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Unit P3: Applications of physics
Topic 1 Radiation in treatment and medicine
Student Notes
Unit P3: Applications of physics
Topic 1
Radiation in treatment and medicine
We Are Learning To
1.1 Demonstrate an understanding of the methods that medical
physicists can employ to help doctors solve medical problems,
including:
a CAT scans
b ultrasounds
c endoscopes
d ionising and non-ionising radiation
The Three Types of Radioactivity
Alpha, Beta, and Gamma radiation are emitted from
unstable nuclei
Alpha
Beta
Gamma
All three types of radiation are known as ionising radiation.
Ionisation is where atoms gain or lose electrons, turning them
into charged particles called ions.
Radiation for
diagnosis
In order to reach a diagnosis, doctors
may use radiation to produce images
that show features inside the body.
Electromagnetic spectrum
PET
scans
X-ray images
& CAT scans
Endoscopes
Ultra-SOUND
- pre-natal checks
- removing kidney stones
Radiotherapy
 A cancerous tumour is exposed to
gamma radiation from lots of different angles.
 This gives normal cells a low dose of radiation,
while the tumour receives a high dose.
 However, levels have to be carefully monitored
so that healthy cells are not damaged as well.
tumour
In some patients radiation treatment cannot destroy the cancer.
Sometimes it is only used to reduce suffering (palliative care).
Unit P3: Applications of physics
Topic 1
Radiation in treatment and medicine
We Are Learning To
1.2 Use the word ‘radiation’ to describe any form of energy
originating from a source, including both waves and particles
1.3 Demonstrate an understanding that the intensity of radiation
will decrease with distance from a source and according to the
nature of the medium through which it is travelling
1.4 Use the equation:
intensity = power of incident radiation / area
I = P/A
Intensity (brightness)
The intensity depends on:
• The distance from the source
• The medium the radiation is travelling through
The word ‘radiation’ is used to
describe any form of energy,
e.g. wave or particle
originating from a source
The intensity of radiation will
decrease with distance from a source
and according to the nature of the
medium through which it is travelling
Different cancer tumours are treated with different intensities of gamma radiation and so
doctors place the source at different distances from the tumour. Intensity is also affected
by the medium the radiation is travelling through. The denser the medium, the weaker
the radiation gets.
Intensity
The strength of a radiation is called its
intensity. This is the power of the radiation
per square metre and is measured in W/m2
Intensity (W/m2) = Power (W)
Area (m2)
Unit P3: Applications of physics
Topic 1
Radiation in treatment and medicine
We Are Learning To
1.15 Explain, with the aid of ray diagrams, reflection, refraction and total internal
reflection (TIR), including the law of reflection and critical angle
1.16 Calculate critical angle using Snell’s Law
1.17 Explain refraction in terms of change of speed of radiation
1.18 Investigate the critical angle for perspex/air or glass/air or water/air boundaries
1.19 Investigate TIR between different media
1.20 Explain how TIR is used in optical fibres
1.21 Explain uses of optical fibres in endoscopes
The Law of Reflection
angle of incidence = angle of reflection
Refraction at a Boundary
The change in direction when a wave moves from
one medium into another is called refraction.
The wave bends because the speed of the wave
changes as it passes from one medium to
another.
Air
Glass
Air
If a wave hits a boundary at
an angle the wave changes
direction (it refracts). The
wave refracts because the
speed of the wave slows
down in the glass.
A wave travelling along the normal does not refract
Air
Direction of
travel of
the wave
wave
normal
glass
Glass
Air
A wave travelling along the normal does not change
direction (it does not refract). However, its speed slows
down.
A ray of light refracts
towards the normal
when it slows down
and away from the
normal when it
speeds up. The
greater the change in
speed the more the ray
deviates from its
original path.
Snell’s law
For waves passing from one medium into another,
Snell’s law links the angle of incidence (i), and angle of
refraction (r).
normal
i
r
medium 1
(air)
n1
medium 2
(glass)
n2
sin i = constant
sin r
The constant in Snell’s law is related to the refractive
index (n) of each material.
n1 sin i = n2 sin r
Refraction and gems
Refractive index = sin i
sin r
Diamond has a very high
refractive index of 2.417 so
it separates colours better
than other substances.
