCD (charge-coupled
devices) produces an immediate digital image
and allows more control over the whole imaging
and data-handling process. Equally as important,
a detection process that requires a lower intensity
will enable hospitals to use lower-power X-rays
and reduce the health hazards associated with
the use of X-rays.
photographic film in
light-proof envelope
Figure 15.59 (a) Arrangement for X-raying a hand; (b) an X-ray of a hand
44 15 Imaging
Figure 15.60 shows the process of having a dental X-ray with the CCD in the patients mouth.
Figure 15.60
Having a dental X-ray
taken
Nature of Science
Risk analysis
The benefits of using X-rays to diagnose illnesses are substantial and obvious. However X-rays are
also a potential health risk because the energy carried by X-ray photons is high enough to cause
ionization and possibly to cause damaging chemical and biological changes in the body (in a
similar way to gamma ray photons, as discussed in Chapter 7).
The risk associated with directing a known amount of a particular kind of radiation into
a particular patient cannot be known with certainty. Controlled experiments that involve
exposing people (or animals) to radiation are clearly not acceptable, so the medical profession
can only deal with statistics gathered indirectly from numerous previous events (medical or
otherwise) in which people have been exposed to known, or estimated, amounts of radiation.
Such data has been repeatedly analysed very extensively in order to assess the risk (the
probability of harm) involved with any particular course of action.
Doctors must balance the risks of exposing a patient to X-rays against the medical benefits to
be gained from a diagnosis or detailed knowledge of the medical problem that they need to treat.
The health of medical staff involved with the use of X-rays in hospitals (radiographers) also needs
to be considered. Standard safety measures include:
using X-rays of as low a power as is consistent with their intended purpose
using X-rays for as short a time as possible
monitoring and limiting the number of X-rays taken of a patient
preventing X-rays from going anywhere else other than the part of the body they are being
used to examine.
The improvement in the technology involved with the production and, particularly, the
detection of X-rays has been so considerable in recent years that the risks are now very well
understood, well controlled and minimal. The required dose of radiation for any particular
purpose is now so much reduced that a long trip in an aircraft (at high altitude) now involves a
greater exposure to ionizing radiation than most X-rays.
Attenuation of X-rays
The amount of absorption depends on the energy carried by the X-ray photons and the type of
material through which they are passing. We will now describe this in more detail, using the
concept of attenuation, which we have already covered in our discussion of optic fibres.
X-rays of higher frequency have higher energy (E = hf) and are more penetrating. They are
often described as having the quality of hard X-rays and they are produced in X-ray machines
that use higher voltages. Soft X-rays, with lower photon energies, are more easily absorbed.
( II )
1
2
1
8
1
16
1
32
Fraction transmitted
1
4
1 2 3 4 5 6 7
Half-value thicknesses
Half-value thickness, x, is defined as the thickness of a medium that will reduce the
transmitted intensity to half its previous value.
Intensity (%)
50
25
12.5
3.1 6.25
0
2x
3x
4x
5x
Thickness, x
46 15 Imaging
X-ray photons with higher energy will be more penetrating, so their half-value thickness of a
particular medium will also be larger.
Worked example
14 a When directed through a homogeneous material, the intensity of an X-ray beam falls to 18 of its initial
value over a distance of 15cm. What is its half-value thickness?
1
b What overall thickness would be needed to reduce the intensity to 16
?
c How would the half-value thickness of this material change if X-rays of longer wavelength were used?
1
or:
( )
( )
x = ln2
This equation is given in the Physics data booklet.
Attenuation coefficients are often called absorption coefficients although, as we have seen,
absorption is not the only cause of attenuation.
Worked examples
15 An X-ray beam is reduced to 93 per cent of its intensity when it passes through a medium of thickness
0.48mm.
a What is the linear attenuation coefficient of the medium?
b What is its half-value thickness?
( )
( ) (
I
a = 1 ln 0
x
I
I0
1
= 0.48 ln
0.93Io
= 0.15mm1
b x = ln2
0.693 = 4.6mm
x =
0.151
16 An X-ray beam of intensity 580mWm2 is passed through parallel layers of two different materials. The
first layer is 4.3cm thick and has a linear attenuation coefficient of 0.89cm1; the second layer is 1.3cm
thick with a linear attenuation coefficient of 0.27cm1.
a Determine the intensity of the X-rays that cross the boundary between the layers.
b What intensity emerges from the second layer?
a I = Ioe x
ln
I = ln580 (0.89 4.3)
I = 13mWm2
b I = Ioe x
ln
I = ln13 (0.27 1.3)
I = 9.1mWm2
Figure 15.63
Mass attenuation
coefficient of water:
variation with X-ray
photon energy
density
103
102
101
100
101
102
103
102
101
100
101
102
Photon energy/MeV
48 15 Imaging
Worked example
17 Use Figure 15.63 to estimate the linear attenuation coefficient of water for photons of energy 100keV.
For 0.1MeV photons, the mass absorption coefficient is approximately 2 10 1cm2g1.
Mass attenuation coefficient = 2 10 1 = , and the density of water is 1.0gcm3.
= 2 10 1 1.0 = 0.2cm1
16 cm
m = 0.12 cm1
thigh
B
100%
15%
X-rays
100%
46%
2%
m = 0.62
cm1
1%
femur
5 cm
18 cm
74 A parallel X-ray beam was reduced in intensity from 100Wm2 to 74Wm2 after passing through 3.0cm
of a medium. Calculate the linear attenuation coefficient in cm1.
