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Physics in the real world - During my visit to Broomfield Hospital I witnessed two aspects of physics in every day use. These were X-Rays and ultrasound; they both do similar jobs, although they both have limitations.

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Introduction

Physics in the real world

During my visit to Broomfield Hospital I witnessed two aspects of physics in every day use. These were X-Rays and ultrasound; they both do similar jobs, although they both have limitations.

A German physicist named Wilhelm Roentgen first discovered X-Rays in 1895, the discovery of X-Rays revolutionised medical science, and it allowed doctors to see through human tissue, to examine broken bones, swallowed objects, and with a modified version, to examine tissues such as the lungs.

They can be used to study the softer tissue of the body, they do this by introducing “contrast media” into the body, this is often a barium compound. If the doctor wants to examine the blood vessels it will be injected into the blood stream.

X-Rays can also be recorded as a moving image by a process called fluoroscopy, in which the X-Rays pass through the body onto a fluorescent screen, creating a moving X-Ray image, which can then be recorded onto a film.

X-Rays are not only used in the medical world, they are also used in airport security, to scan your bags. They are also used to detect flaws in big castings, to study quantum mechanics, crystallography and cosmology.

The only problem with X-rays Is that they are a form of ionising radiation, which means that when it hits an atom, it can knock an electron off, turning the atom into a ion. Ions can then collide with more atoms to create more ions.

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Middle

image00.jpg

Ultrasound uses a principal called Piezoelectric, which means pressure electricity. Pierre and Jacques curie discovered this in 1880. The transducer probe (the handheld probe) emits and receives the sound waves; inside the probe there are a number of quartz crystals called piezoelectric crystals. When current is applied to theses crystals they vibrate rapidly, conversely if pressure or sound waves hit them they produce a small electrical current. image01.jpg

This means that the same crystal can be used to send and receive sound waves.

The wavelength of the ultrasound depends on the frequency, and the material through which it’s travelling; this can be related by the following equations.

v=f λ

v= Speed  of wave in the material its travelling through.

f= Frequency  of the wave.

λ= The wavelength

Light is totally or partially reflected at boundaries, the same thing happens to sound waves; this is what allows ultrasound to work. The more similar the material, the smaller the amplitude of the reflected wave, as compared to the transmitted wave.

Z=pν

v= Speed  of wave in the material its travelling through.

Where Z is the acoustic impedance. Measured in kg-2 m s-1.

The intensity of the reflected sound can be given as a fraction of the intensity of the incident beam. It is given by the formula shown below.

Ir=(Z1-Z2)2

Ii(Z1+Z2)2

Material

Speed of sound/ms-1

Density Kgm-1

Air

330

1.3

Water

1500

1000

Blood

1570

1060

Fat

1450

950

Soft tissue (such as skin)

1500

1050

Muscle

1600

1080

Bone

4000

1500

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Conclusion

The future for ultrasound is improving all the time, with smaller and smaller probes almost to the stage where they can be swallowed, to allow doctors a much better view of the patient’s organs, they are also improving 3D ultrasound making it more widespread. They are also developing systems, which incorporate head up displays, which allow the doctor to look around inside the patient.

The Doppler effect allows police to catch speeding motorists, as it is used to by the radar speed traps to measure the Doppler shift in the radio waves reflected of the passing car.

The same principles are also used for radar, i.e. electromagnetic waves are sent out, those, which are reflected back, are received and are used to create a picture of the object.

Active sonar (as used by submarines and fishermen (to find fish)) works on a very simple principle to ultrasound, as “pings” sound waves into the ocean, the sound hits an object an is reflected back to the transducer, where it is amplified. The distance is worked out with the following equation.

Range= (speed of sound (1500ms-1 in water) x travel time)/2

Bibliography

Pennsylvanian state university website www.xray.hmc.psu.edu

Salters Horners advanced Physics Book

www.Howstuffworks.com  

Philips research

Jill Maskell. Superintendent radiographer at Broomfield hospital.

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