The figure below shows an optical fiber in which a central plastic core of index

Question

The figure below shows an optical fiber in which a central plastic core of index of refraction n1 = 1.60 is surrounded by a plastic sheath of index of refraction n2 = 1.52. Light can travel along different paths within the central core, leading to different travel times through the fiber. This causes an initially short pulse of light to spread as it travels along the fiber, resulting in information loss. Consider light that travels directly along the central axis of the fiber and light that is repeatedly reflected at the critical angle along the core-sheath interface, reflecting from side to side as it travels down the central core. If the fiber length is 400 m, what is the difference in the travel times along these two routes?

Explanation / Answer

Here,
refractive index, n1 = 1.60 n2 = 1.52
Fiber length, l = 400 m

from trignometery
CosA = Base/hypotenuse
h = Base/CosA

Sice
path length = length of fibre / cos(critical angle)

Solving for Critical Angle :
SinC = n2/n1 = 1.52/1.60
C = arcSin(1.52/1.60)
C = 71.805

RPL = 400 / sin71.805
RPL = 421.053 m

Path difference, d = RPL - L = 421.053 - 400 = 21.053 m

Since, n = speed of light in vaccum/speed of light in medium = c/v
Solving for Velcoity, v
v = c/n
v = 3*10^8/1.60
v = 187500000 m/s

Time, t = d/v
t = 21.053 / 187500000
t = 1.123*10^-7 s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects