## College Physics (4th Edition)

We can find the time it takes the signal to arrive through the air: $t_a = \frac{d}{c} = \frac{4.0\times 10^5~m}{3.0\times 10^8~m/s} = \frac{4.0}{3.0}\times 10^{-3}~s$ We can find the time it takes the signal to arrive through the glass fiber: $t_g = (\frac{4.0}{3.0}\times 10^{-3}~s)+(7.7\times 10^{-4}~s)$ $t_g = 2.10\times 10^{-3}~s$ We can find the speed of the signal through the glass fiber: $v = \frac{d}{t_g} = \frac{4.0\times 10^5~m}{2.10\times 10^{-3}~s} = 1.90\times 10^8~m/s$ We can find the index of refraction of the glass fiber: $n = \frac{c}{v} = \frac{3.0\times 10^8~m/s}{1.90\times 10^8~m/s} = 1.6$ The index of refraction of the glass fiber is 1.6