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AQA Physics PHYA5 - 28th June 2016 [Exam Discussion Thread]

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Michelson–Morley experiment

Main aim -> Prove that an aether exists. The aether is a postulated medium through which light travels and so if such an aether exists the speed of light in an observer's frame depends on the relative speed of the observer to this aether.
Result -> Ironically they proved that an aether doesn't exist and thus the speed of light is constant for any observer in any frame of reference and is equal to 3x10^8m/s
Method:
Use a diffractometer consisting of a semisilvered mirror angled at 45, a laser consisting of coherent and monochromatic light is incident on this mirror, approx half of the intensity goes through and travels to a mirror, half reflects up (90 degrees to the straight through beam) and heads towards another mirror which is the same distance away as the the other mirror is to the semi-silvered mirror. The laser beams meet up again at the semi-silvered mirror after reflecting off the mirrors, reflect down to an observing scope and a diffraction pattern is observed.
The diffraction pattern depends on the optical path difference between the straight through beam and the reflected beam and if such an aether exists then the beam travelling parallel to the aether (i.e parallel to the earth orbit direction) will have a different path length to the perpendicular beam. Thus we expect upon rotation of the apparatus by 90 degrees for the path lengths to swap and thus a shift in diffraction pattern. This wasn't observed; there was no change in diffraction pattern. Hence disproving the existence of an aether -> verify c is constant.

(It's hard to describe the apparatus in words, but I'm sure you know what it looks like and can translate my explanation to it)

(edited 7 years ago)

Reply 181

Dann_a

Original post by Protoxylic

Link the question, I did this exam last year so I don't quite remember questions from specific papers

http://filestore.aqa.org.uk/subjects/AQA-PHYA51-W-QP-JUN10.PDF

http://filestore.aqa.org.uk/subjects/AQA-PHYA52A-W-MS-JUN10.PDF

Reply 182

Protoxylic

Original post by Dann_a

http://filestore.aqa.org.uk/subjects/AQA-PHYA51-W-QP-JUN10.PDF

http://filestore.aqa.org.uk/subjects/AQA-PHYA52A-W-MS-JUN10.PDF

If you're referring to Q1. Changes in absolute temperature are numerically equivalent to the changes in temperature measured in Celsius.

Spoiler

It's nothing more than a translation in the scale so whether or not you use K or C for deltaT is arbitrary. Although the units are strictly in kelvin

Reply 183

boyyo

Original post by aksdj

power= E/t
W= J/s
and the activity is the number of nuclei that decay per second so the unit would be s^-1
J*s^-1= W so P=AE

Have you ever done a question where you've had to use it?

Reply 184

particlestudent

Do you guys perform better on your Unit 5 papers or Unit 4?

Reply 185

Mattematics

Could somebody explain why the second option is incorrect?

I thought proper time was the shortest time and that it was the time experienced by the object experiencing both events. Surely therefore the time in the muon's frame of reference will be shorter than the observers, ergo time passes more slowly for the observer?

Reply 186

Protoxylic

Original post by Mattematics

Could somebody explain why the second option is incorrect?

You are misunderstanding the term 'shortest time'. The frame of reference that experiences proper time experiences the shortest time interval. So for every tick in the muon frame, less time has elapsed than for every tick in the observer/rest frame.

Reply 187

aksdj

How do they make this paper hard? I feel like it is more to do with memorising than U4 does anyone have some hard questions they could ask us

Reply 188

ssel17

Original post by Protoxylic

Proper time is defined as the frame that experiences the smallest possible time interval and is the frame for which the person/thing/object is present at both events. E.g a muon travelling between points A and B is present in it's own frame at the origin at both points A and B and thus the muon and it's frame experience proper time - all other time intervals are longer including the observer. So in this case the muon experiences proper time.

The protons experience proper time and also a contracted length for points A and B defined as the edges of the detector, i.e the detector length is shorter in their frame compared to the lab frame.

So basically, in the frame of reference of the muons, the time would be the proper time?

Whilst in the frame of reference of the protons, the protons distance is the contracted distance, not the proper distance?

If that's correct, I think I have got my head around it.