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(a) Describe the terms of reflection, absorption, and transmission with suitable diagrams. Present and explain the relationship between reflection, absorption and transmission coefficients. Describe the differences between materials that are opaque translucent and transparents.
(b) An optical fibre consists of a core of refractive index 1.667 surrounded by a cladding of refractive index 1.523. The refractive index of air is 1.000. (See the question sin the attachment to see figures and questions to calculate).
(c) The visible electromagnetic spectrum is defined to range from blue light of wavelength 400nm through to red light of wavelength 700nm. A light-emitting dioxide display made using GeAs-GaP emits photons with the energy E=1.8eV. Find the wavelength of the light and answer whether it is closer to the blue light or red lights.
(d) A pulse of white light is sent straight down a fibre optic cable 1km long the refractive index to the blue light or red light. (see attachment for further calculations).

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## SOLUTION This solution is FREE courtesy of BrainMass!

4a. Incident radiation energy (Qi) on a surface of a body, partly gets absorbed (Qa) by the body, partly reflected (Qr) and partly transmitted (Qt) through the body (see attached Figure).

Qi = Qa + Qr + Qt

Divided by Qi both the sides,
1 = Qa/Qi + Qr/Qi + Qt/Qi
=> a + r + t =1
where,
a = Qa/Qi == absorption coefficient: fraction of incident radiation energy which gets absorbed by the body.
r = Qr/Qi == reflection coefficient: fraction of incident radiation energy which gets reflected by the body.
t = Qt/Qi == Transmission coefficient: fraction of incident radiation energy which gets transmitted through the body.

Opaque body: No transmission of incident radiation through the body (t=0)
Translucent body: Partially transmit the incident radiation through itself (0<t<1)
Transparent body: 100% transmission of incident radiation through the body (t=1)

4.b.i. core refrective index, nc= 1.667
surrounding cladding refractive index, ns=1.523

refractive index of core with respect to surrounding,
snc = nc/ns = 1/sin(theta_c) [Here, theta_c: critical angle]
=> sin(theta_c) = ns/nc = 1.523/1.667 = 0.9136
=> theta_c = sin-1(0.9136) = 66 degree

4.b.ii. theta_2+theta_c = 90
=> theta_2 = 90 - theta_c = 90-66 = 24 degree

4.b.iii. sin(theta_1)/sin(theta_2) = nc = 1.523
=> sin(theta_1) = 1.523*sin(theta_2) = 1.523*sin(24) = 0.6194
=> theta_1 = sin-1(0.6194) = 38.3 degree

4.c E= 1.9eV == 1.9*1.6*10^(-19) J == 3.04*10^(-19) J == hc/lambda
h = 6.626*10^(-34) m^2 kg/s
c = 3*10^8 m/s
Hence, wavelength,
lambda = h*c/E = 6.626*10^(-34) * 3*10^8/(3.04*10^(-19)) = 6.5388*10^(-7) m == 6.5388 nm == close red colour of the spectrum

4.d. length, L = 1km == 1000 m
refractive index for blue light, nb = 1.639
refractive index for red light, nr = 1.621

(i) speed of blue light, cb = c/nb = 3*10^8/1.639 = 1.830*10^8 m/s
(ii) speed of red light, cr = c/nr = 3*10^8/1.621 = 1.851*10^8 m/s
(iii) time interval, delta_t = t_blue - t_red = L/cr - L/cr = 1000/(1.830*10^8) - 1000/(1.851*10^8) = 6.2*10^(-8) s == 62 ns

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

Â© BrainMass Inc. brainmass.com December 24, 2021, 11:14 pm ad1c9bdddf>
https://brainmass.com/physics/absorbtion/applied-physics-optics-questions-549416