It also has a critical angle of
24.4° so light is internally
reflected many times
before emerging, spreading
out the colours more and
more.
Note: The higher the
refractive index the
slower light travels in the
medium.
Example 1
normal
i
r
medium 1
(air)
air
n1
45
medium 2
(glass)
n1 sin i = n2 sin r
30
glass
n2
1 sin 45 = n2 sin 30
sin 45 = n2
sin 30
0.7071 = n2
0.5
Refractive index of glass = 1.41
Example 2
A light ray approaches a glass
block at 300 to the normal. The
refractive indices of air and glass
are 1 and 1.5 respectively. At what
angle will the light be refracted?
air
n1  1
30
r
glass
n2= 1.5
n1 sin i = n2 sin r
1 sin 30 = 1.5 sin r
sin 30 = 1.5 sin r
sin 30 = sin r
1.5
0.5 = sin r
1.5
0.3333 = sin r
Angle of refraction = 19.50
What happens as the angle changes?
2) Light still gets refracted
1) Light is refracted
n1 sini1 = n2 sinr2
3) Light refracted along surface
4) Light internally reflected
Critical angle, c
sin c = n2/n1
Total internal reflection takes place when:
• The incident substance has a higher refractive index.
• The angle of incidence exceeds the critical angle.
glass
air
Calculation of critical angle
Air n2
n1 sin i = n2 sin r
c
n1 sin c = n2 sin 90
n1 sin c = n2 x 1
n1 sin c = n2
sin c = n2
n1
sin c = 1
n1
air
n2  1
Glass n1
Critical angle question:
Calculate the critical angle for perspex of refractive
index 1.48 when in air.
sin c = 1
n1
c
Sin c = 1
1.48
Sin c = 0.6757
Critical angle = 42.50
Total internal reflection is
used in fibre optics
Fibre optic communications
The transmission of information
by the passage of light through
flexible, glass fibres.
transmitter
>200km
receiver
Endoscopes are used to
look inside the body.
An image of the inside of the
larynx, showing the vocal cords.
Endoscope
An endoscope is a tube which allows a doctor to look into
the passageways of the body without having to operate.
Endoscopes consist of a flexible tube containing glass
fibres called optical fibres. The endoscope allows the
transmission of light into and out of the body. It has a
light source attached, and the light passes along one set of
optical fibres, down the endoscope and out at the end. The
light is reflected off the objects inside the body and then
the light travels back up a different set of optical fibres to
the eyepiece.
The doctor looks at the image through the eyepiece, or it is
displayed on a screen.
Uses of total internal reflection
1. Reflecting prisms in binoculars and
periscopes
2. Optical fibres for
endoscopes, TV, Internet communications
and phone calls
transmitter
>200km
receiver
Unit P3: Applications of physics
Topic 1
Radiation in treatment and medicine
We Are Learning To
1.5 Describe the refraction of light by converging and diverging
lenses
1.6 Relate the power of a lens to its shape
1.7 Use the equation:
power of lens (dioptre, D) = 1/focal length (metre, m)
1.8 Investigate variations of image characteristics with objects at
different distances from a converging lens
Convex lens (converging lens)
Parallel rays of light are refracted
and meet at the focal point
For a diverging lens, the focal point is the
point from which the rays seem to be
coming after passing through the lens.
Concave lens (diverging lens)
Real and virtual images
REAL images can be cast onto a
screen, for example a projector
image. They form where light rays
cross after refraction by a lens.
VIRTUAL images are formed
from where light rays only appear
to come from. A virtual image
cannot be cast onto a screen, for
example the image formed by a
flat mirror or a magnifying glass.
Looking at images
Magnified / enlarged image
close to concave mirror
Diminished image close to
convex mirror
Convex lens (converging lens)
object
F
image
Image description:
Inverted
Diminished
Real
1.
Object
2F
F
F
Image
Convex lens
Object past 2F
2F
Image description:
Inverted
Same size
Real
2.
Object
2F
F
F
2F
Image
Convex lens
Object at F
Image description:
Inverted
Magnified
Real
3.
Object
2F
F
F
2F
Image
Convex lens
Object between F and 2F
Image description:
Upright
Magnified
Virtual
4.
Image
Object
2F
F
Convex lens
Object between F and lens
F
2F
Concave lens (diverging lens)
object
F
image
Image description:
Upright
Diminished
Virtual
Object
image
2F
Concave lens
F
F
2F
Lenses of different powers
Weak
lenses
More
powerful
lenses
The power of lenses
The power of a lens measures how ‘quickly’
parallel rays of light converge to a focus.
focal
length
lens power = 1 / focal length
positive
If the focal length is measured in metres
then the lens power is in dioptres (D).
Converging lenses have positive powers,
diverging lenses have negative powers.
negative
Lens power questions
