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# new AQA AS physics paper 2 desperate help needed !!!!!!

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1. so im doing the first specimen second paper ...and the first question makes nooo sence and iv got the mark sceme up and i just cant see how they got the answer and im freaking out as i thought i knew all my stuff now i just feel like i know nothing ! If anyone can help go onto the new aqa a level physics page and its the specimen paper 2 and its question 1.1 and 1 .5....it makes nooo sence !
2. Are you struggling with all of question one or just specific parts ?
3. If this is the one with the diffraction grating then:-

The percentage uncertainty is worked out as 0.1x10^3 divided by 3.5x10^3 which would give you 0.029 i.e 3%

The grating spacing = (order of maxima) * (wavelength) / (sin angle)

If you look at the interference pattern, order of maxima would be 25. The wavelength is 628nm. At this distance from the laser, the sin of the angle is approximately the same as the tan of the angle which would be 0.135/2.5 (0.135 being the distance that the 25 maxima cover and 2.5 being the distance between the grating and the screen.)

By substituting these values into the equation you get that d = 2.91*10^3 m. The no. of lines = 1/d = 3.43 lines/m.
4. (Original post by Tom22561)
Are you struggling with all of question one or just specific parts ?
all of the question was a mess but the main parts are parts 1.1 and 1.5
5. (Original post by Teenie2)
If this is the one with the diffraction grating then:-

The percentage uncertainty is worked out as 0.1x10^3 divided by 3.5x10^3 which would give you 0.029 i.e 3%

The grating spacing = (order of maxima) * (wavelength) / (sin angle)

If you look at the interference pattern, order of maxima would be 25. The wavelength is 628nm. At this distance from the laser, the sin of the angle is approximately the same as the tan of the angle which would be 0.135/2.5 (0.135 being the distance that the 25 maxima cover and 2.5 being the distance between the grating and the screen.)

By substituting these values into the equation you get that d = 2.91*10^3 m. The no. of lines = 1/d = 3.43 lines/m.
Thank you so much this all makes sence, but just a few questions...
Why did you use 0.1 X10^3 ? surley if its a metre the uncirtancy is 1 metre ?
And how did you find it to be 0.135? how did u calculate that?
6. The grating is defined as having 3.5x10^3 lines per meter so it has a potential error of 0.1x10^3 lines per meter. It is defined to one decimal place If it was less than 3.45x10^3 it would have been written as 3.4x10^3 and if was more than 3.55x10^3 it would have been written as 3.6x10^3. So, the measurement can vary between 3.45x10^3 and 3.55x10^3 i.e 0.1x10^3 lines per meter.

As for the 0.135m, sorry, this should be 0.13m. This is the distance between the first maxima and the last on the interference pattern. The first line is centered around 0.6cm on the ruler and the last is at 13.6cm. If you put this into the equation you get d=3x10^-4m so the no. of lines per meter = 3.34x10^3 - I really should double check things before posting !!
7. (Original post by Teenie2)
The grating is defined as having 3.5x10^3 lines per meter so it has a potential error of 0.1x10^3 lines per meter. It is defined to one decimal place If it was less than 3.45x10^3 it would have been written as 3.4x10^3 and if was more than 3.55x10^3 it would have been written as 3.6x10^3. So, the measurement can vary between 3.45x10^3 and 3.55x10^3 i.e 0.1x10^3 lines per meter.

As for the 0.135m, sorry, this should be 0.13m. This is the distance between the first maxima and the last on the interference pattern. The first line is centered around 0.6cm on the ruler and the last is at 13.6cm. If you put this into the equation you get d=3x10^-4m so the no. of lines per meter = 3.34x10^3 - I really should double check things before posting !!
Thank you so much your a life saver and a genius !!! all that for 2 marks as well ! i think the exam board has gone mad !!!!

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