75 A parallel beam of X-rays passes though a material with a half-value thickness of 3.7cm. If the incident
beam has an intensity of 150Wm2, calculate:
a the linear attenuation coefficient
b the percentage of the incident intensity that emerges from a 4.5cm thickness of the material.
76 The intensity of a parallel X-ray beam is reduced from 195Wm2 to 127Wm2 when it passes through
a medium of thickness 2.10mm.
a What is the total power of the incident beam if it covers an area of 16.0cm by 20.5cm?
b If the wavelength is 2.27 10 11m, how many photons enter the medium every second?
c Calculate the linear attenuation coefficient.
d What is the half-value thickness of the medium?
e If the accelerating voltage producing the X-rays is increased, suggest how the half-value thickness will
change.
77 A medium has a density of 1.9gcm3 and a linear attenuation coefficient of 0.22mm1.
a What thickness of this medium will reduce a parallel X-ray beam to 33 per cent of its incident intensity?
b What is the mass attenuation coefficient of the medium?
78 Explain why it is reasonable to expect that increasing the thicknesses of a medium by equal amounts will
result in equal percentage falls in the transmitted intensities of X-rays.
79 A material has a density of 1.3gcm3 and a mass attenuation coefficient of 2.1cm2g1. Calculate its
half-value thickness.
80 If the linear attenuation coefficient for certain X-rays in soft tissue is quoted at 0.35mm1, what is the
linear attenuation coefficient for bone (with the same X-rays) if bone has a half-value thickness that is 150
times larger than for soft tissue?
81 Confirm the values quoted in Figure 15.64 for the relative intensities of the X-rays passing through the
thigh.
82 Data showing the relative intensity for X-rays passing through a certain material is shown in Table 15.2.
Thickness (cm)
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Relative intensity
1.00
0.78
0.58
0.48
0.33
0.27
0.20
0.17
point
source
P
sharp shadow
extended
source
P
blurred shadow
50 15 Imaging
To increase the brightness of an image formed on
photographic film, an intensifying screen(s) can be used
to enhance the image. These screens contain fluorescent
materials that emit light when struck by X-rays. Photographic
film is much more sensitive to light than X-rays, so the image
patient
is intensified. Figure 15.66 shows a possible arrangement, with
the film placed between two intensifying screens.
oscillating
If X-rays scattered from all parts of the body can reach the
grid
film, the contrast of the image will be reduced because all parts
intensifying
of the film will receive more X-rays than they would if only X-rays
screens
film
travelling directly from the source reached the film. This effect
can be reduced by using a collimating grid as shown in Figure
Figure 15.66 X-ray arrangement with the film placed
15.66. A collimator makes all rays passing through it parallel to
between two intensifying screens
each other. The grid has to be oscillated from side to side during
the exposure to enable all relevant parts of the patient to be imaged.
For some applications, the contrast of an X-ray image can also be enhanced by temporarily
introducing into the patients body something that will affect the absorption of X-rays. For the
digestive system this could be a (non-poisonous!) substance that the patient has to drink. For CT
scans (see below) a contrast medium is sometimes injected into the bloodstream.
The sharpness of an image can also be digitally enhanced later by a computer program.
X-ray source
Figure 15.68
CT scan of head
Figure 15.69
CT scans of slices
within a head
84 Make a list of the advantages of using electronic detectors rather than detecting X-rays photographically.
85 Explain why the detection of scattered X-rays reduces the contrast of an X-ray image.
86 An X-ray imaging system was redesigned to increase the distance between the source and the patient.
a Suggest a reason why this was done.
b If the same source is used, why will the intensity reaching the patient be reduced?
87 Find out what is meant by a barium meal.
88 List two advantage and two disadvantages of CT scans compared to conventional X-rays.
52 15 Imaging
Ultrasound imaging
High-frequency sound waves can be used as an alternative to electromagnetic waves (X-rays) for
obtaining images from inside the body and diagnosing illnesses. Any sound wave that has a frequency
higher than that which can be heard by humans (20kHz) is called ultrasound and is described as
ultrasonic. Frequencies used in ultrasonic imaging are typically between 2MHz and 20MHz.
Unlike X-rays, ultrasound is not an ionizing radiation and has
no known health risk. The equipment is also generally cheaper and
easier to use, however the images produced do not usually have the
same detail (high resolution) as those produced by CT scans because
the longer wavelengths of ultrasound reduce resolution because of
greater diffraction. Ultrasound techniques can produce better images
of some soft tissues than X-rays, however.
The ultrasound waves are usually directed into the patient using a
handheld transducer (often called a probe) which converts electrical
signals into ultrasound waves. Reflections are received back at the
same device (Figure 15.70). In general, some waves will be reflected
whenever they arrive at any boundary between two different media.
This will be discussed in more detail later, but first we will look at
Figure 15.70 Abdominal ultrasound scan
how the ultrasonic waves are produced and detected.
(B-scan)
co-axial cable
acoustic insulator
metal case
backing material
piezoelectric crystal
electrodes
plastic membrane
When reflected ultrasound waves are incident on the same crystal, they can be detected by the
alternating p.d. induced, which has the same frequency as the waves.
