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## Homework Statement

Light of free-space wavelength λ

_{0}= 0.87 μm is guided by a thin

*planar*film of thickness

*d*= 3.0 μm and refractive index

*n*1 = 1.6, surrounded by a medium of refractive index

*n*2 = 1.4

critical angle = 61.04°

n

_{0}= 1.00

(a) Determine (i) the angle of incidence θ and (ii) the propagation constant β of the

*m*= 1 TE mode (you will need to find a graphical or numerical approximate solution here).

(b) What is the wavelength of this mode, measured along the

*z*axis?

## Homework Equations

M = [2dNA/λ

_{0}] + 1

M: number of modes

NA: Numerical aperture

NA = n

_{0}sinα

_{0}

β = n1*(2π/λ

_{0})*sinα

_{0}

## The Attempt at a Solution

substituting NA = n

_{0}sinα

_{0}and rearranging the equation to solve for sinα

_{0}

sinα

_{0}= 8.7*10

^{-7}/ (2*3*10

^{-7})

sinα

_{0}=0.145

arcsinα

_{0}(0.145)

α

_{0}=8.337° (Angle of Incidence)

Now that we have that we can calculate β

β = (1.6)*(2π/8.7*10

^{-7})*sin(8.337)

β = 1.675*10

^{6}(Propagation constant of m=1 mode)

λ / 2d > cosθ

_{c}

λ / (2*3*10

^{-6}) > cos(61.04°)

λ > cos(61.04°)(2*3*10

^{-6})

λ > 2.9μm - Cut off wavelength (wavelength can be no shorter than 2.9μm, otherwise more modes will propagate in the fibre)

I'm pretty sure I've done the question correctly. But I'm not really sure about the propagation constant.