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The Physics PHYA2 thread! 5th June 2013 Watch

  • View Poll Results: What mark do you think you got out of 70?
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    thank you to u too
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    (Original post by StalkeR47)
    No, just 1 is sufficient. There was a question in the exam, 1 was true and the other was not and the mark scheme said TIR should occur. So that is how I know.
    So which 1 needs to be true then?
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    (Original post by Cl28)
    Thank you so much! And for the diffraction grating?
    So the graph of intensity/distance

    Double Slit: fringes of the same width and the same intensity.
    Single Slit: central fringe double the width of other fringes and intensity decreases.

    Thank you!
    diffraction grating is the same as double slit except it has 100 times more slits lool, so all that does it make the maximas sharper (narrower) but yeh intensity stays the same ... how i think of it is say you have a wave of frequency 2hz (intensity graph for double slit) and then a wave of frequency 20hz (intensity graph for diffraction grating) you will see that the wave with a 20hz is alot more sharper.

    and yeh your right about the double slit and single slit
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    (Original post by StalkeR47)
    No, just 1 is sufficient. There was a question in the exam, 1 was true and the other was not and the mark scheme said TIR should occur. So that is how I know.
    Ah crap, everyone's saying different things :/
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    (Original post by StalkeR47)
    No, just 1 is sufficient. There was a question in the exam, 1 was true and the other was not and the mark scheme said TIR should occur. So that is how I know.
    That would mean one can be false right?
    So let's assume that it is moving to a more optically dense medium and therefore n2 > n1. Therefore n2/n1 is a number greater than 1. Can't inverse sin a number greater than 1...
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    (Original post by masryboy94)
    no both HAVE to true !
    Check june 2009 Question 6aii.
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    Hi guys can you help me with something..

    I was wondering what you do for questions about slits and diffraction that say show what you see on the screen. Do you draw the intensity graph type thing or just band or dots


    Also when you diffract white light through a grating how do you show that the inner side is violet and the outer is red. Is this done on the intensity graph?.
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    Check june 2009 Question 6aii. You will find out whether 1 have to be true or both!
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    (Original post by Raimonduo)
    Different to what I would have done, but still correct.
    Ok thanks.
    Could you help me with another small query? Is the normal always a vertical line and do incidence and refraction "cos" into the angle, if that makes sense?
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    (Original post by NabRoh)
    That would mean one can be false right?
    So let's assume that it is moving to a more optically dense medium and therefore n2 > n1. Therefore n2/n1 is a number greater than 1. Can't inverse sin a number greater than 1...
    Yes 1 can be false. Check june 2009 Question 6aii. That is how I know.
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    (Original post by x-Sophie-x)
    Ah crap, everyone's saying different things :/
    Check june 2009 Question 6aii. ..
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    (Original post by StalkeR47)
    No, just 1 is sufficient. There was a question in the exam, 1 was true and the other was not and the mark scheme said TIR should occur. So that is how I know.
    im sure it can't be true because if a light ray hit a boundary (above critical angle) where the medium it is going to is optically less dense it will refract and not T.I.R. and obv if something isn't above critical angle, even if the medium it is going into is optically less dense, it will still refract. it is just the law of refraction. i am 100% certain about it
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    Does anyone have all of the six markers with the mark schemes?
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    (Original post by StalkeR47)
    Check june 2009 Question 6aii.
    That has nothing do with TIR, I suggest you check the exam paper you're on about
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    Any predictions as to what the 6 marker might be guys?
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    (Original post by StalkeR47)
    Yes 1 can be false. Check june 2009 Question 6aii. That is how I know.
    The question proves the point. Light enters the cladding and does not internally reflect BECAUSE the angle of incidence is below crit angle.
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    (Original post by StalkeR47)
    Check june 2009 Question 6aii. ..
    I don't see? :confused:

    More refractive to less refractive: so no TIR

    Angle of incidence does not exceed critical angle so no TIR. :confused:
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    (Original post by Qari)
    That has nothing do with TIR, I suggest you check the exam paper you're on about
    Yes it has. The reason why ray p is refracted is because angle of incident is smaller (not bigger) than the critical angle. But, for TIR, incident substance must have a larger refractive index which it has! So, why TIR does not occur but refraction?
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    (Original post by x-Sophie-x)
    Ah crap, everyone's saying different things :/
    Both have to be true
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    (Original post by NabRoh)
    The question proves the point. Light enters the cladding and does not internally reflect BECAUSE the angle of incidence is below crit angle.
    Yes but incident substance (core) has larger refractive index (1.55) than the cladding (1.45) so TIR should occur. But, since the other one is disobeying TIR rule, TIR does not occur.
 
 
 
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