core26 min studyLesson 209 in Physics

Graded-index rays and fiber guiding

Derive smooth ray bending in a quadratic index profile and contrast it with the acceptance geometry of a step-index fiber.

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What you’ll understand

  • Derive a paraxial ray oscillator from a quadratic transverse index profile.
  • Calculate quarter-pitch ray transport with units and launch conditions.
  • Apply step-index fiber acceptance with the external-medium index included.

A ray can bend without encountering a sharp surface. If refractive index varies smoothly across a transparent material, the ray continuously turns toward the higher-index region. A nearly quadratic profile can bring displaced rays back toward the axis and exchange ray position with angle over a characteristic propagation length. Fibers can also guide through a sharp core-cladding boundary. We will distinguish these two mechanisms while stating when either ray picture is an approximation to the electromagnetic modes of a real waveguide.

Smooth refraction accumulates along the path

For an isotropic medium, stationary optical path gives the vector ray equation d(nt)/ds=∇nd(n\mathbf t)/ds=\nabla n, where t\mathbf t is a unit tangent and s is arclength. A transverse index gradient changes direction as the ray propagates. In a nearly axial system with small slope, replace arclength by z and take n nearly equal to its on-axis value in the derivative coefficient. The transverse equation then relates ray curvature to the transverse index gradient.

Choose the local profile n(y)≃n0(1−g2y2/2)n(y)\simeq n_0(1-g^2y^2/2), where g has units of inverse length. Index is highest at the axis and decreases away from it. Its gradient is approximately −n0g2y-n_0g^2y, giving a restoring curvature. This is a local paraxial model requiring ∣gy∣≪1|gy|\ll1 and small ray slopes. Extending the quadratic expression to arbitrarily large radius would eventually give an unphysical negative index, so the stated profile is never a global material specification.

dds(nt)=∇n,n(y)≃n0(1−12g2y2),d2ydz2+g2y=0\frac{d}{ds}(n\mathbf t)=\nabla n,\qquad n(y)\simeq n_0\left(1-\frac12g^2y^2\right),\qquad \frac{d^2y}{dz^2}+g^2y=0
The first equation uses geometric arclength. The final oscillator equation additionally uses a weak transverse gradient and paraxial propagation about positive z; g is in inverse metres and y in metres.↔ Scroll wide equations horizontally.

Pitch describes position-angle exchange

The paraxial solution depends on both launch height and launch slope. A ray launched parallel to the axis at a positive height reaches the axis after a quarter of the spatial oscillation period, with negative slope. At half a period it returns to zero slope on the opposite side. The full pitch is 2π/g2\pi/g along z, not a time period. A quarter-pitch rod can therefore turn a parallel ray distribution into a focusing distribution at its exit.

In reduced coordinates with u=n0θu=n_0\theta to this order, the GRIN segment is a rotation-like matrix with cosine diagonal terms and sine coupling terms. Its determinant is one, preserving the first-order invariant. It differs from the damping-free oscillator lab because the laboratory oscillator starts at unit displacement with zero velocity and uses time; here the independent variable is propagation distance and physical launch conditions vary. The analogy is useful only after those variables and limitations are declared.

y(z)=y0cos⁡(gz)+θ0gsin⁡(gz),MGRIN(z)=(cos⁡(gz)sin⁡(gz)n0g−n0gsin⁡(gz)cos⁡(gz))y(z)=y_0\cos(gz)+\frac{\theta_0}{g}\sin(gz),\qquad M_{\mathrm{GRIN}}(z)=\begin{pmatrix}\cos(gz)&\dfrac{\sin(gz)}{n_0g}\\-n_0g\sin(gz)&\cos(gz)\end{pmatrix}
Theta zero is the small launch slope in radians. This matrix uses the approximate reduced angle n zero theta and the weak quadratic profile; its off-diagonal entries have the usual length and inverse-length units.↔ Scroll wide equations horizontally.

