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June 2011 G485-Fields, Particles and Frontiers of Physics

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Reply 140

m92singh

Original post by mattslater

I found some awesome revision notes for this online. Also found similar type for G484 if anyone is re-taking.

thanks a lot!!!!! this is great!

Reply 141

CoffeeStinks

I have a suspicion that this paper is going to be heavily essay based.

Reply 142

M_I

Original post by m92singh

On what aspects? what do you require help with?

Pretty much all of it lol.

Mainly ultrasound and x-ray. I don't get anything about ultrasound and don't get some x-ray equations and questions.....thanks.

Reply 143

ChoYunEL

Original post by CoffeeStinks

I have a suspicion that this paper is going to be heavily essay based.

The last one was.. and the one before that..
Wouldn't be surprised if this was too...

Reply 144

anshul95

Original post by Jonsie92

Do we need to know about the lamour frequency in MRI? and the equation for the precission frequency which is f= (yB)/(2pi) because thats no in the formula booklet.

Also what are the equations not in the book that we may need?

we need to know what it is not how to calculate it.

Reply 145

anshul95

Original post by ChoYunEL

The last one was.. and the one before that..
Wouldn't be surprised if this was too...

I think its because most of this unit is theory whereas all the other units including the AS ones (even the electrons, waves and photons) paper has lot more calculations etc.

Reply 146

anshul95

Original post by M_I

Pretty much all of it lol.

Mainly ultrasound and x-ray. I don't get anything about ultrasound and don't get some x-ray equations and questions.....thanks.

ultrasound- sound produced above audible range. The first thing you need to know is how they are produced and received - using the piezoelectric effect. So when a pd is applied to piezoelectric crystal it changes it shape. If an high frequency alternating pd is applied the crystal oscillates. If high enough frequency of pd is used (which is natural frequency of crystal) an ultrasound wave is produced. If a ultrasound is applied to this crystal, then pd is produced, which can then be interpreted by a computer to give information about the reflected ultrasound. The part of the ultrasound equipment that does is called the transducer.

Now when ultrasound is PULSED (short) through the body it is reflected back at the boundaries of tissues. The time delay between the relfected and emitted pulse can be used to calculate distances (more for A scans) and the intensity of the reflected signal can be used to determine tissue types. The amount of reflection taking place depends on the change in acoustic impedance of each material at the boundary. Z(acoustic impedance)=rho(density of material) x c (speed of ultrasound in material). The fraction of reflected intensity is
((Z1-Z2)/(Z1+Z2))^2. Now the air-skin boundary has a large change in acoustic impedance, meaning that if ultrasound where to be used most would be reflected at the air-skin and hardly anything will actually penetrate the body. To solve this, you use a principle called impedance matching by using a coupling medium i.e. a gel is placed on the skin, with approx the same acoustic impedance as the skin so that little reflection takes place (as the gel replaces the air).

A scan: linear transducer is used to measure the time differences in reflected and emitted signals. This can be used to calculate distances of organs/between organs - used to detect abnormalities in the eyes.

B scan: An array of transducers are used to emit a fan shaped ultrasound wave. The transducers measure the relative intensties of the reflected signals which builds up a 2D picture of the tissues beneath the surface. (this one is used in fetal scans).

Doppler: If the ultrasound is reflected by a moving object, most often blood, the frequency of the reflected wave will be different as their is a relative speed between the wave source (transducer) and the blood (as it is moving). This can be used to calculate the velocties of blood flow and can be used to diagnose problems with arteries for example.
adv- non ionsing, relatively cheap, portable.
disadv- poor image quality due to reflected intensity sometimes being very weak but also because it can't finely distinguish between tissues like say CAT or MRI can.

Note that high freq ultrasound=higher quality image but lower penetration. low freq ultrasound=lower quality image but higher penetration.

Reply 147

susan23

what do you guys things is MOST LIKELY to COME UP in the exam?

I pray they dont **** as over as they did with bio and chem.

The Jan 2011 paper was easy...anyone agree?

do you think this is going to be stretch and challenge based? It better not!

how hard do you think it wil be?

Reply 148

Oh my Ms. Coffey

Original post by susan23

Jan 11 paper was harder then the first. So much space, no EMI. I have a feeling the nuclear power plant/Ultrasound will be a big question as MRI/Space have come up already.

Reply 149

susan23

Original post by Oh my Ms. Coffey

Jan 11 paper was harder then the first. So much space, no EMI. I have a feeling the nuclear power plant/Ultrasound will be a big question as MRI/Space have come up already.