The ultrasound waves cannot be emitted and reflected continuously, because then there
would be no easy way of knowing which waves caused which reflections. For this reason, the
ultrasound is transmitted in short pulses, and the time between them should be longer than the
longest time it could take a reflection to be received back at the transducer. Typically the time
between pulses is about 1 104s, which means that the pulses of ultrasound have a frequency
of about 10kHz (pulse repetition frequency). Remember that the ultrasound waves themselves
have a much higher frequency of about 104kHz.
Each pulse might typically contains two waves or more, so that a typical pulse duration is
2 1/f = 2 1/(1 107) = 2 107s see Figure 15.72. This means that the time between pulses
is about 500 times longer than the duration of each pulse. Longer pulse durations improve the
resolution of images.
reflected waves must be
received during this time
Intensity
Time
pulse
duration
Worked examples
18 Ultrasound waves travelling at an average speed of 1600ms1 through a person's body reflect off an
organ and are received back at the probe after 32s.
a What is the distance of the surface of the organ beneath the skin?
b What assumption was made in this calculation?
a distance 2 = speed time = 1600 32 10 6
19 To examine structures relatively far from the surface of the body, the pulse repetition frequency may
need to be reduced to, for example, 2kHz in order to increase the time between pulses.
a What total distance can an ultrasound wave travelling at an average speed of 1580ms1 travel in the
time between pulses (assume that the pulse duration is negligible)?
b Estimate the number of pulses that could be in the body at any one instant.
a distance = speed time = 1580 (1/2000) = 0.79m (79cm)
b The maximum distance a (reflected) wave could travel is approximately twice the width of the body,
which might be about 70cm (depending on orientation). Under those circumstance there would only
be one pulse in the body at any time. That is, the reflected pulse would be received before the next
pulse was emitted.
Acoustic impedance
In general terms, acoustic impedance is a measure of the opposition of a medium to the flow of
sound through it. Knowledge of acoustic impedance is needed to understand ultrasound imaging
because the reflection of ultrasound waves from boundaries between media depends on how
their acoustic impedances compare.
The bigger the difference in impedances, the higher the percentage of incident waves that are
reflected.
54 15 Imaging
Acoustic impedance, Z, can be defined as:
acoustic impedance = density of substance speed of sound in that substance
Or, in symbols:
Z = c
This equation is given in the Physics data booklet.
The units of impedance are kgm2s1. Table 15.3 provides a list of some acoustic impedances
relevant to ultrasound imaging at a typical frequency. (Acoustic impedance is a frequencydependent property.)
Table 15.3 Acoustic properties of parts of the human body
Medium
air
fat
water
soft tissue
kidney
liver
blood
muscle
skin
bone
Speed, v/ms1
340
1460
1480
1500
1040
1060
1575
1580
1730
4080
Density, /kgm3
1.2
950
1000
1050
1560
1570
1057
1080
1150
Acoustic impedance,
Z/10 6kgm2s1
0.000408
1.39
1.48
1.58
1.62
1.66
1.66
1.71
1.99
7.79
Ultrasound imaging clearly depends on the reflection of the waves from various boundaries,
but one place where reflection is definitely not required is at the point where the waves enter
the patients body. From Table 15.3 we can see clearly that the acoustic impedance of air is
very much lower than that of skin, which means that the percentage reflected from the skin
would be very high if there was an air gap. Therefore, the transducer must be in good contact
with the skin and this is helped by the use of a gel (coupling medium) between them. The gel
has an acoustic impedance similar to that of skin. This is an example of a process known as
impedance matching.
The acoustic impedance of steel is about 45 106kgm2s1. When this is compared with air
(0.000408 106kgm2s1), it is clear why sound waves in air reflect off steel well.
Worked example
20 a The speed of ultrasound waves in a particular part of a patients body is 1580ms1. If the tissue has a
density of 1050kgm3, what is its acoustic impedance?
b Use Table 15.3 to determine an average density for bone.
c Apart from air, which pair of media in the table have the highest percentage reflection?
a Z = c = 1050 1580 = 1.66 106kgm2s1
b Z = c
= 1910kgm3
c Bone and fat, because their acoustic impedances have the greatest difference.
89 The acoustic impedance of a certain material is 2.08 106kgm2s1. What is the speed of sound waves in
this material if its density is 1250kgm3?
90 A pulse contains three complete waves that have a frequency of 2.0MHz. They travel at a speed of
1510ms1.
a What is the duration of the pulse?
b If the pulse repetition frequency is 8.4kHz, how far can the waves in a pulse travel before the next pulse
is emitted?
c Are these frequencies suitable for the examination of a part of the body that is 10cm below the surface
of the skin?
91 The ratio of reflected intensity, Ir, to incident energy, Io, at a boundary between two media of acoustic
impedances Z1 and Z2 is given by the following equation (which is not required in this course):
Ir (Z1 Z2)2
=
Io (Z + Z )2
1
2
a Show that this equation predicts that all of the incident energy is transmitted if the two media have
equal impedances.
b What percentage of the intensity is reflected if one medium has twice the impedance of the other?
Does your answer depend on the direction in which the waves are travelling?
c If ultrasound of intensity 0.1000Wcm2 is incident on the boundary between soft tissue and liver, what
intensity is transmitted into the liver?
d Estimate the percentage of intensity transmitted into skin when ultrasonic waves are incident on it from
air. Hence, explain why gels are used with ultrasound transducers.
