Why do bendy glass threads carry far more of your video call across the world without it fading?
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After you watchWhy do bendy glass threads carry far more of your video call across the world without it fading?
The short answer
Optical fibers guide light by total internal reflection at a core boundary with lower refractive index. Each encounter must satisfy its critical-angle condition; an earlier reflection does not guarantee a later one.
Try this next
- What changes when the second angle rises above 42 degrees? Keep the first angle at 60 degrees, change only the second angle, and send another pulse.
The whole story
How it works
The hidden model uses glass index 1.5 and air index 1.0, with incidence angles measured from the normal. The first angle is 60 degrees and the second starts at 30 degrees. Snell law and unpolarized Fresnel reflectance calculate the transmitted and reflected fractions. Above the 41.8 degree critical angle all light reflects in this ideal lossless boundary model. Magnified encounter diagrams are schematic, not a computed bent-fiber ray trace.
What people get wrong
Below the critical angle, not all light escapes: some reflects and some transmits. The critical angle is measured from the normal, not from the wall. Actual telecom fiber has lower-index cladding rather than an air boundary.
The catch
Fiber (light in glass) carries a huge amount of information at once and barely fades over long distances, but the glass must be very pure and you cannot bend it too sharply or even a grazing ray finally hits too steeply and leaks out. Copper (electricity in metal) is cheap, tough, and easy to bend and connect, but the signal fades and picks up noise much faster and cannot carry nearly as much at once.
Questions kids ask
Does one total reflection guarantee the next?
Optical fibers guide light by total internal reflection at a core boundary with lower refractive index. Each encounter must satisfy its critical-angle condition; an earlier reflection does not guarantee a later one.
What happens below the critical angle?
Below the critical angle, not all light escapes: some reflects and some transmits. The critical angle is measured from the normal, not from the wall. Actual telecom fiber has lower-index cladding rather than an air boundary.
Talk about it
- Can the second encounter behave differently from the first? Predict before sending.
- Use the percentages and ray paths to explain where the light went.
For grown-ups
The hidden model uses glass index 1.5 and air index 1.0, with incidence angles measured from the normal. The first angle is 60 degrees and the second starts at 30 degrees. Snell law and unpolarized Fresnel reflectance calculate the transmitted and reflected fractions. Above the 41.8 degree critical angle all light reflects in this ideal lossless boundary model. Magnified encounter diagrams are schematic, not a computed bent-fiber ray trace.
Keep going
What else makes you wonder?
- If light keeps bouncing instead of going straight, does it travel a longer path than a straight beam would?
- Could you trap a beam inside something other than glass, like water or clear jelly?
- How do they get a flash of light to start out at just the right shallow angle to stay trapped?
