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The Physics PHYA2 thread! 5th June 2013

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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.


But that wasn't the point?

We were wondering whether it can TIR if one condition is true.
Reply 1221
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Qari
20
Original post by StalkeR47
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?


Ok, then let me ask you a question,

If angle is greater than critical angle but it is moving from less dense medium to higher dense medium will TIR occur?
Original post by StalkeR47
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?


Because the angle of incidence is smaller than the critical angle?!

What on earth am I missing here?!
for the formation of a stationary wave, why is it important that the reflected waves lose only a small amount of energy?

surely all that matters if for the waves to have the same frequency and wavelength....
Guys what happens when you put white light through the double slit experiement
and why dont you normally ?
Reply 1225
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NabRoh
11
Original post by StalkeR47
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?

Lets look at the question:
- Refractive index of cladding is lower than refractive index of core.
CHECK. This meets the first criteria of TIR
- The angle of incidence is LOWER than the critical angle
CROSS. This does not meet the second criteria of TIR

It met the first criteria but not the second and as you can see it refracts and does not TIR.
If both criteria are not met, TIR does not take place.
Original post by x-Sophie-x
Ah crap, everyone's saying different things :/



okay basically if n2 > n1 you just can't get T.I.R but refraction. because n2 is optically more dense, thats the reason why cladding in optical fibres has a lower refractive index so that T.I.R can occur hence n1 > n2, but when n1 > n2 but the angle of incidence is less than the critical angle you will get still get refraction, therefore BOTH HAVE TO HAPPEN
(edited 10 years ago)
Original post by x-Sophie-x
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:

Yes exactly! More refractive to less refractive -----TIR should occur. But, incident angle is smaller than Critical angle, TIR does not occur. So, refraction occurs. Confusing this is, isn't it?:confused:
Reply 1228
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Qari
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Original post by StalkeR47
Yes exactly! More refractive to less refractive -----TIR should occur. But, incident angle is smaller than Critical angle, TIR does not occur. So, refraction occurs. Confusing this is, isn't it?:confused:


Answer my question?
Original post by StalkeR47
Yes exactly! More refractive to less refractive -----TIR should occur. But, incident angle is smaller than Critical angle, TIR does not occur. So, refraction occurs. Confusing this is, isn't it?:confused:


I get you! yaaaay! :biggrin:

Thank you so much! :smile:
Original post by Jack93o
for the formation of a stationary wave, why is it important that the reflected waves lose only a small amount of energy?

surely all that matters if for the waves to have the same frequency and wavelength....


They have to have simular amplitude so that a node is formed
Original post by yorkshire.lad
Any predictions as to what the 6 marker might be guys?


it was waves in january, so I'm guessing (hoping) it'll be mechanics or materials for the long answer.
Hope we get a nice paper tomorrow... :sigh:

I've sort of slowed down my revision after facing burnout so I haven't done much for this last exam, especially as it's a retake! :redface:
Original post by NabRoh
Lets look at the question:
- Refractive index of cladding is lower than refractive index of core.
CHECK. This meets the first criteria of TIR
- The angle of incidence is LOWER than the critical angle
CROSS. This does not meet the second criteria of TIR

It met the first criteria but not the second and as you can see it refracts and does not TIR.
If both criteria are not met, TIR does not take place.


Yes! So, if only 1 criteria is met, only refraction takes place. Am I right?
Original post by masryboy94
okay basically if n2 > n1 you just can't get T.I.R but refraction. because n2 is optically more dense, thats the reason why cladding in optical fibres has a lower refractive index so that T.I.R can occur hence n1 > n2, but when n1 > n2 but the angle of incidence is less than the critical angle you will get still get refraction, therefore BOTH HAVE TO HAPPEN


I actually thought so too, but StalkeR47 has a point.
Have you had a looked at Question 6 on June 2009? :smile:
Reply 1235
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NabRoh
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Original post by UnknownOrigin
it was waves in january, so I'm guessing (hoping) it'll be mechanics or materials for the long answer.

To be fair the waves was one was very easy so expect a harder 6 marker.
Reply 1236
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OliverG
2
when will I get partial refraction?
Reply 1237
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Qari
20
Original post by x-Sophie-x
I get you! yaaaay! :biggrin:

Thank you so much! :smile:


Trust us both have to occur
Original post by StalkeR47
Yes exactly! More refractive to less refractive -----TIR should occur. But, incident angle is smaller than Critical angle, TIR does not occur. So, refraction occurs. Confusing this is, isn't it?:confused:


EXACTLYYY !!!! therefore you have to have more refractive to less refractive AND greater than critical angle for T.I.R
Original post by StalkeR47
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.


In simpler terms, for TIR to occur:

(n2<n1)and(θi≥θc)(n^2<n^1) and (\theta^{i}\geq \theta^{c})
(edited 10 years ago)

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