The answer for the following section (d) was originally wrong. Thanks to Chris for pointing this out (see the comments below). I had originally drawn the water tank taking up the whole space between the fibre and the screen, but if you read the question you can see that the tank is ‘small’ and the screen is ‘a long distance away from’ the end of the fibre, so the water tank must just take up a small part of the distance. Therefore the beam enters the water tank and exits the other side, before continuing on its way to the screen.

Hi, may I ask why is it in c, equation 2, you assume the reflected angle is 90 degrees? is it due to total internal reflection? but if thats the case how can the light travels along the optical fibre

Not sure I understand your question. Are you possibly misreading my diagram? — I have drawn in a dashed line to represent the normal to the point where the ray hits the edge of the glass. But that dashed line doesn’t represent the path of the ray, which I just didn’t bother drawing (but we know what it is because angle of incidence = angle of reflection). We need to draw the normal in because theta2 (the angle of incidence of the ray when it hits the edge of the glass) is the angle between the ray and the normal.

Does that make sense to you now, or am I missing something in your question?

Hi,

I got it already; there was some gap in my understanding in my previous question.. but thanks!

Hey!

Great blog by the way.

I just had a clarification to make. In Part c), shouldn’t it be cos^-1 > n(clad)/n(core)?

I reasoned this as such. Angle of incidence, i = 90-theta(max). This is because i is measured with respect to the normal.

Thanks — I’m really glad you like it. It’s always good to get nice comments.

I don’t think your point is right — if you look at the diagram in part (c) they have drawn and labelled the angle theta(max), and it is the angle between the ray and the normal. So, as drawn, angle of incidence i = theta(max), not 90 – theta(max). Does that make sense to you or am I missing something in your question?

Incidentally the answer I get is a well-known result (see e.g. https://en.wikipedia.org/wiki/Guided_ray, remembering that the refractive index of air is approximately equal to 1).

http://www.physicsclassroom.com/class/refrn/Lesson-4/Dispersion-of-Light-by-Prisms

I was trying to point out the fact that blue light refracts more than red light so the blue light should be furthest away from the centre.

In the above link, underneath the title the angle of deviation it says

“The angle of deviation is the angle made between the incident ray of light entering the first face of the prism and the refracted ray that emerges from the second face of the prism. Because of the different indices of refraction for the different wavelengths of visible light, the angle of deviation varies with wavelength. Colors of the visible light spectrum that have shorter wavelengths (BIV) will deviated more from their original path than the colors with longer wavelengths (ROY). ” and it has a nice picture showing this.

Yes the violet light gets refracted more than the red, but the physical situation is different from the prism example that you give here. In our case we have a diverging cone of light coming out of the fibre (i.e. the light is spreading out), and when we put a tank of water in front of it the light gets bent inwards, making the resulting spot a bit smaller. The more bending that is done by the tank, the smaller the spot. So in this case, when the refractive index gets bigger, and the light gets bent more, the spot gets smaller.

more bending for the blue light means a bigger angle of refraction which means the light ray is further from the normal and therefore on the screen it is further from the centre than red light?

Thanks for all the really helpful solutions.

I think for this part there will be no rainbow fringe. surely the bending of light at the 3 barriers cancel as the light originates outside the fibre. this means the different frequencies will be travelling at different angles inside the fibre which will all be aligned again when it leaves the fibre. all the barriers are parrellel so it is similar to shining a light through a block where no rainbow is formed.

Thanks for that — you make a very good point. My discussion of this particular answer is pretty weak, as well. I think your answer is probably right, but I would like to find a way of expressing the situation more mathematically. Maybe you can think of one? If so post something here.

