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The Physics PHYA2 thread! 5th June 2013 Watch

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    (Original post by Jimmy20002012)
    In the double slit experiment, how is Maxima and path difference increased or decreased?


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    I think path difference just depends on the wavelength (greater wavelength, greater path difference).
    Fringe spacing (i.e. spacing of maxima) is increased by increasing wavelength, increasing distance of screen from slit or decreasing the slit separation.

    You can tell this from looking at w=(lambda x D)/s
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    Dont get 5d

    http://filestore.aqa.org.uk/subjects...W-QP-JUN11.PDF


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    You need to use some angle rules:

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    See if you can work out what I've done to label the angles like that (think back to GCSE maths).
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    (Original post by usycool1)
    You need to use some angle rules:

    Name:  Capture.PNG
Views: 153
Size:  11.7 KB

    See if you can work out what I've done to label the angles like that (think back to GCSE maths).
    Yeah I get that bit, you know that 180 degrees is a triangle, so you know the critical angle and there is 90 degrees, so you do 180-90-80= 10 degrees which is the refractive angle. But I thought then you would use the equation of n1sin1=n2sin2, but you don't do that, why?


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    (Original post by Jimmy20002012)
    Yeah I get that bit, you know that 180 degrees is a triangle, so you know the critical angle and there is 90 degrees, so you do 180-90-80= 10 degrees which is the refractive angle. But I thought then you would use the equation of n1sin1=n2sin2, but you don't do that, why?


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    You do use it after you've worked out the angle of refraction for when B enters the core. So:

    \dfrac{\sin \i}{\sin \ r} = \dfrac{n_2}{n_1}, where i = \theta

    n_1 is the refractive index of air, which is just basically 1.00
    n_2 is the refractive index of the core, which is 1.47

    So:

    \dfrac{\sin \theta}{\sin 10} = \dfrac{1.47}{1.00}

    \sin \theta = 1.47 \times \sin 10

    \Rightarrow \theta = 14.8

    Make sense?
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    (Original post by usycool1)
    You do use it after you've worked out the angle of refraction for when B enters the core. So:

    \dfrac{\sin \i}{\sin \ r} = \dfrac{n_2}{n_1}, where i = \theta

    n_1 is the refractive index of air, which is just basically 1.00
    n_2 is the refractive index of the core, which is 1.47

    So:

    \dfrac{\sin \theta}{\sin 10} = \dfrac{1.47}{1.00}

    \sin \theta = 1.47 \times \sin 10

    \Rightarrow \theta = 14.8

    Make sense?
    Didn't actually know that equation, why can't you use n1sini=n2sinr?


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    (Original post by Jimmy20002012)
    Didn't actually know that equation, why can't you use n1sini=n2sinr?


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    Erm...what happens when you divide n1 and sin r from both sides of the equation you just gave?

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    They're the same equation, just rearranged.
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    (Original post by usycool1)
    Erm...what happens when you divide n1 and sin r from both sides of the equation you just gave?

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    They're the same equation, just rearranged.
    So can you just always use the equation you have given me above, does it always work?


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    (Original post by Jimmy20002012)
    So can you just always use the equation you have given me above, does it always work?


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    Yup. You can use either, they're both the same thing.
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    Hello guys! I could not find a thread for 'Physics Unit 2 As-Level PHAY2 Exam 05/06/13 exam discussion' so I made 1. Please discuss below anything regarding this exam and ask any questions if you are stuck with anything or any past paper question. All of us will try our best to help those who are stuck. Good luck to all of you very much indeed!
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    Can someone explain to me what multipath dispersion is within an optical fibre?

    From what I've gathered it's something about a signal merging with another causing a data blur/loss. But that's not too clear.
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    (Original post by NedStark)
    Can someone explain to me what multipath dispersion is within an optical fibre?

    From what I've gathered it's something about a signal merging with another causing a data blur/loss. But that's not too clear.
    We really need to know it in detail, but basically its when light is lost (so energy is also lost) in an optical fibre as the light can scatter which can reduce the amplitude of the light that comes out of the optical fibre.


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    (Original post by usycool1)
    Yup. You can use either, they're both the same thing.
    Sorry to bother you again do you know that question yesterday I tried using it with the equation of n1sini=n2sinr, but that end of in you dividing not multiplying when rearranging to find the incidence angle


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    page 186 in the nelson thornes textbook (chapter 12 exam style questions)

    for question 1a ii), how is the uncertainty 12mm as stated in the answers? :confused::confused:

    uncertainty is simply the range/2 isn't it?

    I done (425- 396)/2 = 14.5mm
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    (Original post by fuzzybear)
    page 186 in the nelson thornes textbook (chapter 12 exam style questions)

    for question 1a ii), how is the uncertainty 12mm as stated in the answers? :confused::confused:

    uncertainty is simply the range/2 isn't it?

    I done (425- 396)/2 = 14.5mm
    We came across that question in class. The book is wrong.
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    (Original post by lebron_23)
    We came across that question in class. The book is wrong.
    thought I was doing something wrong then
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    is this another error in the nelson thornes textbook?

    on page 208 (chapter 13 exam style questions)

    question 2a), you only had to draw the rays, but the answer mentions that the angle of incidence of the ray emerging from the base of the glass block into the air, is 40 degrees

    I'm sure its 5 degrees...

    the angle of an incident ray and its refracted (or reflected) ray is always the same between two substance, right?
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    Anyone have the mark scheme for jan13?
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    (Original post by .raiden.)
    Anyone have the mark scheme for jan13?
    Do you happen to have the paper?


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    (Original post by SortYourLife)
    Do you happen to have the paper?


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    I only have the question paper that my school printed out but not the MS :/
 
 
 
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