v=((2*pi*a)/lambda)*sqrt(n1^2 - n2^2) delta=(n1^2-n2^2)/n1^2 x=d/a Example 10.3
ID: 2268170 • Letter: V
Question
v=((2*pi*a)/lambda)*sqrt(n1^2 - n2^2)
delta=(n1^2-n2^2)/n1^2
x=d/a
Example 10.3 The coupling coefficient k is a complex parameter that depends on a variety of factors, such as the wavelength, the refractive indices ofthe fibers, the fiber radius a, and the axes spacing d between the two coupled fibers. A simplified and accurate empirical expression for K in a directional coupler made from two identical step-index fibers is given by?2 A 5.2789-3.663 V + 0.384 1 V2 B-0.7769 + 1.2252V-0.0152V C=-0.0175-0.0064V-0.0009V- with V defined by Eq. (2.27). Consider two fibers for which n|-| .4532, n.-1.4500, and a = 5.0 m. If the spacing between the centers of the fibers is d = 12pm, what is the coupling coefficient rat a 1300-nm wavelength? 2.329 Solutio Using Eq. (2.27) we find that V From the above equation we then find where x = d/a K-20.8 exp [(-1 . 1 693 + 1 . 9945x-0.0373x2)] = 0.694 mm-l for x = 1 2/5-2.4 a) Calculate xfor 1550 nm.Explanation / Answer
a) v=((2*pi*a)/lambda)*sqrt(n1^2 - n2^2)
v= ((2*3.14*5*10^-12)*(sqrt((1.4532)^2-(1.45)^2)))/(1550*10^-9)
v=1.92*10^-6 {for lamda=1550*10^-10 m}
for delta,
delta=(n1^2-n2^2)/n1^2
delta= 4.3992*10^-3
x=d/a
so,
x= 12*10^-6/5*10^-6 = 2.4
A=5.2789-3.663*1.92*10^-6+0.3841*(1.92*10^-6)^2
A=5.27889250939
B=-0.7769+1.2252*V-0.0152V^2
=-0.7769+1.2252*1.92*10^-6-0.0152*1.92*10^-6^2
=-0.77745798
C=-0.0175-0.0064*V-0.0009*V^2
C=-0.017500012
putting values =20.8exp(-(5.27889250939-0.77745798*2.4-0.017500012*(2.4)^2)))
k=0.75787235731mm-1
c) v=((2*pi*a)/lambda)*sqrt(n1^2 - n2^2)
v= ((2*3.14*5*10^-12)*(sqrt((1.4532)^2-(1.45)^2)))/(1480*10^-9)
v=0.00000204494 {for lamda=1480*10^-10 m}
for delta,
delta=(n1^2-n2^2)/n1^2
delta= 4.3992*10^-3
x=d/a
so,
x= 12*10^-6/5*10^-6 = 2.4
A=5.2789-3.663*0.00000204494 +0.3841*(0.00000204494 )^2
A=5.27889250939
B=-0.7769+1.2252*V-0.0152V^2
=-0.7769+1.2252*0.00000204494-0.0152*(0.00000204494)^2
=-0.77689749454
C=-0.0175-0.0064*V-0.0009*V^2
C=-0.01750001308
putting values =20.8exp(-(5.27889250939+-0.77689749454*2.4-(-0.01750001308*(2.4)^2)))
k=0.61866854938mm-1
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