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	2	added requested information from the comment source \| link	edited Oct 8 '13 at 15:40 Kyle Kanos 20.6k104385
inline side-by-side side-by-side markdown We can make a 1st-order approximation by assuming the following: $L=3$ m is the length of the fibre optic cable $d=3\cdot10^{-6}$ m is the diameter of the cable the cable is perfectly straight $\theta=0.785$ rad (~45$^\circ$) is the angle of reflection inside the cable photons are classical balls reflection is perfectly elastic photons still travel at $c$ Simple geometry shows that the particle travels $h=\frac{d}{\sin\theta}=4.24\times10^{-6}$ m over a linear distance of $x=3\cdot10^{-6}$ m. Do this a million times, you find that the photon traveled 4.24 meters instead of 3 meters! Given speed of light in vacuum, it would take 14.1 nanoseconds for the photon to travel the reflected path, whereas it would take 10.0 nanoseconds to travel 3 meters linearly. Both the distance & duration are about 40% increases! Since $L=3$ m and $t=14.1\cdot10^{-9}$ s, then the "linear" photon speed in the fibre optic cable is $v_{\gamma,fo}=2.13\cdot10^8$ m/s, a reduction of about 30%. EDIT As per the request in the comment, using $2c/3$, the reflecting photon would take 21.2 nanoseconds to travel the cable while the linear photon travels the distance in 15 nanoseconds. This would then lead to $v_{\gamma,fo}=1.42\cdot10^8$ m/s (instead of $\sim2\cdot10^8$ m/s) which is still a reduction of about 30%. We can make a 1st-order approximation by assuming the following: $L=3$ m is the length of the fibre optic cable $d=3\cdot10^{-6}$ m is the diameter of the cable the cable is perfectly straight $\theta=0.785$ rad (~45$^\circ$) is the angle of reflection inside the cable photons are classical balls reflection is perfectly elastic photons still travel at $c$ Simple geometry shows that the particle travels $h=\frac{d}{\sin\theta}=4.24\times10^{-6}$ m over a linear distance of $x=3\cdot10^{-6}$ m. Do this a million times, you find that the photon traveled 4.24 meters instead of 3 meters! Given speed of light in vacuum, it would take 14.1 nanoseconds for the photon to travel the reflected path, whereas it would take 10.0 nanoseconds to travel 3 meters linearly. Both the distance & duration are about 40% increases! Since $L=3$ m and $t=14.1\cdot10^{-9}$ s, then the "linear" photon speed in the fibre optic cable is $v_{\gamma,fo}=2.13\cdot10^8$ m/s, a reduction of about 30%. We can make a 1st-order approximation by assuming the following: $L=3$ m is the length of the fibre optic cable $d=3\cdot10^{-6}$ m is the diameter of the cable the cable is perfectly straight $\theta=0.785$ rad (~45$^\circ$) is the angle of reflection inside the cable photons are classical balls reflection is perfectly elastic photons still travel at $c$ Simple geometry shows that the particle travels $h=\frac{d}{\sin\theta}=4.24\times10^{-6}$ m over a linear distance of $x=3\cdot10^{-6}$ m. Do this a million times, you find that the photon traveled 4.24 meters instead of 3 meters! Given speed of light in vacuum, it would take 14.1 nanoseconds for the photon to travel the reflected path, whereas it would take 10.0 nanoseconds to travel 3 meters linearly. Both the distance & duration are about 40% increases! Since $L=3$ m and $t=14.1\cdot10^{-9}$ s, then the "linear" photon speed in the fibre optic cable is $v_{\gamma,fo}=2.13\cdot10^8$ m/s, a reduction of about 30%. EDIT As per the request in the comment, using $2c/3$, the reflecting photon would take 21.2 nanoseconds to travel the cable while the linear photon travels the distance in 15 nanoseconds. This would then lead to $v_{\gamma,fo}=1.42\cdot10^8$ m/s (instead of $\sim2\cdot10^8$ m/s) which is still a reduction of about 30%.

	1	source \| link	answered Oct 8 '13 at 15:31 Kyle Kanos 20.6k104385
We can make a 1st-order approximation by assuming the following: $L=3$ m is the length of the fibre optic cable $d=3\cdot10^{-6}$ m is the diameter of the cable the cable is perfectly straight $\theta=0.785$ rad (~45$^\circ$) is the angle of reflection inside the cable photons are classical balls reflection is perfectly elastic photons still travel at $c$ Simple geometry shows that the particle travels $h=\frac{d}{\sin\theta}=4.24\times10^{-6}$ m over a linear distance of $x=3\cdot10^{-6}$ m. Do this a million times, you find that the photon traveled 4.24 meters instead of 3 meters! Given speed of light in vacuum, it would take 14.1 nanoseconds for the photon to travel the reflected path, whereas it would take 10.0 nanoseconds to travel 3 meters linearly. Both the distance & duration are about 40% increases! Since $L=3$ m and $t=14.1\cdot10^{-9}$ s, then the "linear" photon speed in the fibre optic cable is $v_{\gamma,fo}=2.13\cdot10^8$ m/s, a reduction of about 30%.