Hi. Now that i’ve thought about it more, i think the blue light will be furthest out. if we consider some light that enters the light at a large angle of incidence, the light will disperse so that the blue light takes a more direct route (assuming a perfectly straight fibre, vertical entry, white light at different angles as shown in diagram) as the blue light will bend towards the normal more. the red end of the spectrum that has smaller angles of incidence will be absorbed over the length of the fibre so (at least more than the blue light that is closer to the straight line through) be removed. when the light doing larger angles reaches the end of the fibre, the blue will bend away from the normal the angle back to its original path when outside the fibre but as some of the extreme angles of the red end of spectrum have been removed, this will not be white but blue. i don’t know how light is absorbed but i think the different frequencies over the range will be absorbed at approximately the same angle so there will be a fade from white to blue as frequencies from the red end are gradually ommitted. the light going in will be at the same angle as light coming out according to snells law.

another thought is will there be a a slight diffraction so the single slit diffraction pattern may occur. my initial thought to this was the gap was too wide but after a calculation this isn’t unresonable and the angle is around 8 degrees for the first maxima (smallest diameter, average wavelenth). any thoughts on this? i’m unsure about how the different mediums affect diffraction but a standard single slit diffraction pattern may be seen in a radial pattern.

Hi Harry, thanks for your comment.

I think this is probably going well beyond what the examiners expected, but let’s dig into it a bit more. I have to confess that I am not totally sure of this, so if you have a better argument then do reply and explain your reasoning. Here’s my argument:

Look at the last diagram in the answer above, and imagine you could vary the refractive index of water. If the refractive index of water were reduced down to be the same as air, then this would be equivalent to the case where you didn’t have a tank there at all, so the spot would have width r+δr. Then imagine increasing the refractive index of water a bit — this would be equivalent to having a little tank there and the spot would get smaller. So, using this reasoning, as the refractive index gets bigger the spot will get smaller, so the violet spot (which has the biggest refractive index) should be the smallest.

Hi,

What colour would the spot be around which the rainbow forms- would it be white? If so why because in the refraction of white light through a prism, all the colours split and there is no white spot? Also how would the rainbow look- e.g. circular with red at the furthest point from the centre?

Thanks

Hi James, that’s a really good question. It’s a good question because the lazy answer to the question above is “well it must be a rainbow because that’s what happens with white light and materials with variable refractive indices”, and that is how I answered this question. But I had never thought much about the details, and that’s what Physics is really all about. So here are some more details.

The refractive index of transparent materials tends to increase with frequency of light — there is a nice graph at http://www.telescope-optics.net/chromatic.htm, which shows the actual numbers for several materials including water.

Imagine you had one single frequency of light, and you were able to vary the refractive index of water. Then as the refractive index of water increased, the spot of light would get bigger, and as the refractive index decreased the spot of light of light would get smaller (I’m pretty sure this is the case, but please someone correct me if I’ve got this the wrong way round :-)). But in every case we would still have a spot of light of that particular colour corresponding to that frequency.

So now imagine we have white light (a mixture of all visible frequencies) and the refractive index of water varies with frequency in the way it does in real life. We get a lot of spots of different sizes, but each with the same centre point. The violet spot is the smallest and the red spot is the biggest, because the refractive index is the biggest for violet and the smallest for red. When they were transmitted, the intensity of each of the original frequencies was the same but because the violet spot is smaller and the red spot is bigger, the intensity of the violet spot is a little bit higher, so in most of the spot the spectrum is a tiny bit shifted towards the violet, and at the edge of the spot the spectrum is shifted a long way to the red. And as you move in from the edge, the peak component of the spectrum moves through all the colours from red to violet.

I have to confess I have never seen this in real life, but I think that this means that you get a rainbow fringe, with the outside edge being red.

Surely, the furthest colour from the centre is violet as it is more refracted?

Hi Oxford PAT

Just a quick clarification. In part (a) You don’t seem to address the question’s request for what it means to have total internal reflection. I was wondering if you were being implicit about the phenomenon since you addressed the notion of critical angle. Additionally, your diagram doesn’t seem to address it as well so I’m just a bit wary of the answer.

Thanks

Hi Clemens. Thanks for your comment. Yes, you are absolutely right that I have not actually answered the question here. I have explained what the critical angle is, but I haven’t even mentioned total internal reflection.

As I am sure you know, the definition of total internal reflection should go something like this: A small amount of light is reflected back into the fibre when the ray hits the junction between the two mediums at any angle of incidence. When the angle of incidence goes beyond the critical angle it is no longer possible for light to leave the fibre as we can see from the diagram. The light has to go somewhere, so all of it is reflected back in to the fibre — hence the name ‘total internal reflection’.

Thanks very much for your contribution. This site depends on comments like these to point out errors and make the answer better …

No problem 🙂 I was just quite nervous that the answer does not need to be respond to the question explicitly, so had to clarify.

hi i was thinking what if i leave my answer for part c as (cos^-1)(n clad/ n core ) is this still correct?

Hi Lee, thanks for your question. Your formula is not the value of

theta(max). Looking at my diagram(cos^-1)(n clad/ n core )is the value oftheta1, the angle between the ray of light travelling down the fibre and the axis of the fibre.The answer says ‘refractive index varies with wavelength of light, so there will be a rainbow fringe around the spot’. That is, just like the famous prism experiment, we will see a rainbow effect that results from the variable refraction of different frequencies by the refractive material.

This answer actually depends on a couple of facts that, while true, are a bit more complicated than the simple answer would allow:

1. Lasers only emit a single frequency of light. This is generally true, but some lasers will emit several frequencies.

2. Materials have refractive indexes that vary with frequency. Again this is generally true, but the effect is complex and variable.

What is the answer at part e? Sorry,but i don’t understand your writing.

Hi Harry, thanks for your question.

Theta1 is the angle between the ray and the perpendicular to the interface between the air and the core. Theta2 is the angle between the ray and the perpendicular to the interface between the core and the cladding (aka the angle of incidence of the ray at the interface between the core and the cladding).

In part b we derived a formula for the ‘critical’ angle of incidence, and in this question we are interested in the case where theta2 is the critical angle. This is how we get to the equation

n(core) sin theta2 = n(clad).Finally, in part d the question, the formula with the tangents in it is to find the distance

delta r, which is the amount by which the radius of the sport decreases. It’s not strictly required by the question but it seemed worth adding. To understand it, just notice that the width of the tank iswand it should be easy to spot thatdelta r = w(tan theta(air) – tan theta(water)).thanks

for part c, it’s not clear what theta 1 and 2 are on the diagram so i don’t understand n(cone)sintheta2=n(clad) because doesn’t the angle have to be greater than the critical angle for total internal reflection? also why is theta2=pi/2-theta1. finally for part d, where does tan come into it?

That’s a good question. Yes it is correct. In fact it is a well-known result: see https://en.wikipedia.org/wiki/Numerical_aperture#Fiber_optics. But you are absolutely right to point out that this is not “in terms of the core and cladding refractive indices only” as the question asked.

Of course, in normal engineering situations we generally treat the refractive index of air as being equal to 1 (it’s actually about 1.0003). However, earlier in the question they talked about “the refractive index of air” so it didn’t seem appropriate to eliminate it. Maybe the question is asking us to state that the refractive index of air is approximately equal to 1 and therefore substitute in to eliminate the

n_airterm.Many thanks for that (& also the wikipedia link!). That helped a great deal.

Just one last thought: for part d), do you not have to also take into account the ray exiting the small water-filled glass tank and therefore the water/air interface at that point?

Ah yes you are absolutely right, thanks for pointing this out. I misread the question and thought that the entire space between the fibre and the screen was occupied by the water tank. In fact, if you read it closely, the question clearly does specify that the tank actually only covers a small space next to the fibre.

I’ve changed the answer for part (d) so that it shows the correct scenario.

The answer for part (e) is still the same.

In the above derivation of Theta (max), your final expression still involves the n (air) term. Is your solution correct, therefore?