ppt
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Why will total internal reflection occur when angle of incidence is greater than the critical angle? Anything deal with the particles?
Why does total internal reflection occur above the critical angle?
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hemmul
ppt said:
Because the medias have such properties (resulting in certain refraction index), that there is no transmitted propagating wave. Some part of energy forms a standing wave parallel to the boundary surface in the incidence plane, but the major part is reflected back to the source media.
ppt
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why do the medias have such properties?
hemmul
well, well, well...
) (as they say: "es ist nich leicht ein gott zu sein")who knows? we just observe the nature. it turns out that the properties are like that! However there exist some special investigations dealing with artificial medias (keywords: metamaterials, chirality, nonreciprocity) that are quite perspective...
now if ask what makes up the refraction index - n=sqrt(epsilon*miu) - (square root of relative diel. permittivity and magn. permeability product). So the refraction index depends on EM properties of the matter - what kind of electric or magnetic dipole moment do atoms possesses while interaction with the wave, surely their resonance properties and so on...
it's be better if you state the question more specifically ;)
The_Thinker
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Here is another way of thinking abt it; i pretty much made it up right now but it might help understand it better, p.s this is not to be taken too seriously, but it's an explanation that should suffice for now...
Think of it this way, light has a certain amt of energy given by E = hf. Right, now water is one medium and air is another and they are both horizontal, like a pool water on the bottom and air on the top, water has surface tension,
now light wanting to get out of it will have to go upward, so resolve vectorally, cos90 being zero and cos zero being 1... Therefore the light will have the greatest y-axis component when the light is at 90 and the least when light is at 0 i.e parallel to the surface of water... Therefore light will have least amt of problem getting through the surface tension of water at 0 degrees and the maximum problem of getting out of it 90 degrees... Therefore depending on the refractive indeces or the diffrence in pressure at the join of the two surfaces, we have a critical angle...
I hope this makes some sense...
Guyz please go easy on me...
Think of it this way, light has a certain amt of energy given by E = hf. Right, now water is one medium and air is another and they are both horizontal, like a pool water on the bottom and air on the top, water has surface tension,
now light wanting to get out of it will have to go upward, so resolve vectorally, cos90 being zero and cos zero being 1... Therefore the light will have the greatest y-axis component when the light is at 90 and the least when light is at 0 i.e parallel to the surface of water... Therefore light will have least amt of problem getting through the surface tension of water at 0 degrees and the maximum problem of getting out of it 90 degrees... Therefore depending on the refractive indeces or the diffrence in pressure at the join of the two surfaces, we have a critical angle...
I hope this makes some sense...
Guyz please go easy on me...