The core of fused quartz fiber has an index of refraction of 1.47 and the claddi

Question

The core of fused quartz fiber has an index of refraction of 1.47 and the cladding around it has an index of refraction of 1.46. Calculate the critical angle. 3. 4. (Bonus Question) In question #3 above, what is the shortest distance a light ray can travel after it is reflected in a direction away from the source and remain in the quartz core ifthe diameter of the core is 8 m? Hint: Assume the angle it will be reflected at the next reflection is the critical angle and that the second reflection is on the opposite side of the core.

Explanation / Answer

a)

by using Snell's law,

n1*sin(theta) =n2*sin(theta2)

1.47*sin(thet1)=1.46*sin(90)

===> theta1=83.3 degrees

critical angle, theta1=83.3 degrees

b)

diameter, d=8um

radius,r=4um

use,

d=r/cos(theta1)

d=4*10^-6/(cos(83.3))

d=34.28*10^-6 m

shortet distance, d=34.28 um

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