Calculate:
(a) the power of a
converging lens of focal
length 20 cm.
(b) the power of a
diverging lens of focal
length 50 cm.
(c) the focal length of a
lens of power 4.0 D
lens power = 1 / focal length
(a) power = 1 / 0.20m
= + 5.0 dioptres
(b) power = 1 / 0.50m
= 2.0 dioptres
(c) 4.0 = 1 / f
f = 1 / 4.0
focal length = 0.25 m (25 cm)
Unit P3: Applications of physics
Topic 1
Radiation in treatment and medicine
We Are Learning To
1.9 Use the lens equation:
1/f = 1/u + 1/v
(f = focal length (m), u = object distance (m), v = image
distance (m))
The use of the real is positive sign convention is preferred
and will be used in the exam
The lens equation
1 = 1 + 1
f
u
v
u = object distance along the principal axis from the
centre of the lens
v = image distance of the along the principal axis from
the centre of the lens
f = focal length
By convention the focal distance f is positive for a converging lens
and negative for a diverging lens
u = object distance along
the principal axis from the
centre of the lens
v = image distance of the
along the principal axis from
the centre of the lens
f = focal length
1 = 1 + 1
f
u
v
1 = 1 + 1
2
3
v
1 - 1 = 1
2
3
v
1 - 1 = 1
2F
2
3
v
3 - 2 = 1
6
6
v
1 = 1
6
v
Object
F
F
2F
Image
v =6
Since v is positive, the image is real
u = object distance along
the principal axis from the
centre of the lens
v = image distance of the
along the principal axis from
the centre of the lens
f = focal length
1 = 1 + 1
f
u
v
1 = 1 + 1
2
1
v
Image
1 - 1 = 1
2
1
v
1 - 2 = 1
2F
2
2
v
Object
F
F
2F
- 1 = 1
2
v
v = -2
Since v is negative, the image is virtual
1 = 1 + 1
f
u
v
-1 = 1 + 1
2
4
v
-1 - 1 = 1
2 4
v
-2 - 1 = 1
4 4
v
Object
image
2F
F
u = object distance along
the principal axis from the
centre of the lens
v = image distance of the
along the principal axis from
the centre of the lens
f = focal length
F
2F
-3 = 1
4
v
-3 = 1
4
v
v = -4/3 (or – 1.3)
Since v is negative, the image is virtual
The lens equation: 1 = 1 + 1
f
u
v
Question 1
An object is 10cm from a lens with a focal length of +5cm. Where
will the image be?
Question 2
An object is 5cm away from a converging lens with a focal length
of +10cm. Where will the image be?
Question 3
A pupil sits 100cm from a converging lens with a focal length of
5cm. Where will her image be formed?
Question 4
A magnifying glass with a focal length of 5cm is used to examine a
postage stamp 3cm away.
(1) Where will the image be formed?
(2) What indicates the image is virtual?
Unit P3: Applications of physics
Topic 1
Radiation in treatment and medicine
We Are Learning To
1.10 Identify the following features in a diagram of the eye –
cornea, iris, pupil, lens, retina, ciliary muscles
1.11 Demonstrate an understanding that light is focused on the
retina by the action of the lens and cornea
1.12 Recall that the average adult human eye has a near point at
about 25 cm and a far point at infinity
How do our eyes work?
Iris – makes pupil
larger or smaller
Retina – where
image is formed
Cornea –
window into
eye
Pupil
– gap
that lets
light
enter
eye
Lens
– focuses light
Ring of ciliary muscle – changes shape of lens
Optic nerve
– sends signal
to brain
Forming a sharp image – distant object
To form a sharp image,
light rays must converge
and focus on the retina.
Light rays are refracted
by the cornea and lens
Light rays from a distant
object (parallel rays)
The ciliary muscles are
relaxed pulling the lens
into a thinner shape
Forming a sharp image – closer object
Light rays from an object
that is closer
The ring of ciliary muscles are
contracted, making it smaller and
allowing the lens to relax into a
fatter shape.
How close can you focus?
This is called the near point and it is typically 25cm.
It depends on your eyes and your age (which effects
how flexible your eye lenses are to change).
The far point is at infinity.
Unit P3: Applications of physics
Topic 1
Radiation in treatment and medicine
We Are Learning To
1.13 Explain the symptoms and causes of short sight and long sight
(students will not be expected to draw scaled ray diagrams, but may be expected to
interpret them)
1.14 Compare and contrast treatments for short sight and long sight, including the use
of:
a simple lenses
b contact lenses
c laser correction
(combined lens equation is not required; students will not be expected to draw scaled
ray diagrams, but may be expected to interpret them)
Short- and long-sightedness
• Short-sighted people can focus near
objects but not distant ones.
• Long-sighted people can focus
distant objects but not near ones.
Short-sightedness is
caused by the eyeball being
too long or the cornea too
curved.
near object
distant object


