zed963
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Why does n2>n1 (refractive index) for TIR to occur ?


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Joinedup
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(Original post by zed963)
Why does n2>n1 (refractive index) for TIR to occur ?


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Have you done the lab yet? - I found critical angle & TIR to be pretty self explanatory when you're shining a light beam through a block at different angles. As you increase the angle (to the normal) you'll reach a point where the angle of refraction WOULD BE > 90 degrees... the light just doesn't have anywhere to go outside the higher refractive index material - so it must stay inside.

it doesn't happen the other way round because light crossing a boundary from low to high refractive index is always refracted at a lower angle to the normal than the incident light - which means it must always be less than 90 degrees to the normal.
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zed963
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(Original post by Joinedup)
Have you done the lab yet? - I found critical angle & TIR to be pretty self explanatory when you're shining a light beam through a block at different angles. As you increase the angle (to the normal) you'll reach a point where the angle of refraction WOULD BE > 90 degrees... the light just doesn't have anywhere to go outside the higher refractive index material - so it must stay inside.

it doesn't happen the other way round because light crossing a boundary from low to high refractive index is always refracted at a lower angle to the normal than the incident light - which means it must always be less than 90 degrees to the normal.
Kind of makes sense tbh, I was just wondering if there was some mathematical proof.
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(Original post by zed963)
Why does n2>n1 (refractive index) for TIR to occur ?


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Work it out - use Snell's Law.
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zed963
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(Original post by natninja)
Work it out - use Snell's Law.
Sin i / Sin r ?
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(Original post by zed963)
Sin i / Sin r ?
nisin(i)=nrsin(r)

In air, ni=1 so it simplifies.

Basically anything where the angle of refraction is greater than 90 degrees gives TIR.
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zed963
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(Original post by natninja)
nisin(i)=nrsin(r)

In air, ni=1 so it simplifies.

Basically anything where the angle of refraction is greater than 90 degrees gives TIR.
Why about the angle of 90? Isn't it above the critical angle ?
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(Original post by zed963)
Why about the angle of 90? Isn't it above the critical angle ?
The idea is that if you set the angle of refraction for light leaving a medium then if the angle of refraction is greater or equal to ninety degrees then the light never leaves the medium and TIR occurs. When the angle of refraction is greater than 90 degrees then the angle of incidence is equal to the critical angle.
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zed963
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(Original post by natninja)
The idea is that if you set the angle of refraction for light leaving a medium then if the angle of refraction is greater or equal to ninety degrees then the light never leaves the medium and TIR occurs. When the angle of refraction is greater than 90 degrees then the angle of incidence is equal to the critical angle.
Confusing.

AFAIK if the angle of incidence is greater than the critical angle, then TIR occurs, if it isn't then refraction occurs.
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(Original post by zed963)
Confusing.

AFAIK if the angle of incidence is greater than the critical angle, then TIR occurs, if it isn't then refraction occurs.
Correct, and TIR is also when the angle of refraction is greater than 90​o
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