A step-index boundary uses a different test

For a straight step-index fiber, a higher-index core is surrounded by lower-index cladding. A meridional ray must strike the core-cladding wall beyond the core-side critical angle to remain guided in the ideal ray picture. At a flat input face, Snell’s law links the external launch angle to the internal axial angle. Combining the two conditions gives nextsin⁡βmax⁡=ncore2−nclad2n_{\mathrm{ext}}\sin\beta_{\max}=\sqrt{n_{\mathrm{core}}^2-n_{\mathrm{clad}}^2}, with beta measured from the fiber axis outside the entrance.

The external index must be retained: immersion changes the accepted external angle even though the core-cladding index contrast is unchanged. This elementary expression concerns meridional rays and an ideal straight fiber, not arbitrary skew-ray trajectories, bend losses, or coupling efficiency from a structured field. When the core size is only a few wavelengths, discrete electromagnetic modes replace a dense ray bundle. A graded profile can reduce modal travel-time differences, but a sinusoidal ray drawing alone does not establish a fiber’s dispersion or bandwidth.

A quarter-pitch GRIN segment

A weak quadratic GRIN profile has g=10 m−1g=10\,\mathrm{m^{-1}}. A ray starts at y0=5 mmy_0=5\,\mathrm{mm} with zero launch slope. Find the quarter-pitch length, exit height, and exit slope within the paraxial model.

  1. A quarter pitch requires gz=π/2gz=\pi/2, so z=π/(2g)=π/20 m≃0.1571 mz=\pi/(2g)=\pi/20\,\mathrm m\simeq0.1571\,\mathrm m. This is a propagation distance, not an oscillation time.

  2. The height is y(z)=0.005cos⁡(π/2)=0y(z)=0.005\cos(\pi/2)=0. The ray reaches the axis at the exit, consistent with bending toward the central higher-index region.

  3. Differentiate the solution: θ(z)=−gy0sin⁡(gz)=−10(0.005)=−0.050\theta(z)=-gy_0\sin(gz)=-10(0.005)=-0.050 radians. The largest slope and ∣gy∣|gy| are both 0.05, reasonably small for the stated paraxial illustration.

The result

A 157.1 mm segment sends this ray to the axis with slope −0.050 radians. Its axis crossing does not mean that its direction also vanishes there.

Explore the model

Interactive lab

Interactive lab · oscillator

A spatial-oscillation analogy

Interpret normalized displacement as y/y zero and lab time only as an analogy for propagation distance. Reset, set Damping ratio ζ = 0 and Natural frequency ω₀ = 1, and predict a quarter-cycle zero at pi/2. For a fresh comparison Reset, set ζ = 0 and ω₀ = 2, and predict the first zero at pi/4.

A spatial-oscillation analogyUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe lab is a unit-mass time oscillator with fixed initial displacement 1 and velocity 0. It does not trace a fiber, vary launch angle, calculate total internal reflection, or predict modal dispersion.

Check your understanding

A parallel ray above the axis enters the stated quadratic high-index-center GRIN profile. What happens at one quarter pitch?

Try it yourself

An ideal straight step-index fiber has ncore=1.5n_{\mathrm{core}}=1.5 and nclad=2n_{\mathrm{clad}}=\sqrt2. Find its meridional numerical aperture and maximum external launch sine in air and in an external medium of index 1.25.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Square the two indices before taking their difference, and divide NA by the external index for the launch sine.

Compare with the solution

The acceptance NA is 1.52−(2)2=2.25−2=0.50\sqrt{1.5^2-(\sqrt2)^2}=\sqrt{2.25-2}=0.50. In air, sin⁡βmax⁡=0.50\sin\beta_{\max}=0.50, giving 30 degrees. In index 1.25, sin⁡βmax⁡=0.50/1.25=0.40\sin\beta_{\max}=0.50/1.25=0.40, giving about 23.6 degrees. The external angle shrinks while core-cladding contrast stays fixed. These are ideal ray acceptance boundaries, not promises of unit wave-coupling efficiency.

Take it with you

  • A smooth index gradient bends rays without a discrete interface.
  • The quadratic-profile sinusoid is a local paraxial approximation with a spatial pitch.
  • Fiber acceptance depends on core, cladding, and external-medium indices, while few-mode guiding requires wave theory.

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