Optics A level physics question

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username3477548
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#1
Report Thread starter 9 months ago
#1
In my notes, it says that an advantage of a smaller diameter core (of an optical fibre) is that there's a smaller change of angle between each reflection, so it reflects more time in a given length, which keeps the incident angle greater?
I don't understand why a smaller diameter core means a smaller change in angle which means it reflects more times?😭😭
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Callicious
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#2
Report 9 months ago
#2
(Original post by Qxi.xli)
In my notes, it says that an advantage of a smaller diameter core (of an optical fibre) is that there's a smaller change of angle between each reflection, so it reflects more time in a given length, which keeps the incident angle greater?
I don't understand why a smaller diameter core means a smaller change in angle which means it reflects more times?😭😭
The change in angle thing seems a bit dodgy to me. If you just take a perfectly straight fibre as example, the change in angle for each reflection is the exact same... the incident angle to the interior surfaces remains identical indefinitely... in any case I can't see the relevance in that statement against angle of incidence.

One point that is relevant though is the number of reflections per unit length, which does actually depend on the width of the fibre. Consider some length x of fibre with light travelling inside it, reflecting at some angle \phi against the normal to the internal periphery of the fibre. If the fibre has a width d, you can estimate the average time it takes between the reflections, as

\bar{t}_{TIR} = \frac{d}{c_n\cos\phi}

where c_n is just the speed of the ray inside the fibre medium, which has n optical index. The value of \bar{t} obviously depends directly on d if you fix the denominator, which I'm going to cover now.

The total time spent travelling across the fibre for the setup described is

\bar{t} = \frac{x}{c_n\sin\phi}

and assuming we fix the denominator once more, directly depends on x and should be identical for both fibres if we assume all the angles involved in incidence/etc are the same. Again draw it out if you have to convince yourself.

In any case, since we know the time \bar{t} spent traversing the fibre and also know the average time spent for each reflection event as \bar{t_{TIR}}, we can get an expression for the average number of reflections for each passage,

N_{TIR} = \frac{x}{d} \frac{\cos\phi}{\sin\phi}

Now, consider two fibres, of identical length, and perfectly straight/parallel to each other. Their d differ, though: everything else is the same.

Obviously the wider one will have less reflection events: the total transverse distance traversed by light in it will be the same as for the thinner one, but since the transverse distance per transit is larger for the wider fibre, less transits are required to satisfy this requirement.

Anyway hope that helps... here's a crude diagram... Name:  hoaa.png
Views: 15
Size:  188.4 KB
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username3477548
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#3
Report Thread starter 9 months ago
#3
(Original post by Callicious)
The change in angle thing seems a bit dodgy to me. If you just take a perfectly straight fibre as example, the change in angle for each reflection is the exact same... the incident angle to the interior surfaces remains identical indefinitely... in any case I can't see the relevance in that statement against angle of incidence.

One point that is relevant though is the number of reflections per unit length, which does actually depend on the width of the fibre. Consider some length x of fibre with light travelling inside it, reflecting at some angle \phi against the normal to the internal periphery of the fibre. If the fibre has a width d, you can estimate the average time it takes between the reflections, as

\bar{t}_{TIR} = \frac{d}{c_n\cos\phi}

where c_n is just the speed of the ray inside the fibre medium, which has n optical index. The value of \bar{t} obviously depends directly on d if you fix the denominator, which I'm going to cover now.

The total time spent travelling across the fibre for the setup described is

\bar{t} = \frac{x}{c_n\sin\phi}

and assuming we fix the denominator once more, directly depends on x and should be identical for both fibres if we assume all the angles involved in incidence/etc are the same. Again draw it out if you have to convince yourself.

In any case, since we know the time \bar{t} spent traversing the fibre and also know the average time spent for each reflection event as \bar{t_{TIR}}, we can get an expression for the average number of reflections for each passage,

N_{TIR} = \frac{x}{d} \frac{\cos\phi}{\sin\phi}

Now, consider two fibres, of identical length, and perfectly straight/parallel to each other. Their d differ, though: everything else is the same.

Obviously the wider one will have less reflection events: the total transverse distance traversed by light in it will be the same as for the thinner one, but since the transverse distance per transit is larger for the wider fibre, less transits are required to satisfy this requirement.

Anyway hope that helps... here's a crude diagram... Name:  hoaa.png
Views: 15
Size:  188.4 KB
Hey tysmm this is amazing 💛 have a good day x
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