One ray hits the core-cladding interface at an incident angle of 45o; a second r

Question

One ray hits the core-cladding interface at an incident angle of 45o; a second ray hits the interface at 65o. The critical angle is 40o so both rays are totally internally reflected. The optical fiber is 1.5 m long and the core is 8 m in diameter. What is the difference between the total distance the first ray travels and the second? (Make the simplifying assumption that both rays hit the core-cladding interface at the same point; this difference will be very small compared to the overall distance the two rays travel. Also, do not round the answers to any of the steps in your calculation until you calculate the final difference as this problem is very sensitive to error propagation.) In conclusion, what can you say about the relationship of critical angle and numerical aperture?

Explanation / Answer

ray 1:
distance along the wire covered between two reflections = d*tan(45) = 8*10^-6 m
total reflections = 1.5/8*10^-6 = 1.875*10^5
total distance = 1.875*10^5*8*10^-6/cos(45) = 2.1213 m

ray 2:
distance along the wire covered between two reflections = d*tan(65) = 8*tan(65) = 17.156*10^-6 m
total reflections = 1.5/17.156*10^-6 = 8.7432*10^4
total distance = 8.7432*10^4*8*10^-6/cos(65) = 1.655 m

diff = 0.46624 m

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