# A pulse of white light is sent straight down a fiber optic cable 1 km long. The refractive index for blue light is 1.639 and for red light 1.621. What time interval will there be between the two components when they reach the far end?

Dec 25, 2014

The refractive index is also a direct measure of the factor by which the speed of light is slowed down as compared to the speed of light in a vacuum, $c \left(v a c\right) = 299792458 \frac{m}{s}$

So the blue light is slowed down to

$c \left(b l u e\right) = \frac{299792458}{1.639} = 182911811 \frac{m}{s}$

and the red light will be slowed down to

$c \left(red\right) = \frac{299792458}{1.621} = 184942911 \frac{m}{s}$

At those speeds the light will reach the end of the cable of 1 km (=1000 m) in a time that can be calculated as
time=distance/speed or $t = \frac{s}{v}$

$t \left(red\right) = \frac{1000}{184942911} = 0.000005407$ sec

$\to t \left(red\right) = 5407$ nanoseconds
(nano- meaning one-billionth)

$t \left(b l u e\right) = \frac{1000}{182911811} = 0.000005467$ sec

->t(blue= 5467 nanoseconds

The difference is $5467 - 5407 = 60$ nanoseconds