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AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread] Watch

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    (Original post by Sbarron)
    I had one question where I was expected to work it out first and then put it in my further calculations and it defiantly was not 1.4*10^15 ... That gave the wrong answer
    Was that June 2012? I think I remember that question.


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    Anyone know if there is a difference between heat energy and thermal energy? Which is better to use?
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    (Original post by CD223)
    Yeah - if you try it the other way around you get an answer like -6.8 degrees which doesn't make sense either!


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    Thanks for the help
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    (Original post by Sbarron)
    This is how I've done it
    Attachment 428771
    Thanks I understand it now done loads of questions on inverse square law as its got to come up ! Just got to make sure I use common sense. Eg it moves further away it gets less intense


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    (Original post by slaven123)
    Does anyone know where the 1.22 comes from in this formula for working out the radius with diffraction? And do we need to remember the formula for the exam?
    I don't believe we need to know the derivation but yes, you do need to know that the first minimum appears where

    

\sin \theta = \dfrac{1.22 \lambda}{D}

    Or equally,

    

\sin \theta = \dfrac{0.61 \lambda}{R}

    Where R is the nuclear radius and D is the nuclear diameter.



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    (Original post by MSB47)
    Thanks for the help
    No problem! Thanks for flagging it up - hopefully we don't make that mistake on Thursday now!


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    (Original post by Boop.)
    Anyone know if there is a difference between heat energy and thermal energy? Which is better to use?
    They're equivalent. I'd recommend saying thermal instead of heat though - although they're the same thing, the unit is called "thermal" physics so I suppose AQA will use that terminology.


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    (Original post by MSB47)
    Thanks that question seems pretty decent!
    1)a) calculate the amount of energy that needs to be lost by the diet coke
    b) derive an equation for the amount of energy required to raise the temp of ice to 0C let mass of ice be m
    c) determine an equation for the latent energy of ice
    d) calcluate the mass m of ice that should be added

    Tip: total energy change = 0
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    (Original post by CD223)
    I don't believe we need to know the derivation but yes, you do need to know that the first minimum appears where

    

\sin \theta = \dfrac{1.22 \lambda}{D}

    Or equally,

    

\sin \theta = \dfrac{0.61 \lambda}{R}

    Where R is the nuclear radius and D is the nuclear diameter.



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    Why 1.22? I thought that as n=1 for the first minimum it should just be sinx=y/D
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    (Original post by gcsestuff)
    Thanks I understand it now done loads of questions on inverse square law as its got to come up ! Just got to make sure I use common sense. Eg it moves further away it gets less intense


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    I really don't even know how I done it like that.... I don't get why I can replace k with I and usually when I plug values in to an inverse proportion it tells me that the value is a proportion of the answer not the actual answer.... Like if I double the radius of r and r squared for force between two masses came out as 0.25 that's not the answer, I then have to find a quarter of the original force!!It's so hard not understanding maths!! Lol
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    (Original post by slaven123)
    Why 1.22? I thought that as n=1 for the first minimum it should just be sinx=y/D
    I don't know but if you go back to unit two and we had to find the number of orders n never came out as a whole number
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    (Original post by slaven123)
    Why 1.22? I thought that as n=1 for the first minimum it should just be sinx=y/D
    This equation is known as Rayleigh's Criterion for resolution.
    If we try to look at a very small particle under a microscope, then we won't see just a single spot of the same size. This is because the light from the particle diffracts as it passes through the aperture of the microscope. The diffraction pattern produced by light passing through a circular aperture looks something like this:



    It turns out that the central bright spot spans an angle of about 1.22λ/D, where λ is the wavelength of the light and D is the diameter of the microscope aperture. I will explain the mathematical details of where this formula comes from at the end.

    Now, what if we have two particles shooting light at our microscope aperture? Because the light from each of them spreads out in a pattern like I showed above, if they're too close together we'll have trouble telling that there are two particles rather than one:



    By convention (as an approximate condition) we declare that two particles are too close together to resolve (identify as two separate particles rather than one) if their central diffraction peaks overlap. Since the angle spanned by the central diffraction peak is 1.22λ/D, this means that we say we can't resolve objects unless the angle between them (as measured from the microscope lens) is greater than 1.22λ/D.


