Angle of Incidence, Propagation Constant and Wavelength

[says]

Homework Statement

Light of free-space wavelength λ₀ = 0.87 μm is guided by a thin planar film of thickness d = 3.0 μm and refractive index n1 = 1.6, surrounded by a medium of refractive index n2 = 1.4
critical angle = 61.04°
n₀ = 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/λ₀] + 1

M: number of modes
NA: Numerical aperture

NA = n₀sinα₀

β = n1*(2π/λ₀)*sinα₀

The Attempt at a Solution

substituting NA = n₀sinα₀ and rearranging the equation to solve for sinα₀

sinα₀ = 8.7*10^-7 / (2*3*10^-7)

sinα₀=0.145

arcsinα₀(0.145)

α₀=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*10⁶ (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.

 

[jbsteele]

Is it correct?Also, I am not sure how to calculate the wavelength of the mode. Can someone please help me with this?