Competition - post a photo you've taken of nature >>

A level physics fibre optics

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!

Scroll to see replies

Reply 1

Pangol

Original post by Lizzie Coldrick

There's something about the phrasing of that question that it a bit odd. If the fibre is straight and the ray is incident at the critical angle, it will travel along the core/cladding boundary. If the angle of incidence is 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!)

Reply 2

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.

Reply 3

Pangol

Original post by 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.

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?

Reply 4

Lizancol123

Original post by Pangol

So do you think the answer is correct in that case? The book it came from has been known several times to give the incorrect answer! (Also, sorry for the confusion as to why other numbers are there - there are previous questions that needed that data)

Reply 5

Lizancol123

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

15.44.jpg320.9KB

Reply 6

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.

Reply 7

Lizancol123

Original post by 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.

Thanks! That's what I thought too, but other people in my class seem to have no problem (including my teacher). Do you know by any chance what the correct answer would be?

Reply 8

Lizancol123

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.

Reply 9

ThiagoBrigido

Original post by Lizzie Coldrick

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.

Reply 10

Lizancol123

Original post by ThiagoBrigido

Oh okat then, so the book is right. Thats cool, thanks

Reply 11

ThiagoBrigido

Original post by Lizzie Coldrick

What value have you got for the critical value?

Reply 12

Lizancol123

Original post by ThiagoBrigido

What value have you got for the critical value?

I did thetaC = Sin^-1(1.455/1.472) = 81.28 deg.

Reply 13

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.

Reply 14

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.

Reply 15

ThiagoBrigido

Original post by Stonebridge

The TIR is only possible because the cladding along the core.

https://www.quora.com/What-is-the-function-of-cladding-in-optical-fibre

Reply 16

Lizancol123

Original post by Pangol

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

Reply 17

Pangol

Original post by Lizzie Coldrick

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

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

Reply 18

Lizancol123

Original post by Pangol

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

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

Reply 19

Pangol

Original post by Lizzie Coldrick

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

In part (b), we are taking the ray to act in a straight line parallel to the fibre, so the total distance it has to travel is 3000 m.

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.