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The Physics AS-Level Thread

Mutleybm1996

c)ii) You know the sensitivity of the touchscreen is 20mVmm^-1, if the V resolution is 5mV, what is the mimimum distance the screen can detect?

Oh 1.2/(60x10^-3)

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Reply 1181

jaylim95

hi I have a question on wave interference. Please help me, i can think of an answer by guessing, but i can't provide a proof or explain this to myself.

in single or double slit experiment, what change will happen to the interference pattern when the slit width is increased or decreased (other variables held constant)? This is not about the slit separation, but rather the size of the slit itself. by the way, does the pattern becomes less bright than before if slit width is decrease and vice versa?

from what i learned, the smaller the size of slit, the larger the diffraction. hence when the slit width decrease, there should be an increase in fringe separation or number of order, but i dont know why. however, given the equation n(lambda)=dsin(theta) where n is number of order, lambda is wavelength and d is slit width(for single slit) or slit separation for (double slit), a decrease in slit width should lead to a decrease in the number of order, shouldn't it? then there is a contradiction with the first statement. can someone explain this to me?

Reply 1182

Mutleybm1996

19

Original post by jaylim95

hi I have a question on wave interference. Please help me, i can think of an answer by guessing, but i can't provide a proof or explain this to myself.

in single or double slit experiment, what change will happen to the interference pattern when the slit width is increased or decreased (other variables held constant)? This is not about the slit separation, but rather the size of the slit itself. by the way, does the pattern becomes less bright than before if slit width is decrease and vice versa?

from what i learned, the smaller the size of slit, the larger the diffraction. hence when the slit width decrease, there should be an increase in fringe separation or number of order, but i dont know why. however, given the equation n(lambda)=dsin(theta) where n is number of order, lambda is wavelength and d is slit width(for single slit) or slit separation for (double slit), a decrease in slit width should lead to a decrease in the number of order, shouldn't it? then there is a contradiction with the first statement. can someone explain this to me?

To an extent yes, however you should have learnt that if the slit is smaller than the wavelength then it just bounces back off and doesn't go through.
Someone else should be able to help you more easily, but this is just a starting point to get you thinking

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Reply 1183

JackB784

7

Original post by jaylim95

hi I have a question on wave interference. Please help me, i can think of an answer by guessing, but i can't provide a proof or explain this to myself.

in single or double slit experiment, what change will happen to the interference pattern when the slit width is increased or decreased (other variables held constant)? This is not about the slit separation, but rather the size of the slit itself. by the way, does the pattern becomes less bright than before if slit width is decrease and vice versa?

from what i learned, the smaller the size of slit, the larger the diffraction. hence when the slit width decrease, there should be an increase in fringe separation or number of order, but i dont know why. however, given the equation n(lambda)=dsin(theta) where n is number of order, lambda is wavelength and d is slit width(for single slit) or slit separation for (double slit), a decrease in slit width should lead to a decrease in the number of order, shouldn't it? then there is a contradiction with the first statement. can someone explain this to me?

There is more diffraction the closer the wavelength is to the slit size I thought.

Reply 1184

ayesha_17

1

ImageUploadedByStudent Room1400097519.826406.jpg

Please could someone help me with the steps for working out part c (last bit at the bottom)? The answer is 30 000 ohms.

My working:
voltage of LDR: 5000x(6.6x10^-5) = 0.33 V
Voltage of R: 0.75 V
Hence, voltage of variable resistor: 6V (the emf) - (0.33+0.75) = 4.92 V
So, R of variable resistor using V=IR: 4.92/6.6x10^-5 = 75000 ohms

Thanks!

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Reply 1185

MangoFreak

17

Original post by ayesha_17

Please could someone help me with the steps for working out part c (last bit at the bottom)? The answer is 30 000 ohms.

My working:
voltage of LDR: 5000x(6.6x10^-5) = 0.33 V
Voltage of R: 0.75 V
Hence, voltage of variable resistor: 6V (the emf) - (0.33+0.75) = 4.92 V
So, R of variable resistor using V=IR: 4.92/6.6x10^-5 = 75000 ohms

Thanks!

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I have no idea what you did, but any voltage across any components will be split according to the ratio of the resistances of those components (so if you have 3 volts across two resistors of resistance 10ohm and 20ohm, then the voltage across each of those will be split according to the ratio 10:20, 1:2, so 1 volt and 2 volts, respectively).

In that question, you have one resistor of 5 kilohms, another of 5 kilohms, and one of unknown resistance. 0.75V is 1/8th of 6V. Since the LDR is also 5 kilohms, it must be the case that the voltage across it is the same. So the voltage is shared between the two fixed resistors and the variable resistor in the ratio 2:6 (because 2/8ths of the total voltage is being put into the fixed resistors, that leaves 6/8ths for the variable). That must also mean that the total resistance is shared in the same ratio.

So the combined resistance of the two fixed resistors is equal to 2/8ths of the total resistance of the circuit, and hence the resistance of the variable resistor is 6/8ths of the total resistance (I'm sure you can work out how to convert 2/8ths of X into 6/8ths of X).

