The critical angle for an optical fibre is 60°. Light travels through the fibre 2.0x10^8 ms-1 and the length of the fibre is 20 km. Calculate the difference in the time taken for the light to travel through the fibre by a light beam travelling the maximum distance and a light beam travelling the shortest distance. (5 marks)
x1 is the distance along the axis and x2 is the distance at an angle
sin60 = x1 / x2
so x2 = x1/sin60 = 20 / sin60 = 23.09 km
t1 = s /v = 20x10^3 / 2x10^8 = 1x10^-4 s
t2 = s /v = 23.09x10^3 / 2x10^8 = 1.155x10^-4 s
i.e. time difference = 0.155x10^-4 s
#waves #opticalfibres #refraction #light #physics
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