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2016 Official AQA New Spec AS Level Physics Paper 1 - 24th of May 2016

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Original post by aurora.98
"If a gap is much larger than the wavelength, there is no diffraction.
If a gap is a few wavelengths wide, there is a little diffraction.
If the gap is the same size as the wavelength, there is maximum diffraction." [from a revision guide]

What if a wavelength is much smaller than the gap? I know the answer to the question on set 2 states that there is diffraction but why is this the case? I said there wouldn't be any diffraction but I'm clearly wrong :frown: Any help would be much appreciated!


When my physics teacher showed us a simulation of waves passing through a large slit, it showed the waves largely passing straight through, however it also showed some diffraction on the edge of the wave front, where the wave had diffracted around the slit.
Original post by Chickenslayer69
That is from Set 2.

I'm assuming you use this equation: http://prntscr.com/b70q2y

Note that the L in m / L is the total length rather than the length from X and Y (The other L is from X and Y). If you rearrange this equation, you should get the correct answer.


I did that, but I got it wrong, and when I checked the mark scheme, they muitlpied it by 4, which is what I don't get?
(edited 7 years ago)
Original post by imjustaloner
i did that, but i got it wrong, and when i checked the mark scheme, they muitlpied it by 4, which is what i don't get?


(2l)^2 = 4l^2
Original post by Chickenslayer69
(2l)^2 = 4l^2


How well did you do on the paper! Thanks btw!
Original post by Chickenslayer69
(2l)^2 = 4l^2


I still don't know where I'm going wrong, I got 136.5367?
Original post by Chickenslayer69
(2l)^2 = 4l^2


I still don't know where I'm going wrong, I got 136.5367?
Why do you use the length 0.66? And not 0.91?
Original post by Imjustaloner
I still don't know where I'm going wrong, I got 136.5367?
Why do you use the length 0.66? And not 0.91?


http://prntscr.com/b70y44

I do not know the reason why you use 0.66 instead of 0.91, you just 'do'.
Original post by Chickenslayer69
http://prntscr.com/b70y44

I do not know the reason why you use 0.66 instead of 0.91, you just 'do'.


How was the paper?
Original post by Imjustaloner
How was the paper?


I didn't do all of it, I did the ones that look troubling, it is a hard paper :unimpressed:
Original post by Chickenslayer69
I didn't do all of it, I did the ones that look troubling, it is a hard paper :unimpressed:


Same, how are you feeling about the exam?
Original post by Imjustaloner
Same, how are you feeling about the exam?


Not sure. I don't really know what to expect :frown:

How about you?
Original post by Chickenslayer69
Not sure. I don't really know what to expect :frown:

How about you?


Same will they add some uncertainties and experimental questions?
Original post by Chickenslayer69
Not sure. I don't really know what to expect :frown:

How about you?


I'm just worried in general...
Im really worried. If I fail this paper would I be able to resit it in Year 13??
Reply 134
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Original post by 11234
When were looking at the uncertainty do we take the smallest value it can measure e.g. with a ruler +-1mm or do we take HALF of the smallest measurable quantity i.e. +-0.5mm


Help pls???
Original post by 11234
Help pls???


You take half the measurement, but times by 2, as the length is from one end to another
Reply 136
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Original post by Chickenslayer69
http://prntscr.com/b70y44

I do not know the reason why you use 0.66 instead of 0.91, you just 'do'.


Imagine the whole string is vibrating and that points X and Y are just bits of the whole string. The WHOLE string has mass 3.1x10-3kg and the WHOLE string has length 0.91m. Therefore mass per unit length is m/l=3.4x10-3. This value does not change its like density kinda. Since the string is vibrating in the third harmonic the formula is 3x1/2l x sqrt(tension/mass per unit length) = 330. Now the BETWEEN points X and Y is 0.66m and between here they are vibrating at 330Hz. So finally use the value 0.66m in this formula with your previous mass per unit length i.e.

m/l = 3.4x10^-3 Therefore: 330 = 3/(2x0.66) x sqrt(tension/3.4x10-3) Now rearrange and you should get 71.7N
Original post by DatComicalManiac
Im really worried. If I fail this paper would I be able to resit it in Year 13??


you dont need to.. this paper is just for as level so when you did the alevel it's all written off and replaced with that grade :tongue:
Reply 138
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Original post by Imjustaloner
You take half the measurement, but times by 2, as the length is from one end to another


But the question in the spec paper set 2 used 2mm as the absolute uncertainty so what do I do
Reply 139
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11234
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Original post by 11234
When we use micrometers, verniers and stopwatches do we take 2 x the uncertainty because of possible zero error or only once. Because i think you have to do that for a ruler but not sure about these


Could anyone help pls?? Much appreciated

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