92 Why would you expect X-ray imaging to produce better resolution than ultrasonic imaging?
gel
Ultrasound
intensity
probe
probe
The simplest type of ultrasound scan is an A-scan. At each boundary between different
media (for example, fatmuscle or musclebone) some of the waves are reflected and
some are transmitted. The reflected waves are
abdomen
received by the piezoelectric transducer and a p.d.
wall
is induced that can be displayed as a p.d.time
graph. Figure 15.73 shows a typical example.
organ
The amplitudes (intensities) of the reflected
bone
pulses received at the transducer depend on the
distance that the waves have travelled, the type
of boundary from which they have been reflected
and the number of other media boundaries that the
waves have crossed. It is called an A-scan because
it displays information in the form of varying
amplitudes.
A-scans are useful for obtaining accurate
measurements of a known situation. By moving the
transducer to different locations the exact position,
size and shape of an organ can be determined. There
are a wide range of applications of this technology
outside of medicine for example in the detection of
Time
faults in pipes and railway lines.
Worked example
21 Consider Figure 15.73. Calculate the width, s, of the organ if the time delay between reflections from the
two boundaries is 73s.
2s = vt = 1040 73 10 6 (assume a wave speed of 1040ms 1)
s = 0.038m (3.8cm)
56 15 Imaging
B-scans are more widely used than A-scans in hospitals. These display the information in the
form of varying brightness in a two-dimensional, real-time video image such as that shown
in Figure 15.70 (A-scans are one-dimensional). They are called B-scans because they display
information in terms of brightness.
The information is obtained in essentially the same way as in an A-scan, except that the
amplitude of the reflected wave is represented by the brightness of a dot on a screen. The picture
is constructed by a computer program using the information from one or more transducers inside
the ultrasound probe, transmitting waves in a slightly different direction while the probe is
moved to different positions.
Attenuation/dB cm1
The range of frequencies used in ultrasound imaging has already been mentioned (between
2MHz and 20MHz), but why are those frequencies used? The choice is affected by two wave
properties, which are frequency dependent in opposite senses diffraction and attenuation.
1 It is important that the ultrasound beam emerging from the transducer is parallel and can be
directed towards a particular location on the patients body. This means that there should not
be much diffraction of the emerging beam. Therefore, the wavelength needs to be significantly
less than the aperture on the transducer from which it is transmitted, because the amount of
diffraction depends on the ratio /b, where b is the width of the structure causing the diffraction.
For similar reasons, for good resolution the wavelength also needs to be much smaller than the
parts of the body being examined (Chapter 9).
So, in order to reduce the possible effects of
muscle
10
diffraction and to improve resolution, a high
frequency (short wavelength) is preferable.
8
Because the dimensions of the transducer
and parts of the body may be typically a few
6
liver
millimetres or more, the chosen wavelength needs
4
to be shorter than approximately 1mm, which
corresponds to a minimum frequency of about
2
2MHz (using v = 1600ms1), although higher
frequencies will be preferable in this respect.
2
4
6
8
10
2 It is also important that as little of the
Ultrasound frequency/MHz
ultrasound energy is absorbed as possible.
Figure 15.74 How attenuation of a
Because attenuation increases with frequency,
parallel ultrasound beam varies with
lower frequencies are preferable in this respect.
frequency in two parts of the body
See Figure 15.74, which is a simplification but
(simplified)
broadly represents the situation.
For any particular examination, the ultrasound frequency used depends on the acoustic
impedances of the part(s) of the body being scanned. The duration of the pulse is another factor
affecting the resolution of the image (longer pulses are preferred). The pulse repetition frequency
may also need to be adjusted for parts of the body that are at different distances from the surface
of the skin. The ultrasound technician, or the doctor, may need to make adjustments depending
on the particular circumstances.
Disadvantages:
Poor resolution because of the relatively long wavelengths used.
Ultrasonic waves do not transmit well through bone.
Cannot be used effectively with spaces that contain gas (such as the lungs and stomach)
because the waves are strongly reflected at the boundaries.
Utilizations
1 Sketch the structure of the eye seen in Figure 15.76 and label the different parts (research
any information you may need).
ToK Link
We often only see what we expect to see
Its not what you look at that matters, its what you see. Henry
David Thoreau. To what extent do you agree with this comment on
the impact of factors such as expectation on perception?
An untrained observer looking at images that represent the inside of
the human body often find it difficult to interpret the true meaning
of the picture they are looking at. Doctors need to be carefully
trained in these skills. But we all tend to see in any image what we
expect to see because the brain does not have the time to fully
evaluate all the information that is available, and it may jump to
(sometimes incorrect) conclusions based on previous observations.
Figure 15.77 provides a simple example. Of course, when we are
alerted to the fact that an image requires special scrutiny, we give it
much more attention than it would otherwise receive.
58 15 Imaging
93 Consider Figure 15.73.
a Explain how it is possible for the reflected waves that have travelled the longest distance to have the
largest amplitude.
b Explain why the time scale is not regular.
c If the width of the organ was 2.2cm and the wave speed in it was 1030ms1, what was the time
difference between the second and third reflected pulses?
d Determine the average acoustic impedance of the organ if its density was 1540kgm3.
