# Refractive index inside a fibre

The refractive index $n$ of a fibre is calculated as; $$n = c_0/c_m$$

Where $c_0\approx 300000km/s$ is the speed of light in a vacuum, and $c_m$ is the speed of light in the fibre in question. How can I calculate the speed of light in a fibre?

I am a complete physics novice so answers need to be in laymen's terms; Is there a simple equation that can be used to calculate $c_m$ given a set of properties of a fibre's manufacturing perhaps?

UPDATE

The reason I am asking this (after reading the answers I am getting) is because from the refractive index I can then go on to calculate other things like the Numerical Aperture. Perhaps I should be going around this another way?

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It's normally the other way around - you know the fibre's refractive index from the mix of glass types used to make it. This is what you need to control very-very carefully to make commercial fibre.

Edit: Thinking about it - the best way to calculate 'n' for a finished fibre is probably to measure the speed of light. Measuring the time between pulses arriving to parts in 10^9 is fairly easy with regular lab equipment, measuring 'n' by a refractometer is a lot less accurate.

My guess is you would use a laser pulse as the source and a splitter which put the light down both a short length of fibre (eg 1m) and a longer known length. Then measure the time difference in arrival at the end of each - the rise time of the source and response of the detector would then cancel out.

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Ah! I see, I was approaching this from a purely theoretical point not a practical one. As I have now updated my question, perhaps my intention is clearer. I thought I needed to start by calculating the refractive index before I could then go on to calculate the NA, V number et all, so this was like a fundamental starting point? –  jwbensley Nov 18 '12 at 23:55

There are other relationships involving the refractive index which are more practical than using the 'speed of light'. For example, since the velocity of a wave is related to its wavelength, then

$n=\frac{c_{0}}{c_{m}}=\frac{\lambda_{0}}{\lambda_{m}}$

Another one is Snell's law, which gives the relationship between the incident angle and refraction angle as light passes from one transparent medium to another (eg: air to water).

The formula is:

$n_{1}sin{\theta_{1}}=n_{2}sin{\theta_{2}}$

where $n_{1}$, is the refractive index of medium 1, $\theta_{1}$ is the angle of incidence (measured from the normal, which is perpendicular to the surface), $n_{2}$ is the refractive index of medium 2 and $\theta_{2}$ is the angle of refraction (measured from the normal).

When light travels from a medium of higher refractive index into a lower refractive index medium (such as water to air), there is a critical angle beyond which all the incident light is reflected. This is known as Brewster's angle, given by:

$\theta_{B}=arctan(\frac{n_{2}}{n_{1}})$

By measuring this angle, the relative refractive index can be determined.

A refractometer is often used to measure the refractive index of an ionic liquid (such as sea water) using the angle of refraction. A drop of liquid is placed on a lens and when viewed through the eyepiece, a line appears on a graduated scale indicating the refractive index.

For more accurate measurements, microscopes especially designed for measuring angles of incidence and refraction are available.

Of course, other techniques exist for determining the refractive index of other transparent media, such as glass, crystals, polymers, oils or gases. The appropriate technique depends on the nature of the medium and the level of accuracy required.

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I see, this is all very interesting. I have updated my question please see the updates, thanks for the info so far though, its great! :) –  jwbensley Nov 18 '12 at 23:50
Brewster's angle is an angle where there is complete transmission, not complete reflection. What you described is the critical angle of total internal reflection. –  The Photon Oct 5 '13 at 3:14