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# The Physics PHYA2 thread! 5th June 2013 watch

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1. (Original post by SAS18)
so how do you know how many sig figs answer you should give? cos it keeps changing in the markscheme
Rule of thumb is always 2 s.f. BUT, be very careful, if it asks to an appropriate number of s.f, you wanna look at the s.f of the numbers they've supplied in the question, and then adjust the s.f of your answer accordingly.
2. (Original post by masryboy94)
so for the white light being diffracted, you don't get no dark fringes? just a continuous spectrum, right?
You do get dark fringes.
3. Cannot believe I am stuck on the first question . I don't get it, got the answer but got 10 to the power 6 instead of 10 to the power 5??

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4. (Original post by SAS18)
so how do you know how many sig figs answer you should give? cos it keeps changing in the markscheme
if they don't state to how many s.f. you should always have it to 2. if they say state to an appropriate s.f. then look at your question, if say they had blah blah blah 180N blah blah blah 46.5 degrees. you would know you need to have your answer to 3 s.f. (show full answer though in the step before)
5. (Original post by x-Sophie-x)
For the optics topic, in the core of an optical fibre, if total internal reflection/refraction doesn't happen, then does the reflected ray always reflect at the same angle as the angle of incidence?

For example, question 4, June 2012.

Angle of incidence=Angle of refraction=85 degrees. Is this always the case?
Angle of incident is always the same as angle of reflection. Always.
6. (Original post by StalkeR47)
You do get dark fringes.
but in that picture you posted, there was no dark fringe, it was the rainbow continuously you went from violet to red and straight back to violet and so on. ?
7. In an experiment, a narrow beam of white light from a filament lamp is directed at normal incidence at a diffraction grating. Complete the diagram in Figure 5 to show the light beams transmitted by the grating, showing the zero-order beam and the first-order beams.

can anyone help me here please!
8. Does anybody have notes going through everything we need to know?
Yup, I'm in panic, haven't done any solid revision for physics, just past papers and I would be VERYYYYYYYYYYYYY thankful If somebody could share with their revision notes.

p.s. have anybody posted all unit 2 experiments in this thread ? pleaaase , somebody help, I feel so miserable in first place because of todays chemistry.

9. (Original post by Jimmy20002012)
Cannot believe I am stuck on the first question . I don't get it, got the answer but got 10 to the power 6 instead of 10 to the power 5??

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Using w=fs, you've found out f, but need to multiply this by s (the distance), i.e. 1000m
10. (Original post by StalkeR47)
Angle of incident is always the same as angle of reflection. Always.
Ah right, thanks
So how would you know when total internal reflection happens?
I know that TIR happens when the angle of incidence equals (?) the critical angle, but what if you're not given this?
11. (Original post by masryboy94)
yeppp
Oh I know what you mean. You do get max and min for the interference pattern. However, each fringe is a continuous spectrum.
12. (Original post by x-Sophie-x)
Ah right, thanks
So how would you know when total internal reflection happens?
I know that TIR happens when the angle of incidence equals (?) the critical angle, but what if you're not given this?
When I get questions on this, I never know why it TIR. I need some clarity as well.
13. (Original post by Jimmy20002012)
Cannot believe I am stuck on the first question . I don't get it, got the answer but got 10 to the power 6 instead of 10 to the power 5??

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You are given 1km distance which is a horizontal distance. so, Work done = force times distance in the direction of the force. By finding the horizontal force, you can use the equation. Is that ok?
14. (Original post by StalkeR47)
You do get dark fringes.
You don't get dark fringes with a continuous spectrum?!?!
15. Anybody have a link to the Jan 2013 paper? Thanks
16. (Original post by StalkeR47)
You are given 1km distance which is a horizontal distance. so, Work done = force times distance in the direction of the force. By finding the horizontal force, you can use the equation. Is that ok?
Ohhhh? Thanks, what would I do without you

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17. (Original post by x-Sophie-x)
Ah right, thanks
So how would you know when total internal reflection happens?
I know that TIR happens when the angle of incidence equals (?) the critical angle, but what if you're not given this?
No probs! So, TIR happens if the angle of incident is bigger than the critical angle. Or, when the incident substance has a larger refractive index than the emergent substance. If you are not given the either, you will have to use one of the equations. sini(critical)=n2/n1
18. (Original post by Jimmy20002012)
Cannot believe I am stuck on the first question . I don't get it, got the answer but got 10 to the power 6 instead of 10 to the power 5??

Posted from TSR Mobile
we already found out in the previous question to be

so sub that value in with the velocity to give
19. (Original post by Raimonduo)
Rule of thumb is always 2 s.f. BUT, be very careful, if it asks to an appropriate number of s.f, you wanna look at the s.f of the numbers they've supplied in the question, and then adjust the s.f of your answer accordingly.

(Original post by masryboy94)
if they don't state to how many s.f. you should always have it to 2. if they say state to an appropriate s.f. then look at your question, if say they had blah blah blah 180N blah blah blah 46.5 degrees. you would know you need to have your answer to 3 s.f. (show full answer though in the step before)
Thanks so much guys!
20. Pleaseeee can some one help me on the following question from the June 2010 paper , 3cii,5b,6ai,6aii, & 7d I really appreciate it thanks ( I'm going to fail this

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