94 a Explain why the use of a gel with an ultrasound probe can be considered as an example of impedance
matching.
b Suggest a suitable value for the acoustic impedance of the gel.
95 Consider Figure 15.74. Estimate the percentage of the incident ultrasound intensity that is transmitted
after passing through 2cm of liver if the frequency is:
a 4MHz
b 8MHz.
96 a Suggest a reason why ultrasound may be of little use in the diagnosis of problems with the brain.
b Give two reasons why ultrasound is used in preference to X-rays for pre-natal scanning.
97 Find another widespread use of ultrasound for medical diagnosis. Why is ultrasound used for this (rather
than other imaging techniques)?
98 Suggest why ultrasound is not used for imaging lungs.
99 Research the use of Doppler ultrasound for diagnosing some heart conditions.
In NMR, single spinning protons in the nuclei of hydrogen atoms are made to precess when
the patient is placed in a very strong uniform magnetic field (typically between 1T and 3T).
This is known as the primary magnetic field. Hydrogen atoms are found throughout the body,
especially in water molecules. Such magnetic fields maybe be 50000 times stronger than the
Earths magnetic field, but they are not known to cause any harmful effects (although there are
a few well-understood exceptions). Magnetic fields of this strength can only be produced using
very high electric currents circulating in coils of wire. This requires that the coils are at very low
temperatures, so that they can become superconducting.
Use of RF signals
The spinning hydrogen nuclei can be made to precess together, in phase, by excitation from a resonant
radio-frequency (RF) magnetic field. The RF causes resonance because it has the same frequency
as the Larmor frequency, which will have a value somewhere within the radio wave section of the
electromagnetic spectrum, typically about 60MHz depending on the strength of the magnetic field.
Because the nuclear magnetic fields of the protons are now precessing in phase with each
other, they will generate a rotating magnetic field strong enough to be detected as an oscillating
voltage by coils placed around the patient.
After the excitation, the spins relax back to their original distribution at a characteristic
rate called the relaxation rate. This rate varies with the type of tissue, so that determination
of relaxation times leads to information about the type of tissue which, in turn, leads to more
detailed and better-quality images.
100 a Suggest reasons why some people may find having an MRI scan an unpleasant experience.
b Why would an MRI scan not be used to diagnose a broken arm.
c Use the internet to research whether having an MRI scan has any adverse health effects.
101 Explain what is meant by the Larmor frequency.
102 Why are radio frequencies used to excite hydrogen atoms in MRI?
103 Explain what is meant by the term relaxation time and why it provides useful information in MRI.
104 a What are the essential differences between a CT scan and an MRI scan?
b Why are powerful computers essential for both of these procedures?
c Make a list of the advantages and disadvantages of these two types of scan.
60 15 Imaging
Summary of knowledge
15.1 Introduction to imaging
When light rays/wavefronts spreading out from a point object are incident on a lens that is
thicker in the middle than at its edges, the rays will be refracted and converged to form a real
image at the point where the rays cross (unless the object is at the focal point, or nearer to
the lens). This kind of lens is called a converging (convex) lens.
If the rays are incident on a lens that is thinner in the middle than at its edges, the rays will
be refracted and diverged and a virtual image can be seen when looking through the lens at
the point from which the rays appear to have come. This kind of lens is called a diverging
(concave) lens.
The principal axis of a lens is defined as the imaginary straight line passing through the
centre of the lens, which is perpendicular to its surfaces.
The focal point of a lens is defined as the point through which all rays parallel to the principal
axis converge after passing through the lens (or the point from which they appear to diverge).
The focal length of a lens is defined as the distance between the centre of the lens and the focal
point. Its value depends on the refractive index of the material and the curvature of its surfaces.
The optical power of a lens is defined as 1/focal length; P = 1/f. Optical power is measured in
dioptres, D. Power (D) = 1/focal length (metres).
The paths of three rays from the top of any extended object, which pass through a lens and
then go to the top of the image, can be predicted. Using these rays, diagrams can be drawn
to determine the position and nature of the image formed when objects are placed at various
distances from a lens.
In diagrams and calculations throughout this topic we have assumed that the lens is thin and
that the rays are close to the principal axis. If this is not true, the image will not be formed
exactly where predicted and the focus/image will not be as well defined.
Real images are formed where rays actually cross. Virtual images are formed when rays
diverge into the eye the image is formed where the rays appear to have come from.
The linear magnification of an image is the ratio of the height of the image, hi, divided by
the height of the object, ho. m = hi/ho = v/u.
The angular magnification of an image is the angle subtended at the eye by the image
divided by the angle subtended at the eye by the object. M = i/o. When referring to optical
instruments it is common to refer to angular magnifications rather than linear magnifications.
If the object is placed further away from a converging lens than the focal point, the image
formed will always be real and inverted.
The thin lens formula is 1/f = 1/v + 1/u. This formula can be used to determine the position
and nature of an image. When using this formula it is important to remember that a distance
to a virtual image and the focal length of a diverging lens are always negative. A positive
magnification indicates that the image is upright; a negative magnification indicates that the
image is inverted (m = v/u).
If the object is placed at the focal point, or closer to a converging lens, the image will be
magnified, upright and virtual. Used in this way the lens is described as a simple magnifying glass.
The nearest point to the human eye at which an object can be clearly focused (without
straining) is called the near point. It is accepted to be 25cm from a normal eye and often
given the symbol D. The furthest point from the human eye that an object can be clearly
focused (without straining) is called the far point for a normal eye it is at infinity.
