Topic 1 Radiation in treatment and medicine Unit P3: Applications of physics Topic 1 Radiation in treatment and medicine Student Notes

Unit P3: Applications of physics Radiation in treatment and medicine W.A.L.T We Are Learning To Unit P3: Applications of physics Topic 1 Radiation in treatment and medicine 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 X-ray images & CAT scans PET scans Endoscopes Ultra-SOUND - pre-natal checks - removing kidney stones

path of rotating gamma source 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. path of rotating gamma source 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 Radiation in treatment and medicine W.A.L.T We Are Learning To Unit P3: Applications of physics Topic 1 Radiation in treatment and medicine 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 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 (W/m2) = Power (W) Area (m2) 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 Radiation in treatment and medicine W.A.L.T We Are Learning To Unit P3: Applications of physics Topic 1 Radiation in treatment and medicine 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

angle of incidence = angle of reflection The Law of Reflection angle of incidence = angle of reflection 12

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 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. Glass Air

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.

n1 sin i = n2 sin r 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 medium 1 (air) n1 i sin i = constant sin r medium 2 (glass) n2 r The constant in Snell’s law is related to the refractive index (n) of each material. n1 sin i = n2 sin r

Refractive index = sin i sin r Refraction and gems 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. Refractive index = sin i sin r Note: The higher the refractive index the slower light travels in the medium.

Example 1 n1 sin i = n2 sin r 1 sin 45 = n2 sin 30 sin 45 = n2 sin 30 normal 45 i medium 1 (air) air n  1 medium 2 (glass) glass n2 r 30 n1 sin i = n2 sin r 1 sin 45 = n2 sin 30 sin 45 = n2 sin 30 0.7071 = n2 0.5 Refractive index of glass = 1.41

Example 2 n1 sin i = n2 sin r 1 sin 30 = 1.5 sin r sin 30 = 1.5 sin r 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? n1 sin i = n2 sin r 1 sin 30 = 1.5 sin r sin 30 = 1.5 sin r 30 air n1  1 sin 30 = sin r 1.5 glass n2= 1.5 0.5 = sin r 1.5 r 0.3333 = sin r Angle of refraction = 19.50

What happens as the angle changes? 1) Light is refracted 2) Light still gets 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. 20

Calculation of critical angle glass air Air n2 n1 sin i = n2 sin r Glass n1 c n1 sin c = n2 sin 90 n1 sin c = n2 x 1 n1 sin c = n2 sin c = n2 n1 air n2  1 sin c = 1 n1

sin c = 1 n1 Sin c = 1 1.48 Sin c = 0.6757 Critical angle question: Calculate the critical angle for perspex of refractive index 1.48 when in air. sin c = 1 n1 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 receiver >200km

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 Reflecting prisms in binoculars and periscopes Optical fibres for endoscopes, TV, Internet communications and phone calls transmitter receiver >200km 26

Unit P3: Applications of physics Radiation in treatment and medicine W.A.L.T We Are Learning To Unit P3: Applications of physics Topic 1 Radiation in treatment and medicine 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) Concave lens (diverging 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

1. Image description: Inverted Diminished Real Convex lens Object 2F F F 2F Image Convex lens Object past 2F

2. Image description: Inverted Same size Real Convex lens Object at F

3. Image description: Inverted Magnified Real Convex lens Object 2F F F 2F Image Convex lens Object between F and 2F

4. Image description: Upright Magnified Virtual Convex lens Object 2F F F 2F Convex lens Object between F and lens

Concave lens (diverging lens) object F image

Image description: Upright Diminished Virtual Concave lens Object 2F F F 2F Concave lens

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. lens power = 1 / focal length 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 = 1 / focal length 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 Radiation in treatment and medicine W.A.L.T We Are Learning To Unit P3: Applications of physics Topic 1 Radiation in treatment and medicine 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 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

Since v is positive, the image is real 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 1 = 1 + 1 2 3 v 1 - 1 = 1 2 3 v Object 1 - 1 = 1 2 3 v 2F F F 2F Image 3 - 2 = 1 6 6 v 1 = 1 6 v v = 6 Since v is positive, the image is real

Since v is negative, the image is virtual 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 1 = 1 + 1 2 1 v Image 1 - 1 = 1 2 1 v Object 1 - 2 = 1 2 2 v 2F F F 2F - 1 = 1 2 v v = -2 Since v is negative, the image is virtual

Since v is negative, the image is virtual 1 = 1 + 1 f u v - 1 = 1 + 1 2 4 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 - 1 - 1 = 1 2 4 v Object image - 2 - 1 = 1 4 4 v 2F F 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 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 Radiation in treatment and medicine W.A.L.T We Are Learning To Unit P3: Applications of physics Topic 1 Radiation in treatment and medicine 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 Optic nerve – sends signal to brain Pupil – gap that lets light enter eye Lens – focuses light Ring of ciliary muscle – changes shape of lens

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 Radiation in treatment and medicine W.A.L.T We Are Learning To Unit P3: Applications of physics Topic 1 Radiation in treatment and medicine 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. Long-sightedness is caused by the eyeball being too short or the lens too thin.   near object near object   distant object distant object diverging lens converging lens   corrected view of distant object 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 Radiation in treatment and medicine W.A.L.T We Are Learning To Unit P3: Applications of physics Topic 1 Radiation in treatment and medicine 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. Ultrasound probe Skin of pregnant woman’s belly ***SAME GAPS TASK AS ON WORKSHEET*** Words: picture, reflected, depth, medium 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 Pictures from http://www.gehealthcare.com/uken/

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? MAIN - *: * **Ref: AQA GCSE p