Continuing to think out loud here. This helps me organise by thoughts and opens them up to correction from you lot, before they veer too far of course!
I have purchased a cheap telescope for the forthcoming test. This will be easier to point accurately before inserting the receptor than a wide-field pair of binoculars would be.
Well, err... maybe so, but the whole thing (as seen with this post) could be a red herring.
Using a short (condenser-style) lens of the same aperture as the telescope should provide an image on the photo sensor just as bright but because of its wider field of view
it would not be necessary to point it so accurately.
Because the sweet spot on this low capacitance photodiode is so tiny, I already am finding it difficult to use the telescope as a light gathering device. This is working without its eyepiece, as a long focus lens, and the slightest jitter on the tripod mount means reception comes and goes.
In contrast, I was able to set up the short focus condenser-lens-based receptor box by hand and simply prop it on an All-Bran packet.
Possible disadvantages of the short focus unit could be greater difficulty in finding the transmission source while first pointing up, and the unknown way the photodiode will respond to a microscopic poInt image. However the rough optical figuring of typical condenser lenses could mitigate this latter effect.
The large field of view of the condenser lens admitting other, interfering, light sources is unlikely to be a problem, since the small sensitive area on the photodiode drastically cuts down the working field of view.
So - unless I've missed something - no need for fancy optical sighting instruments at receiving locations - just a standard condenser-lensed box.
Anyway I had the telescope based outfit working in the garden yesterday and worked out that the minimum usable flux at the receiving end for a noise-free picture was about 5 microwatts. I did this as follows:
Beam contains (a bit less than) 1 milliwatt. Diameter of circular cross-section beam at end of garden = 16cm. Therefore area of beam (Πr²) = 201 sq cm. Receiving area with telescope stopped down to minimum usable = 1 sq cm. There proportion of total flux required = 1/201 of 1 milliwatt or 5 microwatts.
This sounds to me rather more than I had hoped.
Knowing this enables one to work out what the transmitting power of a future optical broadcasting station would need to be, based on say, a 5-mile (8km) service radius, 2-inch (50mm) condenser lenses on the receptors and current signal-to-noise performance. The final figure would be greatly affected by the compactness of the vertical radiation pattern of the omni-directional 'doughnut' h.r.p - if indeed this were to be used.
Steve O