The angular magnification, M, of a simple magnifying glass varies between D/f, for the image
at infinity, to (D/f) + 1 for the image at the near point.
Summary of knowledge 61
Lens aberrations (especially with higher-power lenses) are the principal limitations on the
magnification achievable by optical instruments that use lenses.
Spherical aberration produces distorted images. It is the inability of a lens having spherical
surfaces to bring all rays incident on it (from a point object) to the same focus. It may be
reduced by adapting the shape of the lens, or by only using the centre of the lens.
Chromatic aberration is the inability of a lens to bring rays of different colours (from a
point object) to the same focus. It occurs because refractive index varies slightly with colour
(wavelength). It can be reduced by combining lenses of different shapes and refractive indices.
Diagrams can be drawn to represent these aberrations and how they can be reduced.
Mirrors with curved surfaces can also be used to focus images. The terminology and the
principles involved are very similar to those concerning lenses. Curved mirrors can have
spherical aberration problems.
The objective lens of a compound microscope forms a real magnified image of an object that
is placed just beyond its focal point. The eyepiece then acts as a magnifying glass to produce
a final image, which is inverted, magnified and virtual.
Ray diagrams can be constructed to represent a microscope in normal adjustment with
the final image at (or near to) the near point. Angular magnification equals the linear
magnification of objective multiplied by the angular magnification of the eyepiece.
Resolution is often more important than magnification in optical instruments. Good
resolution can be considered to be the ability to see points as being separate. Magnifying an
image can improve resolution, but not if the resolution is already poor.
In general, resolution is improved by using better quality lenses, large apertures and small
wavelengths (minimising diffraction effects). Large apertures also have the advantage of
collecting more light and producing brighter images.
Two objects are considered to be just resolvable if the angle, , that they subtend at the eye or
optical instrument is larger than 1.22/b (Rayleighs criterion), where b is the diameter of the
receiving aperture.
The objective lens of a telescope forms a diminished, real and inverted image of a distant
object at its focal point. The eyepiece acts as a magnifying glass to produce a final image
at infinity (in normal adjustment), which is inverted, diminished and virtual. The linear
magnification is less than one, but the telescope produces an angular magnification, M = fo/fe.
Ray diagrams can be constructed to represent a telescope in normal adjustment when the
distance between the lenses is fo + fe.
Reflecting telescopes use converging mirrors as their objectives (rather than converging lenses).
Newtonian mountings use plane mirrors to reflect light into an eyepiece lens at the side.
Cassegrain mountings use diverging mirrors to increase magnification and enable the
observer to look through the telescope directly towards an object.
Optical astronomical telescopes on the Earths surface receive light that has been affected by
passing through the Earths atmosphere. Some radiation is absorbed or scattered, and some is
refracted irregularly. Placing satellites on orbiting satellites above the atmosphere overcomes
these limitations.
Radio waves (including microwaves) are much less affected by the atmosphere (than light) and
radio telescopes can be terrestrial. Many astronomical objects emit radio waves. The simplest
radio telescopes have an aerial placed at the focal point of a single parabolic dish reflector.
As with other waves, resolution is limited to angles larger than 1.22/b. Because radio waves
from space might have a typical wavelength of about 1m, good resolution using a single dish
can only be achieved if it has a large diameter.
Higher resolution when receiving radio waves is possible using interferometry techniques,
in which the signals from two or more synchronized telescopes are combined electronically
62 15 Imaging
and made to interfere. The spacing and centre of the interference pattern can be used to
accurately determine the direction to the source of radiation. The maximum resolution
achieved with two telescopes can be calculated using 1.22/b, with b equal to their
separation. Using many telescopes in an array can improve resolution further.
Most data are sent along cables using either electrical pulses in copper wires, or infrared
pulses in optic fibres.
Data are sent using digital pulses, rather than continuously varying analogue signals. Digital
data are transferred as a very large number of pulses, each of which can have only one of two
possible levels (commonly called 0 or 1).
As pulses travel along a cable they attenuate and disperse. Attenuation is the gradual loss of
intensity of a signal as it passes through a material. Dispersion is the broadening of the width
of a pulse and the associated decrease in intensity.
These effects limit the distance that data can be transferred (before they need to be
amplified and reformed) and the amount of data that can be sent in a given time through a
particular cable.
These effects are significantly lower with infrared pulses in optic fibres than they are with
electrical pulses in copper wire.
Electrical pulses also produce changing electromagnetic fields that can spread away from the
cable and cause interference by inducing tiny e.m.f.s in other cables. These random unwanted
signals are often called electronic noise. Signals in optic fibres do not have this problem.
Electronic noise can be significantly reduced in copper cabling by using twisted pairs or
co-axial cables.
Data are transferred in digital form because when pulses are affected by dispersion,
attenuation and noise, they can still usually be distinguished as 0s or 1s because there only
two distinct levels. (Whereas analogue signals may become too distorted.)
Optic fibres use the effect known as total internal reflection (Chapter 4). The radiation is
reflected internally because the angle of incidence inside the fibre is always larger than the
critical angle. In general n1/n2 = 1/sinc or, if air is the external medium: n = 1/sinc.
