# Why are bend losses higher at higher wavelengths in fiber optic cables?

I have been researching the effect of wavelength on macro-bend losses for my extended essay (a 4000 word paper on a subject of your choice, high school level) on fiber optic cables. I asked this question already about two months ago, and this was the reply:

An intuitive explanation for why this occurs is as follows. If you think of the wavelength of the light as the limiting length scale for it to resolve changes in the guiding structure then as the wavelength increases resolution decreases. Since the resolution has decreased the bend in the guide appears to become more abrupt and causes a larger perturbation in the propagating mode resulting in more radiated power from the optical fiber.

Consider how the radius of curvature (R) would appear to change as the wavelength increases. To do this we can normalize (R) to the resolving ability of the light, the wavelength (λ). Looking at R/λ it is apparent that this normalized version of the radius of curvature becomes smaller at larger wavelengths.

I had trouble understanding the explanation. If anyone would be so kind as to explain it to me further I would greatly appreciate it. Thank you :).

• What is relevant is the homogeneity of the wave-guide; fibers are designed with the assumption that they are perfectly straight. But the homogeneity is relevant only on the scale of the wavelength: at optical frequencies you do not have bend losses if your bending radius is of 1 meter, because over a few wavelength of propagation length, the fiber can be considered perfectly straight and the approximation holds: imagine considering a 1 micrometer long section of a 1 meter radius circle, it will appear as a line. – EigenDavid Jan 7 '19 at 10:40
• As the radius of the bend decreases, or the wavelength increases, the approximation becomes less accurate and the light can escape. – EigenDavid Jan 7 '19 at 10:40