# A level physics fibre optics

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Hi. I've never asked a question on TSR before so I don't know if it's in the right place or not.

Basically, in my physics book the question is:

An optical fibre core is made of glass of refractive index 1.472 and is surrounded by a cladding of refractive index 1.455. [then there is a diagram showing a ray bouncing from side to side through the core]. The ray travels through step-index propagation and bounces at the critical angle. Take c=2.997x10^8

Calculate how long energy takes to travel along a 3000m fibre in this mode of propagation.

The answer in the back of the book is: 14.56 microseconds. I have no idea how to get there because I have tried so many times and each time I got a completely different answer.

Any help would be much appreciated!

Basically, in my physics book the question is:

An optical fibre core is made of glass of refractive index 1.472 and is surrounded by a cladding of refractive index 1.455. [then there is a diagram showing a ray bouncing from side to side through the core]. The ray travels through step-index propagation and bounces at the critical angle. Take c=2.997x10^8

Calculate how long energy takes to travel along a 3000m fibre in this mode of propagation.

The answer in the back of the book is: 14.56 microseconds. I have no idea how to get there because I have tried so many times and each time I got a completely different answer.

Any help would be much appreciated!

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#2

(Original post by

Hi. I've never asked a question on TSR before so I don't know if it's in the right place or not.

Basically, in my physics book the question is:

An optical fibre core is made of glass of refractive index 1.472 and is surrounded by a cladding of refractive index 1.455. [then there is a diagram showing a ray bouncing from side to side through the core]. The ray travels through step-index propagation and bounces at the critical angle. Take c=2.997x10^8

Calculate how long energy takes to travel along a 3000m fibre in this mode of propagation.

The answer in the back of the book is: 14.56 microseconds. I have no idea how to get there because I have tried so many times and each time I got a completely different answer.

Any help would be much appreciated!

**Lizzie Coldrick**)Hi. I've never asked a question on TSR before so I don't know if it's in the right place or not.

Basically, in my physics book the question is:

An optical fibre core is made of glass of refractive index 1.472 and is surrounded by a cladding of refractive index 1.455. [then there is a diagram showing a ray bouncing from side to side through the core]. The ray travels through step-index propagation and bounces at the critical angle. Take c=2.997x10^8

Calculate how long energy takes to travel along a 3000m fibre in this mode of propagation.

The answer in the back of the book is: 14.56 microseconds. I have no idea how to get there because I have tried so many times and each time I got a completely different answer.

Any help would be much appreciated!

*greater*than the critical angle,

*then*it will bounce from side to side through the core in the way you describe. We also don't need to know the refractive index of the cladding, unless there are other parts to the question that might need that (for example, if you are asked to calculate the critical angle).

I expect they are wanting you to work out the speed of light in the core and then work out how long it would take for light moving at this speed to travel 3000 m. Do you know how to find the speed of light in the core?

(When I do this, I get 14.73 microseconds, so close to their result but not quite the same, which is surprising considering the number of significant figures they are using. Not at all impossible I'm missing something!)

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#3

It is a strange question. And the answer given is what you get if you use the refractive index of the cladding rather than the core.

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#4

(Original post by

It is a strange question. And the answer given is what you get if you use the refractive index of the cladding rather than the core.

**Stonebridge**)It is a strange question. And the answer given is what you get if you use the refractive index of the cladding rather than the core.

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(Original post by

So it does! And I wonder if this gives a hint to what they are on about, and raises a question I had never thought about before. If the ray hits the boundary at exactly the critical angle, which medium is it now travelling in? The core or the cladding?

**Pangol**)So it does! And I wonder if this gives a hint to what they are on about, and raises a question I had never thought about before. If the ray hits the boundary at exactly the critical angle, which medium is it now travelling in? The core or the cladding?

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I think (I hope) I have attached the file with a picture of the question if it helps...

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#7

The diagram implies the ray travels through the core, so we can only assume that whoever worked out the answer for the book publisher made a very elementary mistake.

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(Original post by

The diagram implies the ray travels through the core, so we can only assume that whoever worked out the answer for the book publisher made a very elementary mistake.

**Stonebridge**)The diagram implies the ray travels through the core, so we can only assume that whoever worked out the answer for the book publisher made a very elementary mistake.

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I just found this: the critical angle is the angle at which the refracted ray will travel along the boundary between the two media. So technically, the refracted ray is kind of in between the two media. However, remember there is also a partially reflected ray (which will obey the law of reflection i.e. angle of incidence = angle of reflection). The partially reflected ray will be much stronger than the refracted ray at the critical angle. The pip question c) is asking about the partially reflected ray.

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#10

**Lizzie Coldrick**)

Hi. I've never asked a question on TSR before so I don't know if it's in the right place or not.

Basically, in my physics book the question is:

An optical fibre core is made of glass of refractive index 1.472 and is surrounded by a cladding of refractive index 1.455. [then there is a diagram showing a ray bouncing from side to side through the core]. The ray travels through step-index propagation and bounces at the critical angle. Take c=2.997x10^8

Calculate how long energy takes to travel along a 3000m fibre in this mode of propagation.