    Now, where does this particular formula come from? 1.22 sounds like a pretty strange number. It turns out that the intensity of diffracted light from a circular aperture (in the Fraunhofer limit) is given by

    I=I0(2J1(kDsin(θ))kDsin(θ))2,

    where k=2π/λ and J1(x) is the "Bessel function of the first kind, of order one". This is a special function that turns up a lot in the solutions to partial differential equations with some sort of cylindrical symmetry. You don't need to really know anything about it, except where it is equal to zero - since the first place where the intensity is zero is the edge of the central diffraction spot. It turns out that the first zero of this bessel function is at kDsin(θ)=3.8317.... If you expand the sine for small angles and rearrange, you can find the desired expression.


    (Source: http://www.quora.com/Where-does-this...its-derivation)
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    Do we need to know about diffraction when λ=2r or something like that?
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    (Original post by Boop.)
    Anyone know if there is a difference between heat energy and thermal energy? Which is better to use?
    Well according to this heat is energy transfer so it looks like they're the same thing
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    (Original post by CD223)
    This equation is known as Rayleigh's Criterion for resolution.
    If we try to look at a very small particle under a microscope, then we won't see just a single spot of the same size. This is because the light from the particle diffracts as it passes through the aperture of the microscope. The diffraction pattern produced by light passing through a circular aperture looks something like this:



    It turns out that the central bright spot spans an angle of about 1.22λ/D, where λ is the wavelength of the light and D is the diameter of the microscope aperture. I will explain the mathematical details of where this formula comes from at the end.

    Now, what if we have two particles shooting light at our microscope aperture? Because the light from each of them spreads out in a pattern like I showed above, if they're too close together we'll have trouble telling that there are two particles rather than one:



    By convention (as an approximate condition) we declare that two particles are too close together to resolve (identify as two separate particles rather than one) if their central diffraction peaks overlap. Since the angle spanned by the central diffraction peak is 1.22λ/D, this means that we say we can't resolve objects unless the angle between them (as measured from the microscope lens) is greater than 1.22λ/D.


    Now, where does this particular formula come from? 1.22 sounds like a pretty strange number. It turns out that the intensity of diffracted light from a circular aperture (in the Fraunhofer limit) is given by

    I=I0(2J1(kDsin(θ))kDsin(θ))2,

    where k=2π/λ and J1(x) is the "Bessel function of the first kind, of order one". This is a special function that turns up a lot in the solutions to partial differential equations with some sort of cylindrical symmetry. You don't need to really know anything about it, except where it is equal to zero - since the first place where the intensity is zero is the edge of the central diffraction spot. It turns out that the first zero of this bessel function is at kDsin(θ)=3.8317.... If you expand the sine for small angles and rearrange, you can find the desired expression.


    (Source: http://www.quora.com/Where-does-this...its-derivation)
    Great explanation, Cheers! Thank god we don't need to know it to that detail haha. Its much easier to understand when you know where they plucked these random numbers out from
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    (Original post by 000alex)
    Do we need to know about diffraction when λ=2r or something like that?
    We don't need to know anything bar the equation for the first minimum posted above


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    (Original post by slaven123)
    Great explanation, Cheers! Thank god we don't need to know it to that detail haha. Its much easier to understand when you know where they plucked these random numbers out from
    Yeah it's often the case when you don't know the reason behind something it turns out to be degree level physics haha!


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    Just casually having a sneaky little revision break!!Name:  image.jpg
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    Just couldn't resist sharing lol
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    Anyone doing core 4 edexcel Tuesday is it morning or afternoon?


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    (Original post by Sbarron)
    Just casually having a sneaky little revision break!!Name:  image.jpg
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    Just couldn't resist sharing lol
    That's not good for a healthy break


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