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AQA Physics PHYA5B (Medical Physics) - 28th June 2016

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It was awful. I also ran out of time so I didn't even make a start on the X-ray question so that's an instant 6 mark gone.
Original post by CourtlyCanter
It was awful. I also ran out of time so I didn't even make a start on the X-ray question so that's an instant 6 mark gone.


Really? I thought it was okay apart from the last question... which parts did you struggle with?
Reply 23
What rhe **** was that mrad-1 question?
Original post by particlestudent
Really? I thought it was okay apart from the last question... which parts did you struggle with?


Bits from everywhere. For example, the piezoelectric one, I forgot the word 'polarity' so I just left it blank and forgot to go back to it so now my answer reads like 'There is an electrode on either side of the piezoelectric crystal and when an alternating current is applied across it. At a BLANK, the crystal expands which pushes the air in front of it. The crystal contracts when the BLANK is reversed. This sequence is repeated millions of times per second and this creates a pressure wave which is the ultrasound wave.
Original post by dnan
What rhe **** was that mrad-1 question?


Exactly, I don't even remember covering this in the damn specification.

The diameter I got came to 6 (or something else) x10^-6 metres
Original post by CourtlyCanter
Bits from everywhere. For example, the piezoelectric one, I forgot the word 'polarity' so I just left it blank and forgot to go back to it so now my answer reads like 'There is an electrode on either side of the piezoelectric crystal and when an alternating current is applied across it. At a BLANK, the crystal expands which pushes the air in front of it. The crystal contracts when the BLANK is reversed. This sequence is repeated millions of times per second and this creates a pressure wave which is the ultrasound wave.


I don't think you need to talk about polarity :biggrin:

June 2013- Outline, with reference to the diagram, the process by which the transducer produces ashort pulse of ultrasound. (4marks)

Alternating potential difference applied across the crystal/causes crystal to expand and contract/creating pressure waves in the crystal / plastic membrane /frequency of alternating pd equal to that of crystal /resonant frequency ofcrystal which is above 20 kHz/short application of ac to produce short pulse /use of backing material to damp and stop vibration of crystal
I mentioned points 1,2,4,5
(edited 7 years ago)
Reply 27
Original post by particlestudent
Exactly, I don't even remember covering this in the damn specification.

The diameter I got came to 6 (or something else) x10^-6 metres



I had no clue at all. Not even a general equation i could think of, so I just did the mrad-1 value divided by 2pi (couldnt remember how to convert radians to angles) then did sin(ans)/ diameter. Defo wrong.
Original post by dnan
I had no clue at all. Not even a general equation i could think of, so I just did the mrad-1 value divided by 2pi (couldnt remember how to convert radians to angles) then did sin(ans)/ diameter. Defo wrong.


I done sintheta * that number they gave us in radians mode. But oh well, it was only 2 marks
Original post by particlestudent
I don't think you need to talk about polarity :biggrin:

June 2013- Outline, with reference to the diagram, the process by which the transducer produces ashort pulse of ultrasound. (4marks)

Alternating potential difference applied across the crystal/causes crystal to expand and contract/creating pressure waves in the crystal / plastic membrane /frequency of alternating pd equal to that of crystal /resonant frequency ofcrystal which is above 20 kHz/short application of ac to produce short pulse /use of backing material to damp and stop vibration of crystal
I mentioned points 1,2,4,5


How many marks do you think I got for that?
Original post by particlestudent
Exactly, I don't even remember covering this in the damn specification.

The diameter I got came to 6 (or something else) x10^-6 metres

Did it want us to treat it like the visual acuity calculation? I also forgot the 'x2' which means I've got it wrong...
Reply 31
Original post by CourtlyCanter
Did it want us to treat it like the visual acuity calculation? I also forgot the 'x2' which means I've got it wrong...


A what?
Is anyone going to make an unofficial mark scheme for the Medical Physics paper?
Reply 33
Original post by CourtlyCanter
Is anyone going to make an unofficial mark scheme for the Medical Physics paper?


56.7% or something for percentage one. Spatial resolution for first question.
Lens gets fatter when looking near, thinner when far. Ciliary muscles vhange it.
Pupil dilates for low light, gets smaller for bright light.
Original post by CourtlyCanter
How many marks do you think I got for that?

2 i'd say
Original post by CourtlyCanter
Did it want us to treat it like the visual acuity calculation? I also forgot the 'x2' which means I've got it wrong...


I dont even know what that is
Original post by dnan
56.7% or something for percentage one. Spatial resolution for first question.
Lens gets fatter when looking near, thinner when far. Ciliary muscles vhange it.
Pupil dilates for low light, gets smaller for bright light.

I got the percentage, and i said for the first one there must be one receptor inbetween the rods/cones :s-smilie:
Original post by dnan
56.7% or something for percentage one. Spatial resolution for first question.
Lens gets fatter when looking near, thinner when far. Ciliary muscles vhange it.
Pupil dilates for low light, gets smaller for bright light.


For the first question I just said There must be a one-cell separation between two cells on the retina that are each signalling an impulse.

For the same object further away question, I just said The muscles(didn't say ciliary because I forgot what the word was...) contracts which causes the lens to curve less and thus becomes less converging. This process is known as accommodation.

For the same object in dim light I just said The iris muscles cause a change in the amount of light into the pupil which acts as an aperture. The circular muscles relax and the radial muscles contract which enlarges the pupil. A greater area of the pupil means more light reflecting off the object goes into the eye.
(edited 7 years ago)
Original post by particlestudent
I dont even know what that is


2 x (average separation of light-sensitive cells on retina) / distance from retina to optical centre of the eye.) = Visual Acuity (radians)
Original post by CourtlyCanter
2 x (average separation of light-sensitive cells on retina) / distance from retina to optical centre of the eye.) = Visual Acuity (radians)


Is this on the AQA book for medical :s-smilie:

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