光纤通信计算题 英文题 但计算简便.1、consider a fiber with a 25um core radius,core index n1=1.48,and the core cladding index difference Δ=0.01,if ƒ=1320nm,what are the numerical aperture NA,the value of normalized frequency V and
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![光纤通信计算题 英文题 但计算简便.1、consider a fiber with a 25um core radius,core index n1=1.48,and the core cladding index difference Δ=0.01,if ƒ=1320nm,what are the numerical aperture NA,the value of normalized frequency V and](/uploads/image/z/9563114-2-4.jpg?t=%E5%85%89%E7%BA%A4%E9%80%9A%E4%BF%A1%E8%AE%A1%E7%AE%97%E9%A2%98+%E8%8B%B1%E6%96%87%E9%A2%98+%E4%BD%86%E8%AE%A1%E7%AE%97%E7%AE%80%E4%BE%BF.1%E3%80%81consider+a+fiber+with+a+25um+core+radius%2Ccore+index+n1%3D1.48%2Cand+the+core+cladding+index+difference+%CE%94%3D0.01%2Cif+%26%23402%3B%3D1320nm%2Cwhat+are+the+numerical+aperture+NA%2Cthe+value+of+normalized+frequency+V+and)
光纤通信计算题 英文题 但计算简便.1、consider a fiber with a 25um core radius,core index n1=1.48,and the core cladding index difference Δ=0.01,if ƒ=1320nm,what are the numerical aperture NA,the value of normalized frequency V and
光纤通信计算题 英文题 但计算简便.
1、consider a fiber with a 25um core radius,core index n1=1.48,and the core cladding index difference Δ=0.01,if ƒ=1320nm,what are the numerical aperture NA,the value of normalized frequency V and how many modes propagation in the fiber?
please explain the calculating results.
光纤通信计算题 英文题 但计算简便.1、consider a fiber with a 25um core radius,core index n1=1.48,and the core cladding index difference Δ=0.01,if ƒ=1320nm,what are the numerical aperture NA,the value of normalized frequency V and
.建议你去看一本光纤通信方面的书.
题目的意思是:光纤的纤芯折射率1.48,包层折射率与之相差0.01,问你数值孔径是多少?归一化截止频率是多少?有几个模式可以传播?