The Physics PHYA2 thread! 5th June 2013

Hey I am fine! I think this is from jan 2013 right? I have got the past paper but not the mark scheme. Just give me few minutes and I will post the answer to those question. Revision is going really well and I am very confident on unit 2 now. But I am just a bit lazy now and I can not be bothered anymore with the revision. HAHA. 2 mins m8, I will help you. What about you, How is your revision going and how many exams have you got now? I have 3 more to go and I will finish all my exams on 6th of june. Also m8, could you please tell me how to move on to another paragraph? I hate writing lods in 1 paragraph. Thanks.

Hey I am fine! I think this is from jan 2013 right? I have got the past paper but not the mark scheme. Just give me few minutes and I will post the answer to those question. Revision is going really well and I am very confident on unit 2 now. But I am just a bit lazy now and I can not be bothered anymore with the revision. HAHA. 2 mins m8, I will help you. What about you, How is your revision going and how many exams have you got now? I have 3 more to go and I will finish all my exams on 6th of june. Also m8, could you please tell me how to move on to another paragraph? I hate writing lods in 1 paragraph. Thanks.

Haha, you seem to be well prepared for the exam anyway , how how have you been revising.
Yes the question is from the Jan13 paper, don't really get what path difference is and how to do the last question?
Got 5 exams left
They are all next week, got two on Monday


Posted from TSR Mobile

OK. I am going to try my best to solve this for you. So, for the Q7bii) You know that the wavelength is 3.2x10^-2m. The path difference is the difference in distances from two coherent sources to an interference fringe. So we know that the interference fringe is towards the detector. You know that the ray on the right arriving at the detector has to cover less distance that the other ray on the left. However, from part 7bi) we calculated the wavelength which is the difference in distance between the two coherent sources. So the answer is 3.2x10^-2m. Am I right? And in other words, The path difference is the distance between the two rays arriving at the detector. Q7e) This question is quiet hard. I think I will be able to get 1 or 2 marks in this question. ANSWER: Remember, increasing the frequency decreases the wavelength since c=f times lambda. So, you can say, the wavelength will be halved since the frequency is doubled. For the second mark, you can say that the maximum is still shown since changing the frequency does not change the pattern shown. Hope this helps. Hey, could you please tell me how can I move on to the next paragraph? If you are still unsure, ask me again.

That has helped a lot, thanks
So basically the path difference is the same as the wavelength as they both have the same length, what causes the path difference to change? and the with increasing frequency does not change the maxima of a wave, am I right?
What do you mean by moving to the next paragraph?

I know this is asking for much, but could you quickly go over the conservation of energy topic, I understand the basics it just when does energy convert to heat totally?


Posted from TSR Mobile

That has helped a lot, thanks
So basically the path difference is the same as the wavelength as they both have the same length, what causes the path difference to change? and the with increasing frequency does not change the maxima of a wave, am I right?
What do you mean by moving to the next paragraph?

I know this is asking for much, but could you quickly go over the conservation of energy topic, I understand the basics it just when does energy convert to heat totally?


Posted from TSR Mobile

No! Not at all. Ask me as much as you like and I will be more than happy to help you So, do not mind at all asking me anything at all. + It also helps me to understand the context better. The two coherent sources does not have the same length. Remember, one has to travel lass distance towards the detector since the slit to the right is closer to the detector and the slit to the left is a bit further away from the detector. If they have no path difference, the answer is 0m. The wavelength just tells you the lengths of both which is just the difference in distance between them two rays. The wavelength of course will cause the change in the path difference. Oh yes! I think for the question 7e) you can also say that the path difference will change since the wavelength has been changed. And since the wavelength of the ray was halved (as the frequency was doubled), the path difference will also double. Does it makes sense? The conservation of energy: In the exam, you may be asked to state the principle of conservation of energy. Answer: Energy can not be created of destroyed, but it can only be transformed from one form to another. You need to be able to use the equation, work done= force times the perpendicular distance in the direction of the force. You need to know that the area under the graph of force-distance graph gives the work done. You also need to know that the work done to stretch the spring to extension is given by, 1/2 times force times Delta extension = work done. There is nothing else that you need to know for this topic. Further topics are the kinetic energy and potential energy.

And yeah, you know I write a big long paragraphs? I can not move to another paragraph by pressing 'enter'. How can I do it? Thanks.

So why is the path difference the same if the two waves are not the same? Also with calculating the word done under the graph, say the graph isn't linear, apparently you cannot do a triangle and work it out from there, instead you are to count the squares, how do you actually do that though??

Thanks for all this help by teh way
What did you get in the Unit 1 exam?


Posted from TSR Mobile

Two waves have different wave length. Say for example, first wavelength has a wavelength of 100nm. The other one has 300nm. The path difference is therefore 300-100 = 200nm. If you calculate the wavelength of 2 waves when the frequency of both waves is given, the wavelength would be the path difference between 2 waves. Does this make sense? If the graph does not have a constant gradient, they will either give you the trapezium strips (from mathematics c2) or they will give you the 1cm grid from which you will have to count and estimate the number of blocks which will give the area as work done. Amm.. I think we will get the results for everything in August. Why? Did you get yours already?

Did unit 1 in Jan, would you count the number of small squares and that would be the work done?


Posted from TSR Mobile

Did unit 1 in Jan, would you count the number of small squares and that would be the work done?


Posted from TSR Mobile

Alright! I did not do the exam in January. What did you get?

Got a B, but resit it in may


Posted from TSR Mobile

what do you think the 6 markers gonna be on ?? Thanks in advance!

It could be on the young's double slit experiment or the reflection/refraction and how do they occur etc. EDIT: It is more likely to be on waves eg, progressive waves such as transverse and longitudinal.