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The Physics PHYA2 thread! 5th June 2013 Watch

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    Stalker from what I see, I think it's better if you do my exam for me lol
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    (Original post by StalkeR47)
    No! All I was saying that why TIR does not occur if incident substance has larger refractive index than the other? Can you answer this? HUH?
    it would occur BUT ONLY if it exceed the critical angle too ! hence why in optical fibres the core has a larger refractive index than cladding allowing T.I.R if above critical angle, and if it is not above critical angle you get refraction hence loss of signal causing the fibre to not be secure.
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    Can someone sketch out how partial internal reflection would look like? i.e for Question 5)b)i) in Jan 13'.
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    (Original post by Qari)
    So do you understand that both have to be met for TIR to occur
    Indeed I do
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    (Original post by UnknownRoyalist)
    Stalker from what I see, I think it's better if you do my exam for me lol
    Amm sorry. I did not get you. Can you answer this please? Anyone please? why TIR does not occur if incident substance has larger refractive index than the other? Ca
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    (Original post by vengeance111)
    They have to have simular amplitude so that a node is formed
    oh yeah didn;t think of that lol, thanks
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    (Original post by mejdoub.wassim)
    Can someone sketch out how partial internal reflection would look like? i.e for Question 5)b)i) in Jan 13'.
    A dotted line reflected off the boundary, (when you have a ray grazing the boundary).

    I think
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    (Original post by StalkeR47)
    Amm sorry. I did not get you. Can you answer this please? Anyone please? why TIR does not occur if incident substance has larger refractive index than the other? Ca
    i answered you bro
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    (Original post by masryboy94)
    it would occur BUT ONLY if it exceed the critical angle too ! hence why in optical fibres the core has a larger refractive index than cladding allowing T.I.R if above critical angle, and if it is not above critical angle you get refraction hence loss of signal causing the fibre to not be secure.
    But how do you know that if the incident substance has larger index than the other. And, incident angle is smaller than critical angle. TIR does not occur. How do you know this? Any explanations?
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    (Original post by StalkeR47)
    Amm sorry. I did not get you. Can you answer this please? Anyone please? why TIR does not occur if incident substance has larger refractive index than the other? Ca
    I implied that you're smart but nevermind lol. Sorry. Don't know. I'm already lost as it is..
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    Does anyone have any good notes for waves?
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    (Original post by masryboy94)
    it would occur BUT ONLY if it exceed the critical angle too ! hence why in optical fibres the core has a larger refractive index than cladding allowing T.I.R if above critical angle, and if it is not above critical angle you get refraction hence loss of signal causing the fibre to not be secure.
    Signals would not be secure if the ray is refracted by the cladding.:confused:
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    (Original post by StalkeR47)
    But how do you know that if the incident substance has larger index than the other. And, incident angle is smaller than critical angle. TIR does not occur. How do you know this? Any explanations?
    but you clearly quoted that the incident substance has larger refractive index in other words the core (incident substance) is of larger refractive index than the cladding
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    (Original post by x-Sophie-x)
    A dotted line reflected off the boundary, (when you have a ray grazing the boundary).

    I think
    What would the angle look like? Is it the same as when it hits the boundary? Also, where would the dotted line begin, exactly at the point at which the ray makes contact with the boundary?
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    Why des intensity of light decrease in single slit?


    Posted from TSR Mobile
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    For PIR, it's basically TIR but dotty, and aswell as the refracted wave.
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    Stalker, I'm not sure what you're trying to say exactly but I think you should accept that those guys are right haha
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    (Original post by UnknownRoyalist)
    I implied that you're smart but nevermind lol. Sorry. Don't know. I'm already lost as it is..
    Alright! Thanks man. Not really smart. I was just trying to clarify what I thought as in the exam paper question. And, I am still not sure. I know the rules properly but I need an explanations to why is that. Thanks.
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    (Original post by StalkeR47)
    Signals would not be secure if the ray is refracted by the cladding.:confused:
    ofc not, because since the waves refracted and didnt T.I.R the signal would escape hence why it needs to exceed critical angle. BUUUUUT since refractive index of cladding is more than air (given it exceeds critical angle at that point) it would then T.I.R. back. into the core
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    What does everyone think the 6 marker will be?
 
 
 
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