diverging lens
corrected view of
distant object
Long-sightedness is caused
by the eyeball being too short
or the lens too thin.
near object
distant object


converging lens

corrected view
of near object

Correcting sight problems practical
Short-sightedness – distant
rays focus in front of retina
Long-sightedness – rays from
a near object focus behind retina
distant object
near object

eye lens

retina
corrected vision
corrected vision

diverging lens

converging lens
Contact lenses
• These are an alternative to
glasses. They are placed in
front of the cornea.
• Some are softer than others,
but all allow oxygen to
permeate to the eye.
• It is important to clean them properly to
prevent infection – although some are now
disposable.
Laser correction of sight
• This uses a finely controlled laser to vapourise a
portion of the middle part of the cornea and
reshape it.
• A permanent change is made to
the point at which light rays
meet inside the eye.
• Cost: £400 - £1500 per eye.
• Advantages: ‘Instant’ permanent
change to eyesight problems
(no more glasses).
• Disadvantages: Changes cannot
be reversed. Technically complex
procedure – so it can go wrong.
Unit P3: Applications of physics
Topic 1
Radiation in treatment and medicine
We Are Learning To
1.22 Explain uses of ultrasound in diagnosis and treatment
Pre-natal scanning
. Pre-Natal Scanning
X-rays can be used to see inside the body - (unsafe for a baby)
Ultrasound can create images and is safer.
Passes through new substance (skin, muscle, bone) > waves are
reflected as echoes.
The reflected waves (echoes) are detected by a computer.
These build up a picture from each echo.
Ultrasound waves are partially reflected when they meet a
boundary between two different media. The time taken for the
reflections to reach a detector is a measure of how far away such
a boundary is.
For scans in early pregnancy you'll be asked to drink lots of water so
your bladder pushes the uterus upwards for a better picture
How does ultrasound work?
Ultrasonic waves are partly
_________ at the boundary as
they pass from one _______
to another. The time taken for
these reflections can be used
to measure the _______ of the
reflecting surface and this
information is used to build up
a __________ of the object.
Words: picture, reflected,
depth, medium
Ultrasound
probe
Skin of
pregnant
woman’s belly
Body tissue
(e.g. muscle)
Ultrasound is
used to treat
kidney stones
Absorption of ultrasound
energy can also be used
by physiotherapists to
treat injured muscles.
Imaging the heart
atrium
heart valves
ventricle
Advantages & disadvantages of
ultrasound, X-rays & CT scans
X-ray images
& CAT scans
These use harmful ionising radiation – so
exposure should be kept to a minimum & doctors
monitored to check their cumulative dose.
Ultra-SOUND
Less likely to harm
the patient but give a
less clear picture
X-rays, CT scans and ultrasound can all
produce real-time moving images.
Ultrasound
Ultrasound is SOUND that has a frequency
above 20 000Hz (so we can’t hear it).
This is reflected each time it reaches a
boundary between different substances
(media).
Medical uses for ultrasound include:
– Pre-natal scanning
– Monitoring blood flow in the heart
– Destroying kidney stones
Non-medical uses include:
– Cleaning jewellery
– Detecting flaws or cracks
Why is ultrasound for scanning foetuses
instead of X-rays, which would give a cleaner picture?