The core optic fibre(s) are protected from damage by cladding. The refractive index of the
cladding material must be less than that of the core. The cladding also prevents different
fibres from coming in contact with each other (crosstalk).
Dispersion in optic fibres has two main causes waveguide dispersion and material
dispersion.
Waveguide dispersion is due to the fact that different rays (that started together) travel
along slightly different paths. This problem can be limited by using graded-index fibres, in
which the refractive index increases progressively towards the centre. This has the effect of
confining rays to curved paths close to the centre of the fibre.
Step-index fibres have cores of constant refractive index.
Material dispersion can occur if radiation of different wavelengths is used. This is because
they travel at different speeds (so they have slightly different refractive indices). This can be
overcome by using monochromatic light (from a laser or infrared LED).
The intensity of a signal confined to an optic fibre decreases exponentially with distance
along the cable. If the intensity decreases from I0 to I, then the attenuation (in dB) is
10log10I/I0. It is usual to quote an attenuation per unit length of cable (e.g. 1.5dBkm1). A
similar equation can be used for power instead of intensity.
The decibel (dB) scale is a logarithmic scale commonly used to compare an intensity (or
power) to a reference level, especially where there are large differences involved.
Compared with twisted pair and co-axial cables, optic fibres have lower attenuation, greater
data transfer rates, do not produce noise, are more secure and are smaller and lighter.
Summary of knowledge 63
When an X-ray beam is directed at a human body some of the radiation will be absorbed
and scattered in the body and some will be transmitted directly, so that some X-rays can
be detected on the other side of the body. It is this variation that makes X-rays so useful in
medical imaging.
Different parts of the body will absorb X-rays by different amounts and the intensity of the
detected beam will show variations representing the presence of parts of the body with
different densities and absorption rates.
The X-rays that are transmitted can be detected either photographically or by the use of
CCDs (charge-coupled devices, as used in digital cameras). The use of CCDs allows the
electronic storage and manipulation of images.
The intensity, I, of a parallel beam of X-rays (not spreading out) decreases exponentially
with distance, x, due to absorption and scattering: I = Ioex. is a constant called the
linear attenuation coefficient. It represents the amount of attenuation per unit length in a
particular medium (for radiation of a specified wavelength). The usual unit is cm1.
Attenuation can also be represented in the same way as for attenuation in an optic fibre:
attenuation (dB) = 10log(I1/I0).
Absorption due to the photoelectric effect is the principal means of attenuation of X-rays
and it is largely dependent on the proton number, Z, of the atoms present. For example,
bone contains elements with a higher average proton number than soft tissue, and therefore
absorbs a higher percentage of X-rays.
The attenuation of X-rays is often characterized by the half-value thickness of a particular
medium, x, which is defined as the thickness of a medium that will reduce the transmitted
intensity to half its previous value.
Linear attenuation coefficient and half-value thickness are inversely related: x = ln2.
The mass attenuation coefficient is used to compare the attenuation in unit masses of
different materials: mass attenuation coefficient = linear attenuation coefficient/density =
/. The usual unit is cm2g1.
Attenuation of X-rays is greater for lower frequencies (photons of lower energy). Such beams are
produced by lower voltages and are often called soft X-rays. Hard X-rays are more penetrating.
High-quality X-ray images should have high intensity and contrast. The edges of different
areas of the image should be sharp and well resolved. But at the same time it is important, for
safety reasons, that the power of X-rays used should be as low as possible.
Techniques for improving the quality of the images include: using a small source that is not
too close to the patient; placing an oscillating collimating grid between the patient and the
detector; using intensifying screens containing fluorescent materials.
Computed tomography (CT) uses computer-controlled X-rays and machinery to obtain sharp
images of planes of the patient (scans) with good resolution. These scans can be combined to
present a three-dimensional image.
There is a health risk associated with all uses of X-rays. Ultrasound imaging has no known risk,
but the images have disappointing resolution because of the relatively long wavelengths used.
Ultrasound waves (sound with frequencies higher than can be heard by humans) are directed
into a patients body and reflect back off boundaries between different media.
Acoustic impedance, Z, is a measure of the opposition of a medium to the flow of sound
through it. Z = c, where is the density of the medium and c is the speed of the wave in the
medium. The units of acoustic impedance are kgm2s1.
The bigger the difference in impedances, the higher the percentage of incident waves that
are reflected at a boundary between two media.
Ultrasound waves are produced using the piezoelectric effect, in which an alternating voltage
applied across a crystal transducer makes it vibrate at the same frequency, sending waves into
64 15 Imaging
the surroundings. When reflected waves are received back at the probe, oscillating voltages
are produced and detected.
The transducer (probe) is placed next to the skin of the patient with a gel between them
(eliminating air). The gel has an acoustic impedance chosen to transmit the waves into the
body efficiently.
The ultrasound waves are transmitted in pulses, with sufficient time between the pulses for
the reflected waves to be clearly detected. Resolution is improved by having several complete
ultrasound waves in each pulse.
The simplest types of ultrasound scans are known as A-scans (amplitude scans). The
amplitude of the waves reflected from different boundaries in the patients body are displayed
as an amplitude time graph. Information from the graph can be used to determine the
position and size of various parts of the body.
B-scans are widely used in hospitals. The information is obtained in essentially the same
way as in an A-scan, except that the amplitude of the reflected wave is represented by
the brightness of a dot on a screen. A two-dimensional real-time video image picture is
constructed by a computer programme using the information from one or more transducers
inside the ultrasound probe, transmitting waves in a slightly different direction, often while
the probe is moved to different positions.
