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I was reading about fibre optic cables and it was mentioned that ,the individual "light pipes" are coated with a material whose refractive index is less than that of that of the glass. My question is why a material with smaller $\mu$ ? According to sin $\theta$=1/$\mu$, if I decrease $\mu$ then $\theta$,which is the critical angle,will increase.Hence data loss will increase!(since these cables are used in communication) Then why are we doing so? or is my logic incorrect?

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According to sin θ=1/μ, if I decrease μ then θ,which is the critical angle, will increase.

Your formula is for propagation from an optically dense medium into air ($n=1$). For optical fiber you should consider the case where both media have non-unity refractive index.

The simplest case is step index fiber with a core large enough to allow solution by ray optics. This kind of fiber works on the principle of total internal reflection. Total internal reflection occors when the incident angle at the interface is greater than the critical angle. The critical angle occurs when

$$\sin\theta = \frac{n_2}{n_1}$$

If $n_1 < n_2$, you'd need $\sin\theta > 1$ to solve this equation. Since the sine function is always in the range [-1, 1], there is no such solution. And indeed, total internal reflection doesn't occur when $n_1 < n_2$, and optical fiber are designed with the core index ($n_1$) slightly higher than the cladding index ($n_2$).

Even if you talk about fiber small enough that you must consider the waveguide properties of the structure, rather than simply ray optics, you'll find that there is no guided wave solution unless $n_2<n_1$.

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It sounds like your $\mu$ is $n_1 / n_2$ where light is being transmitted from a medium with index of refraction $n_1$ to an index of refraction $n_2$.

Decreasing $n_2$ therefore increases $\mu$, which appears to be your fundamental misunderstanding. Yes, you want $\mu$ as big as possible, and the way you do that (if you cannot increase $n_1$) is by decreasing $n_2$.

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