Topic: Optoelectronics
For a 3-$\mu m$ diameter optical fiber with core and cladding indexes of refraction of 1.545 and 1.510, respectively. The cut-off wavelength is
- A. $2.3~\mu m$
- B. $1.29~\mu m$
- C. $1.5~\mu m$
- D. $3.24~\mu m$
Answer: B. $1.29~\mu m$
Explanation: $\lambda_{c}=\frac{2\pi a n_{1}}{V_{c}}(2\Delta)^{1/2}$ where $V_{c}=2.405$ (constant), $a=$ radius of optical fiber ($1.5~\mu m$), $n_{1}=$ core refractive index (1.545). $\Delta=1-\frac{n_{2}}{n_{1}}=0.022$. $\lambda_{c}=\frac{2\pi\times1.5\times10^{-6}\times1.545}{2.405}\times(2\times0.022)^{1/2} = 1.28\times10^{-6}m=1.28~\mu m$.