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

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 I ye.

@CourtlyCanter @dnan


For the power question on the ear... did you multiply the intensity worked out in the previous section by 4pi(500)^2 or whatever the distance was? I done this because it gave me units of Watts

Unofficial ms :

Qs1
One unstimulatd cell btw two stimulated cells.
Cones as only cones at fovea
Dont know how to do the calculation only two marks.
Ciliary muscles relax, suspensory ligaments taut, thinner lens/less power/more focal length, less refraction of light
Pupil dialates, circular muscles relax/radial muscles contract, more light enters eye, rods stimulated cones not.

Qs 2
Why was there variations in dBA scale for diff freq~ not sure just wrote it mimics the ear response , ear less sensitive at high and low frequencies.
Age related hearing loss increases as freq increases
Noise induced hearing loss increases till 4000 where it is maximum after that becomes normal
Using formula of 92=10log (I/1*10^-12)
Power = answer above*4(pi)(400^2

Qs3 i think this was the six marker
A- evacuate glass tube for removal of air molcules that would other wise collide with electrons and slow them
B - heated filement for thermonic emission
C- anode for accelerating electrons, pd btw anode and filament, at anode electons slowed down , lose KE by releasing x ray photons called bremstrahlung radiation, could talk about rotating and bevelled nature of anode, chaacteristic photon energies due to ionisation of atoms in anode.
Lead shield- attenuate/ stop x ray photons that escape tube and enter living tissue.
( I may have got A,B,C the wrong way round as i cant remeber them)

Qs4
Half value thickness is the thickness of a material that will reduce the intensity of specific energy x ray photons to half its original value.
Proof by ln2/ 15
Something like 57% by finding E=50e^-0.046*12 what ever you ge divid by 50 and times by hundred

Last question i got 57% by doing I=Ioe^ux with the answer for I over Io x100 is %age answer.

I did something x400^2. Think it was answer from previous question. 400 being distance

Unofficial ms :

Qs3 i think this was the six marker
A- evacuate glass tube for removal of air molcules that would other wise collide with electrons and slow them
B - heated filement for thermonic emission
C- anode for accelerating electrons, pd btw anode and filament, at anode electons slowed down , lose KE by releasing x ray photons called bremstrahlung radiation, could talk about rotating and bevelled nature of anode, chaacteristic photon energies due to ionisation of atoms in anode.
Lead shield- attenuate/ stop x ray photons that escape tube and enter living tissue.
( I may have got A,B,C the wrong way round as i cant remeber them)

@mohdstudent
@veldt127


How was it?

Mostly ok, think I forgot some things as per when it comes to (mostly) written exams. Definitely didn't get that weird eye angle one.

For the eye, it formed a right angled triangle with an angle of theta mrads (cant remember the angle's value) and adjacent side of length 55mm. You had to find the opposite side in metres, which was equal to (55*10^-3)*tan(theta)

I struggled with the eye ones too. Hopefully it's 53 for an A* like June 12 and 13, this one involved unusual calculations