Shadow20074
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#1
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Hi there, to anyone who has this paper...
I've done fine up until question 7(c).
It asks me to calculate the Critical Angle of the glass-liquid boundary.
But I can't for the life of me see how to do this without knowing both Reflective Indexes.
I know the R.Index of the glass, but not of the liquid.
The refractive angle would be 90.....but how do I work this out?:confused:
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na93ice
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(Original post by Shadow20074)
Hi there, to anyone who has this paper...
I've done fine up until question 7(c).
It asks me to calculate the Critical Angle of the glass-liquid boundary.
But I can't for the life of me see how to do this without knowing both Reflective Indexes.
I know the R.Index of the glass, but not of the liquid.
The refractive angle would be 90.....but how do I work this out?:confused:
Sin(criticle angle) is refractive index of 1st medium divided by the refractive index of 2nd medium

refractive index of glass =1.5
refractive index of water = 1.33

sin (CritA) = 1.33/1.5

sin^-1(1.33/1.5) = 62.5 degrees
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na93ice
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#3
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btw you got phya2 today?
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Zuzuzu
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This was a really ****ing stupid question that epitomised the June 2010 paper. IIRC, the 'calculation' was 90 - 27 giving an answer of 63.
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Shadow20074
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woo thanks!!
yeah, pretty nervous
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Shadow20074
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(Original post by Zuzuzu)
This was a really ****ing stupid question that epitomised the June 2010 paper. IIRC, the 'calculation' was 90 - 27 giving an answer of 63.
90-27 :eek:
I really hate some of these questions
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na93ice
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ive got this paper too, best of luck
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Shadow20074
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(Original post by na93ice)
ive got this paper too, best of luck
Good luck to you too!
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LucyO_103
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#9
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Can someone post a pic of how 7(a) is supposed to look? I'm really stuck, thaaaaanks
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StalkeR47
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(Original post by LucyO_103)
Can someone post a pic of how 7(a) is supposed to look? I'm really stuck, thaaaaanks
Hi! Sure! I have got that exam on 5th of June. Good luck to you!
Attached files
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LucyO_103
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(Original post by StalkeR47)
Hi! Sure! I have got that exam on 5th of June. Good luck to you!
THANKS!!
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StalkeR47
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#12
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(Original post by LucyO_103)
THANKS!!
No problem!
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R2D2.
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#13
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I dont understand how this question works. It says we have to spot that it passes the critical angle at that stage. But how are we meant to know that the refractive index of water is 1.33?

edit* never mind. The start of the question says it gets internally reflected for the first time at that angle so we have to assume that's the critical angle :P

I hate questions that require you to think like this
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DenisDaniel
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Thats not true coz it doenst say the liquid is water
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DenisDaniel
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#15
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(Original post by na93ice)
Sin(criticle angle) is refractive index of 1st medium divided by the refractive index of 2nd medium

refractive index of glass =1.5
refractive index of water = 1.33

sin (CritA) = 1.33/1.5

sin^-1(1.33/1.5) = 62.5 degrees
(Original post by Zuzuzu)
This was a really ****ing stupid question that epitomised the June 2010 paper. IIRC, the 'calculation' was 90 - 27 giving an answer of 63.
Ye sure????
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Wazzz108
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#16
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How did u do that? Like i mean why do it, whats the significance of subtracting and the first time part
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