MRI has come up really? is this in the june paper? so has that magnetic resonance electron **** come up then....and larmors frequency thing...?

Reply 150

CoffeeStinks

Could anyone help me.

How does x ray diffraction show that the atomic radius is around 10^-10m?

Reply 151

Oh my Ms. Coffey

Original post by susan23

MRI has come up really? is this in the june paper? so has that magnetic resonance electron **** come up then....and larmors frequency thing...?

MRI was an 8+2 marker in June, there was 13 marks of Ultrasound in Jan but no big questions. The specimen paper also had 6 marker for MRI with pros/cons.

Reply 152

ChoYunEL

A transformer question hasn't come up yet but those aren't too difficult (*cross my fingers*)
Something on Nuclear Power, wouldn't be surprising. Be similar to the Solar,
lots of mass has been lost, e=mc^2

They've had something about x-ray intensifiers and the other thing twice
Could come up again, it seems popular.
Maybe a big marker on how a x-ray works and for the reason it had spikes of energy.

(edited 12 years ago)

Reply 153

Oh my Ms. Coffey

Original post by ChoYunEL

A transformer question hasn't come up yet but those aren't too difficult (*cross my fingers*)
Something on Nuclear Power, wouldn't be surprised. Be similar to the Solar,
lots of mass has been lost, e=mc^2

They've had something about x-ray intensifiers and the other thing twice
Could come up again, it seems popular.
Maybe a big marker on how a x-ray works and for the reason it had spikes of energy.

Contrast media came up last time, maybe the actual image intensifiers will.

Reply 154

ChoYunEL

Original post by Oh my Ms. Coffey

Have you got all all of those little pointers on one document?

Not personally but yes :colondollar:

Reply 155

random man5

Original post by sulexk

Hello,

Image intensifiers and intensifying screens are different

Image intensifiers: Xrays are incident on a phosphor screen. The xray photons produce a greater number of visible light photons
The visible light photons strike a photocathode(where by photoelectric effect) electrons are released.
The electrons are accelerated and focused by the postively charged anode so they strike another phosphor screen, which then gives out visible light. The final image can be viewed on a monitor

Intensifying screen- used to reduced exposure time for an xray image by a factor of 100 or more. The screen consists of a film sandwiched between to intensifier screens. Each screen is made from a material such as phosphor- phosphor is a scintillator(means it changes the xrays to visible light photons). The energy of a single x-ray photon incident on the scintillator is changed to several thousands of visible light photons, which produces a brighter image.

Thanks

But then surely when the question says: "A student suggests an image intensifier uses the photoelectric effect. Explain why this suggestion is incorrect." then the question is wrong?? :confused:

(That's the exact wording of the question from Jan 11)

Reply 156

ChoYunEL

Original post by random man5

Thanks

But then surely when the question says: "A student suggests an image intensifier uses the photoelectric effect. Explain why this suggestion is incorrect." then the question is wrong?? :confused:

(That's the exact wording of the question from Jan 11)

The answer to this is...
[INDENT]X-rays produce visible light
or
In photoelectric effect electrons are emitted[/INDENT]

I didn't understand it at first..
(from the mark scheme.)

1. Intensifier used as X-ray would pass through film
2. Intensifier converts X-ray photon to many visible (light) photons (which are absorbed by film)
3. *Lower exposure / fewer X-rays needed

(edited 12 years ago)

Reply 157

Martik

Was getting confused about Lenz's law... as simple as it may be to learn.. it talks about the conservation of energy is applied therefore the emf induced is the opposite of the change that caused it... as energy was put in to move it against the resistance force?

Im guessing the resistance force is the force that the induced current will experience which is upward if you are moving the wired downward through the mag field ... so its kinda like a drag effect, as it only is in effect when motion of the wire is in effect.

Someone set me straight please :smile:

(edited 12 years ago)

Reply 158

TobeTheHero

Original post by Martik

you're right :smile:

lenz's law is all about conservation of energy.

'An induced current is always in such a direction as to oppose the motion or change causing it'

(edited 12 years ago)

Reply 159

Terror

Original post by TobeTheHero

you're right :smile:

lenz's law is all about conservation of energy.

'An induced current is always in such a direction as to oppose the motion or change causing it'

Thats exactly it, it's just a simple induction, and the fact that the induced current is in the *Opposite* direction gives the reason why the equation for magnetic flux linkage is given a negative sign: