# If someone can tell me about light

1. Apr 14, 2005

### Tsunami

...that would be welcome.

I'm a third year Engineering student, majoring in Physics. This year I have an introductory course in Photonics. Unfortunately, introductory says it: as soon as I try to figure something out in depth, I have to look it up on the net.

I'm having trouble finding info about the following:

when light is transmitted through a wave conductor (in real life situations this wave conductor will usually be an optical fiber), the light always stays inside the conductor because there is total optical reflection at the walls. This is because the refractive index is higher in the inner medium, than it is in the outer medium (that's still pretty straightforward).

An important quality of wave conductors is that it provides a good way to make light take a curve : it's perfectly possible to make a wave conductor in a shape of a quarter-circle, and thus make the light take a smooth curve of 90°. However, my course notes say, there will always be a certain loss of light in this curve , and you can minimize this loss by making the radius of the circle greater (I mean 'the curving ray', not the radius; the property that describes the distance from the geometrical middle of the curving surface to the edge - I don't know the correct English term). Also, it says that when the difference in refractive index between outer and inner medium (n_inner-n_outer) is greater, the radius can also be choosen a little smaller.

I wouldn't know what that is based on. So my question is: how does the radius (or curving ray) and the difference in refractive index relate to the amount of light that is lost (read: gets transmitted) through the walls of the wave conductor?

I hope I'm posting in the right category, this is my first visit here. Hopefully I'll be able to answer some questions for other people too instead of merely asking them.

Whatever the case, I thank you for at least reading my question in full, and for whatever help you wish to give me in addition to that.

2. Apr 14, 2005

### ZapperZ

Staff Emeritus
You posted in the wrong section. You want the Homework help section. Instead, you went into Academic and Career advise sectioin.

Zz.

3. Apr 14, 2005

### DaveC426913

Are you looking for a quantitative (formulaic) answer, or merely a conceptual understanding of why it is so?

4. Apr 14, 2005

### Tsunami

Usually, if someone passes me the formula, then I can figure out why it is so as well.

But if you can give me either, that would be great.

Zapper : I wasn't sure, since my question is of academic level... I'm always confused when it comes to comparing foreign systems of education (I assumed college level was something like what we call 'secondary school').

Maybe a moderator can move this thread? Or can I do it myself?

5. Apr 14, 2005

### Monique

Staff Emeritus
This forum is for career guidance, I will move it to homework as ZapperZ suggested.

6. Apr 14, 2005

### GCT

I've read through your post rather quickly (due limited time) so the following may not apply

draw a ray from the inner surface/lower end of the quarter circle straight up until it hits the outer surface. From this point draw a line toward the center of this circle, a radial line. Connect these two lines and you should have a triangle.

Note that the radius is the normal for the incident ray. You should now able able to relate this incident angle to the critical angle so that you can obtain a relationship between the indices of refraction and the radius.

7. Apr 14, 2005

### GCT

$$(R-d)/R=n_{out}/n_{in}$$

8. Apr 15, 2005

### Tsunami

Thank you very much.