How does the flat sides of a corner reflector effectively channel the radio signal onto the waveguide?
That’s a neat question — it sounds a bit counter-intuitive at first!
https://www.astronomy.me.uk/wp-content/uploads/2025/10/1000036614-161025.png
Here’s why a corner reflector antenna doesn’t just bounce the hydrogen-line radio waves back out into space, but instead concentrates them into the feed:
1. Geometry of the 90° corner
- A flat mirror (one reflecting plane) would indeed send most of the incoming wave back in the general direction it came from.
- But a corner reflector has two mutually perpendicular reflecting planes (like the corner of a cube).
- Electromagnetic waves that strike both planes get reflected twice, which reverses their direction and channels the energy back toward the bisector line — where you place the dipole feed.
So instead of scattering back outward, the geometry makes the reflected wavefronts reinforce at the dipole.
2. Feed in the “focal line”
- The driven element (dipole) sits in front of the apex, about 0.25–0.35 λ away.
- At this spot, the direct wave from the sky and the waves that bounce off the corner surfaces arrive in phase.
- That constructive interference boosts the effective aperture and directs the energy into the dipole rather than letting it re-radiate randomly.
3. Reciprocity
- Antennas are reciprocal devices: if a corner reflector transmits a focused beam outward, it will also receive energy most strongly from that same direction.
- When receiving, the reflecting surfaces corral radiation coming from the desired direction and cancel much of it from behind/other angles (good front-to-back ratio).
4. Why not “back to space”?
- Some energy is scattered back outward — no antenna is perfect. That’s why we talk about aperture efficiency (η ~ 0.5 for a simple corner).
- But because of the geometry and feed placement, a significant portion of the intercepted wave is redirected coherently into the dipole, not wasted.
✅ In short:
The corner geometry makes reflected waves reinforce at the feed. It’s like a parabolic dish in two flat pieces: the dish “bends” waves to the focus, the corner “flips” them to line up at the feed.
Would you like me to sketch (or generate) a ray-diagram style illustration showing how a plane wave at 1420 MHz reflects in a corner and converges on the dipole? That might make it clearer than equations alone.
Of course! Here’s a clean ray diagram sketch showing how plane waves reflect off a 90° corner reflector and focus onto the dipole feed positioned in front of the apex.
https://www.astronomy.me.uk/wp-content/uploads/2025/10/corner_reflector_ray_wavefronts.png
Diagram explanation (step-by-step):
- Incoming plane wave (e.g. 1420 MHz hydrogen line) arrives from far field.
- Rays strike two perpendicular reflecting planes, each at 45°.
- Each ray undergoes two reflections, reversing its direction toward the bisector axis.
- The dipole feed is placed at ≈ 0.25 – 0.35 λ from the apex.
- Direct and reflected rays arrive in phase at the dipole, reinforcing the signal.
- Some power is scattered back outward (aperture efficiency < 100 %), but most energy is funneled to the feed.
📐 Key geometry notes
- Reflector angle: 90°
- Feed distance: typically about 0.25 λ from the apex for maximum gain
- Plane wavefront after reflection is effectively reversed and aligned with the feed axis.
Would you like me to also overlay phase fronts (wavefront surfaces) on the diagram to make the constructive interference clearer?
Here’s the enhanced ray diagram with wavefronts overlaid — this shows not just the reflection paths, but also how the phase fronts align at the dipole, producing constructive interference:
🔸 What’s happening in the diagram:
- Incoming plane wavefronts (shown as parallel lines) approach the 90° corner.
- The wavefront strikes both reflector planes, and each point on the wavefront is reflected twice.
- After reflection, the emerging wavefronts are reversed and aligned along the bisector axis.
- At the dipole feed position, the direct path and the reflected paths have equal optical length (or integer multiples of λ), so their phases match.
- The result is constructive interference, concentrating energy at the dipole.
- The reflected phase fronts are nearly planar again, but redirected and phased correctly for efficient coupling.
📏 Typical geometry (at 1420 MHz, λ ≈ 21 cm):
- Feed distance from apex: ~0.25 – 0.35 λ (≈ 5 – 7 cm)
- Reflector panels at 90°
- Dipole centered on the bisector line.
This phase alignment is the real secret of why a corner reflector works — much like a parabolic dish, but using planar surfaces to flip the direction of the wavefronts instead of bending them.
Would you like me to annotate a version of this diagram with path lengths and phase angles (e.g. marking λ/4 distances and in-phase points at the feed)?
Exactly — that’s the key insight. A 90° corner reflector doesn’t have a single geometric focus (like a parabola). It creates an in-phase region along the bisector where the doubly reflected fields arrive with (approximately) equal phase. Your dipole samples that region.
I’ve updated the sketch to make this explicit:
https://www.astronomy.me.uk/wp-content/uploads/2025/10/corner_reflector_ray_wavefronts.png
What this shows
- Multiple two-bounce rays are drawn terminating at different points along the dipole, not a single point.
- A shaded “equal-phase strip” indicates the small region along the bisector where the path-length differences are ≲ a small fraction of λ, so fields add coherently.
- The reflected wavefronts near the feed are nearly planar, aimed toward that strip — not converging to a point.
Why that’s correct (and useful)
- By image theory, the dipole with two perpendicular planes behaves like the dipole plus its images; reception is strongest where those sources sum in phase.
- The optimal feed spacing (~0.25–0.35 λ from the apex) is where the direct and double-reflected contributions are closest to in-phase across the dipole’s length, giving good coupling and pattern (gain & F/B).
- Practically, the dipole is ~λ/2 long, so it naturally “collects” energy across a finite in-phase segment, not a point.
If you like, I can add iso-phase contours (loci where the double-bounce optical path differs by n·λ) to quantify the width of that coherent region for your exact geometry at 1420.405 MHz.
Great—here are the iso-phase contours that quantify the coherent region, built from the corner’s double-reflection geometry:
https://www.astronomy.me.uk/wp-content/uploads/2025/10/corner_reflector_isophase_bands.png
What you’re seeing
- For a plane wave incident along −x onto a 90° corner, two specular reflections are equivalent to a point inversion about the apex.
- Hence the equal-phase contours are vertical lines (every (\Delta L=\lambda/2) i.e. (\Delta\phi=\pi)), and the dipole samples an equal-phase strip across its length. That’s why it’s not a single “focus”, but a coherent region along the bisector.
How to use this
- Put the dipole at a spacing where (\Delta\phi=2kx) is a constructive value for your chosen reference (e.g., align the double-reflected and direct contributions or include any single-bounce term you’re optimizing for).
- Because (\Delta\phi) is flat in (y), the phase across the dipole is nearly uniform; the element factor/polarization then dominates how efficiently you couple.
If you’d like, I can regenerate this plot for your exact feed spacing, and overlay gain vs. spacing (simple aperture-efficiency model) at 1420.405 MHz.