Calculate optical path in SELFOC cylindrical fiber optic

[Bromio]

Hi.

This is my first message in this forum. I'm not English, so sorry my spelling.

Homework Statement

Calculate the optical path done by a meridional ray, supposing it covers a horizontal distance, d, in z-axis direction. $$\gamma_0$$ is the launch angle (with z-axis).

Homework Equations

Optical path equation:
$$L = \displaystyle\int^d_0 n(\rho) dl$$

Because we work with SELFOC fiber optic, refraction index is: $$n^2(\rho) = n^2_0\left(1-\alpha^2\rho^2\right)$$

The trajectory equation is:
$$\rho(z) = \displaystyle\frac{\sin\gamma_0}{\alpha} \sin\left(\displaystyle\frac{\alpha z}{\cos \gamma_0}\right)$$

$$dl = \sqrt{d\rho^2+dz^2}$$ (infinitesimal calculus).

The Attempt at a Solution

I've tried to write optical path equation in function of variable z. So, the resultant integral is:
$$\displaystyle\int_{0}^{d}n_0\sqrt[ ]{1-\sin^2\left(\gamma_0\right) \sin^2\left(\displaystyle\frac{\alpha z}{\cos \gamma_0}\right)}\sqrt[ ]{1+\displaystyle\frac{\sin^2\gamma_0}{\cos^2\gamma_0}\cos^2\left(\displaystyle\frac{\alpha z}{\cos\gamma_0}\right)}dz$$

How can I solve this integral?

Thanks!

 

[mark726]

Hello and welcome to the forum! It's great to have you here and don't worry about your spelling, we're all here to learn and help each other.

To solve this integral, we can use the substitution method. Let u = sin(αz/cosγ0) and du = (α/cosγ0)cos(αz/cosγ0)dz. Then we can rewrite the integral as:

∫n0√(1-sin2(γ0)u2)√(1+(sin2(γ0)/cos2(γ0))u2)(cosγ0/α)du

= n0/cosγ0∫√(cos2(γ0)-sin2(γ0)u2)√(cos2(γ0)+sin2(γ0)u2)du

= n0/cosγ0∫√(cos2(γ0)-sin2(γ0)u2)√(1+u2)du

= n0/cosγ0∫√(1-(sin2(γ0)/cos2(γ0))u2)√(1+u2)du

Now we can use the trigonometric substitution method to solve this integral. Let u = tanθ, then du = sec2θdθ. Substituting this in, we get:

n0/cosγ0∫√(1-tan2(γ0)θ2)√(1+tan2θ)sec2θdθ

= n0/cosγ0∫√(1-tan2(γ0)θ2)secθsecθdθ

= n0/cosγ0∫sec2θdθ - n0/cosγ0∫sec2θtan2(γ0)θ2dθ

= n0/cosγ0tanθ - (n0/cosγ0)(1/2)(tan2(γ0)θ2-θ) + C

= n0/cosγ0tan(tan-1u) - (n0/cosγ0)(1/2)(tan2(γ0)(tan-1u)2-(tan-1u)) + C

= n0/cosγ0tan(tan-1(sin(αz/c