Higher ultrasound frequencies (smaller wavelengths) have less diffraction so that the beams
are more directional and the images have better resolution. However, higher frequencies also
undergo more attenuation. The frequency being used may need to be changed depending on
the particular circumstances.
Despite its poor resolution, ultrasound imaging provides a quick, safe, economical and mobile
way of examining inside the body, especially when soft tissues are involved. Ultrasound cannot
penetrate into bone effectively and cannot be used for spaces that contain air (e.g. lungs).
Nuclear magnetic resonance (NMR) provides an alternative to CT scans for providing
images of sections through the body. Medical applications of NMR are also commonly called
magnetic resonance imaging, MRI. Because MRI does not involve ionizing radiation, it is
considered safer than X-ray processes.
MRI uses the spins of protons in hydrogen atoms. Hydrogen atoms are found throughout the
body, particularly in water molecules.
Resonance is the name given to the effect in which a system (that can oscillate) absorbs
energy from another external oscillating source.
Protons spin and behave like tiny magnets. These spins are usually randomly orientated
so that they will not produce any net observable magnetic effect. During an MRI scan the
patient is placed in a very strong primary magnetic field. This causes the spinning protons to
precess around the direction of the external field.
The rate of precession is called the Larmor frequency and it is proportional to the strength
of the applied magnetic field. The Larmor frequency is in the radio-wave (RF) section of the
electromagnetic spectrum.
When protons are also subjected to an oscillating electromagnetic field of the same frequency
(provided by the RF coils), resonance occurs and the protons begin to spin together, in phase.
This affects the overall magnetic field strength.
After the external RF signal is stopped the protons return to their earlier state at a rate that
depends on the kind of tissue in which they are located. The changing magnetic field can
be detected by the RF coils. The different relaxation rates provide information about the
type of tissue.
As well as the primary magnetic field, by imposing gradients of magnetic field in three
perpendicular directions (x, y and z), signals from different parts of a patient can be made
to resonate at different frequencies, allowing reconstruction of the three-dimensional
distribution of protons.
Examination questions 65
white light
principal axis
white light
On a copy of the diagram, after refraction in the lens, draw the paths for the rays of red light and blue
light present in the white light.
(2)
b Use your diagram in a to explain chromatic aberration.
(3)
c State one way in which chromatic aberration may be reduced.
(1)
d An object is placed 5.0cm from the lens and is illuminated with red light. The focal length of the lens
for red light is 8.0cm. Calculate the:
i position of the image
(2)
ii
linear magnification.
(1)
IB Organization
Q2 a
Draw a ray diagram to show how a converging mirror can be used to produce an inverted, magnified,
real image.
(3)
b i Describe the images formed by diverging mirrors.
(2)
ii
Give one everyday use for diverging mirrors.
(1)
Q3 a The diagram shows two rays of light from a distant star incident on the objective of an astronomical
telescope. The paths of the rays are also shown after they pass through the objective lens and are
incident on the eyepiece lens of the telescope.
objective lens
light from
a distant star
eyepiece lens
F0
The principal focus of the objective lens is Fo. On a copy of the diagram, mark the position of the:
i principal focus of the eyepiece lens (label this Fe)(1)
ii
image of the star formed by the objective lens (label this I).
(1)
b State where the final image is formed when the telescope is in normal adjustment.
(1)
c Complete the diagram in a to show the direction in which the final image of the star is formed for the
telescope in normal adjustment.
(2)
66 15 Imaging
d The eye ring of an astronomical telescope is a device that is placed outside the eyepiece lens of the
telescope at the position where the image of the objective lens is formed by the eyepiece lens. The
diameter of the eye ring is the same as the diameter of the image of the objective lens. This ensures that
all the light passing through the telescope passes through the eye ring.
A particular astronomical telescope has an objective lens of focal length 98.0cm and an eyepiece lens of
(4)
focal length 2.00cm (i.e. fo = 98.0cm, fe = 2.00cm). Determine the position of the eye ring.
IB Organization
Q5 a A signal of power 53mW enters an optic fibre of length 10.4km. If the power of the signal at the end
of the cable has reduced to 32mW, calculate the attenuation per km along the cable.
(2)
b Dispersion is a cause of attenuation in the cable. Distinguish between waveguide dispersion and material
dispersion.(3)
c Explain how the use of graded-index fibres reduces waveguide dispersion.
(2)
(2)
(2)
Q8 a
Computed tomography (CT) scans can provide much more useful information than individual X-rays.
Outline the techniques used in CT scans that produce this improvement.
b i Give two advantages that CT scans have compared with ultrasound scans.
ii
Give two advantages that ultrasound scans have compared with CT scans.
(3)
(2)
(2)
Examination questions 67
(3)
e Suggest, using your answer to d, why is this beam is suitable for identifying a bone fracture.
(1)
IB Organization
Q10 Nuclear magnetic resonance (NMR) provides an alternative to CT scans for obtaining images from inside the
body.
a Explain in general terms what is meant by resonance.(2)
b During an NMR scan, what parts of the patient are made to resonate, and how is this resonance
produced?(3)
c Why is NMR usually considered to be safer than the use of X-rays?
(1)