The answer in the back of the book is: 14.56 microseconds. I have no idea how to get there because I have tried so many times and each time I got a completely different answer.

Any help would be much appreciated!

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reply

(Original post by

Note that total internal reflection happens when light meets the cladding at which refractive index is lower than the glass. Essentially you need to find the velocity at which it travels through the medium. n=c/v ( use the cladding refractive index), rearranging v=n/c. Use the formula for velocity to find the time. 1.455x10^-5 or 14.56 microseconds.

**ThiagoBrigido**)Note that total internal reflection happens when light meets the cladding at which refractive index is lower than the glass. Essentially you need to find the velocity at which it travels through the medium. n=c/v ( use the cladding refractive index), rearranging v=n/c. Use the formula for velocity to find the time. 1.455x10^-5 or 14.56 microseconds.

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#12

(Original post by

I just found this: the critical angle is the angle at which the refracted ray will travel along the boundary between the two media. So technically, the refracted ray is kind of in between the two media. However, remember there is also a partially reflected ray (which will obey the law of reflection i.e. angle of incidence = angle of reflection). The partially reflected ray will be much stronger than the refracted ray at the critical angle. The pip question c) is asking about the partially reflected ray.

**Lizzie Coldrick**)I just found this: the critical angle is the angle at which the refracted ray will travel along the boundary between the two media. So technically, the refracted ray is kind of in between the two media. However, remember there is also a partially reflected ray (which will obey the law of reflection i.e. angle of incidence = angle of reflection). The partially reflected ray will be much stronger than the refracted ray at the critical angle. The pip question c) is asking about the partially reflected ray.

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(Original post by

What value have you got for the critical value?

**ThiagoBrigido**)What value have you got for the critical value?

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#14

I repeat.

The speed of the light that is travelling through the

(The cladding index (together with the core) is used to find the critical angle.)

The answer in the book is wrong. It happens a lot. They used the wrong value.

I would be interested to hear any logical reason (or get a link to an article) explaining why the cladding index would be used to find the speed of light in the core, and not the core index.

The speed of the light that is travelling through the

**core**of the fibre (as shown bouncing around in the question) due to internal reflections, is determined using the refractive index of the**core**. (Not the cladding).(The cladding index (together with the core) is used to find the critical angle.)

The answer in the book is wrong. It happens a lot. They used the wrong value.

I would be interested to hear any logical reason (or get a link to an article) explaining why the cladding index would be used to find the speed of light in the core, and not the core index.

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#15

Here's an interesting thing. I thought I recognised the format of the question, and I managed to find where I recalled it from. It's from Practice in Physics by Akrill, Bennet and Millar. I have the fourth edition, where it is question 8.35, so the OP either has an earlier or later edition than I do. The question is identical. But the answers at the back of the book seem to differ from those the OP sees in her edition.

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#16

(Original post by

I repeat.

The speed of the light that is travelling through the

(The cladding index (together with the core) is used to find the critical angle.)

The answer in the book is wrong. It happens a lot. They used the wrong value.

I would be interested to hear any logical reason (or get a link to an article) explaining why the cladding index would be used to find the speed of light in the core, and not the core index.

**Stonebridge**)I repeat.

The speed of the light that is travelling through the

**core**of the fibre (as shown bouncing around in the question) due to internal reflections, is determined using the refractive index of the**core**. (Not the cladding).(The cladding index (together with the core) is used to find the critical angle.)

The answer in the book is wrong. It happens a lot. They used the wrong value.

I would be interested to hear any logical reason (or get a link to an article) explaining why the cladding index would be used to find the speed of light in the core, and not the core index.

https://www.quora.com/What-is-the-fu...-optical-fibre

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(Original post by

Here's an interesting thing. I thought I recognised the format of the question, and I managed to find where I recalled it from. It's from Practice in Physics by Akrill, Bennet and Millar. I have the fourth edition, where it is question 8.35, so the OP either has an earlier or later edition than I do. The question is identical. But the answers at the back of the book seem to differ from those the OP sees in her edition.

**Pangol**)Here's an interesting thing. I thought I recognised the format of the question, and I managed to find where I recalled it from. It's from Practice in Physics by Akrill, Bennet and Millar. I have the fourth edition, where it is question 8.35, so the OP either has an earlier or later edition than I do. The question is identical. But the answers at the back of the book seem to differ from those the OP sees in her edition.

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#18

(Original post by

Yes, my book is practice in physics (my teachers call it pip for short). I have the third edition

**Lizzie Coldrick**)Yes, my book is practice in physics (my teachers call it pip for short). I have the third edition

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(Original post by

This strongly suggests that the answer we have been debating in your edition is wrong and was later corrected!

**Pangol**)This strongly suggests that the answer we have been debating in your edition is wrong and was later corrected!

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#20

(Original post by

So the answer in you'd is 14.91 microseconds? How would you get to that?

**Lizzie Coldrick**)So the answer in you'd is 14.91 microseconds? How would you get to that?

In part (c), the ray bounces from side to side, so it has to cover a total distance greater than 3000 m. We can work out the distance it has to travel using a bit of basic trigonometry, taking the angle of incidence to be the critical angle, and that leads to the given answer.

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