I hope that helps (it probably didn't; I'm verbose :emo:

).

(edited 9 years ago)

Reply 1186

sneakbo2

10

Original post by ayesha_17

Please could someone help me with the steps for working out part c (last bit at the bottom)? The answer is 30 000 ohms.

My working:
voltage of LDR: 5000x(6.6x10^-5) = 0.33 V
Voltage of R: 0.75 V
Hence, voltage of variable resistor: 6V (the emf) - (0.33+0.75) = 4.92 V
So, R of variable resistor using V=IR: 4.92/6.6x10^-5 = 75000 ohms

Thanks!

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Resistance for R is 5000ohms, Resistance of LDR is 5000 ohms. Therefore voltage across them is the same as they will have the same current (they are in series).

So all you have to do is I = V/R which is 0.75/5000 to get (1.5x10^-4) Amps.

That will be the same across all components in series, so that is the current across the variable resistor.

Next, you should know each component takes a certain amount of voltage from the total, so you have (6 - 0.75 - 0.75= 4.5V) will give you the remaining voltage across the variable resistor.

Now just R= V/I; So therefore the 4.5/(1.5x10^-4) to get 30,000 ohms.

That's how to do it using the standard Kirchoff's laws, the ratio method is a bit more complicated I think

(edited 9 years ago)

Reply 1187

ayesha_17

1

Got it, thanks so much!

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Reply 1188

Hudl

16

Does anyone know where I can get more practise material for AQA Physics if I have done all the past papers :/

Reply 1189

ChordaeTendineae

Original post by AhmedDavid

Does anyone know where I can get more practise material for AQA Physics if I have done all the past papers :/

If you've done the new spec papers you can try the old spec papers which can be found here: http://www.tomred.org/physics-unit-1.html
But it gets a bit difficult to do a whole paper because the syllabus has changed quite a bit, so the topics you need are spread over about 3 different papers.

Also here (if you scroll down all the way to the bottom) there's are document with a lot of past paper questions- more than you'll probably have time to do: http://www.tomlinsonhall.com/PolesworthSchool6thForm/subjects/physics/asalevel.html

Reply 1190

katt123

1

Hey guys can you please help me understand this question, can't seem to get my head around it! Thank you soo much!!x
http://www.school-portal.co.uk/GroupDownloadFile.asp?GroupId=13924&ResourceId=38945

Reply 1191

JackB784

7

Original post by katt123

Hey guys can you please help me understand this question, can't seem to get my head around it! Thank you soo much!!x
http://www.school-portal.co.uk/GroupDownloadFile.asp?GroupId=13924&ResourceId=38945

Which question number?

Reply 1192

katt123

1

Original post by JackB784

Which question number?

1- all parts!sorrry!that was soo stupid of me

Reply 1193

Dalek1099

21

Original post by katt123

1- all parts!sorrry!that was soo stupid of me

Luckily I have done this one in class so I can explain it where the metre is connected determines where the circuit is split up into separate branches of resistors in parallel it is very confusing so I would understand you finding it difficult I struggled loads.so for 1a) the reciprocal resistance is 1/(50+50)+1/(50+50)=1/100+1/100=2/100 so resistance in part one is 50ohms and in reciprocal of the resistance part 2 is 1/50 +1/(50+50+50)=1/50+1/150=3/150+1/150=4/150 so resistance in part two=37.5 ohms.

Reply 1194

PhRose

2

can someone explain this multi-choice question for me; its q4???

Reply 1195

Bean97

Hi
Can someone please explain to me how you would obtain an emission line spectrum please?

Thanks

Reply 1196

PhRose

2

can someone explain q6 on http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/june2010-qp/6PH01_01_que_20100527.pdf
its multichoice- I though it would have been A because it is thrown form rest whereas B shows an initial velocity

Reply 1197

PhRose

2

Original post by Bean97

Hi
Can someone please explain to me how you would obtain an emission line spectrum please?

Thanks

something giving off light (like in an electric discharge tube or any light source like the sun) gives off light then you look at it through a diffraction grating which shows up funky discontinuous lines of different colours. these are formed as electrons transit form a higher energy level to another one; the energy is equivalent to the band gap which in turn corresponds to a frequency which corresponds to a certain colour of light.
...i think... :smile:

Reply 1198

Bean97

Thanks for replying!

Just wanna check, do you cause the vapour to fluoresce by firing fast moving electrons at it ?

Ive read another version in some notes given to me by my teacher which say ud excite the vapour by shining white light on it
But that dosent seem right to me because then the visible light photons wud either not have enough energy to excite the electrons and/or theyd interfere with the spectrum

Reply 1199

Mutleybm1996

19

For Frequency= 1/period if the period is measured in mV does the period need to be converted e.g. Frequency= 1/(0.9x10-3 ) or can i use the 0.9? In the mark-scheme it says 0.9 but i want clarification as to why if it's possible

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