[Rinehart]

Ah very good, thank you for that. As I said, I suspected that I had made a fundamental mistake, and there it was: the electrons don't pass through the insulator. I had always envisaged it as something like what you get with a spark plug or spark-gap transmitter, when the voltage rises high enough the electrons jump from one electrode to the other. Of course this produces a lot of heat, but through the miracle of magic thinking, I assumed that didn't matter.
So let me see if I've understood:
1) in targets with photoemissive materials, light falling on them creates free electrons which escape from the material altogether, the number of which depends on the intensity of the light.
2) these electrons eventually do settle back on the material, almost all of them at some other point.
3) this process is constantly occurring all over the plate, with the result that on average the distribution of electric charge over the plate as a whole corresponds to the image. (Perhaps the expression "electron image" comes from this?)
4) the purpose of the electron beam is to deliver a constant electric current to the mosaic, one that can move across it without needing a physical conduit, like an electric wire.
5) at the moment that the beam strikes the mosaic, emitted electrons from other points on the mosaic can't settle on the point the beam strikes. This causes a discontinuity; the amount of charge that can't settle there corresponds to the number of photoelectrons released at that point.
6) this causes current flow to the signal plate, and because the beam moves over the mosaic in a very precise way, the signal is a representation of the electron image.
7) unfortunately the scanning beam itself can cause electrons to be liberated from the mosaic, and these are called secondary electrons. These also are sent to the signal plate, resulting in a lot of signal noise.
8) to reduce the number of secondaries sent to the signal plate, there is a secondary electron collector which will rout them to the anode to complete the circuit instead of being sent to the amplifier and then to the transmitter.
Is this correct?
In terms of the non-uniform peak sensitivity, I think it was in William Eddy's or Thomas Hutchinson's book. Somewhere in preparing for Christmas I've misplaced a bunch of my notes, so I'm not sure where I read it.
Also, thanks for explaining why the mosaic was given a final treatment with evaporated silver. I had heard they did this, but didn't know why.
Can I ask you a couple more questions? The first concerns an unimportant problem in an oscilloscope and doesn't pertain to a camera tube, but which is a vital problem for a picture tube: how did they get around the problem of varying the intensity of the electron beam without changing the cross-sectional area of it?
And secondly, what kind of oscillator produces the sawtooth wave that drives the deflection circuits?

To answer your question about the RCA amateur iconoscope tube, the insulator was in fact transparent, and I should have mentioned that it was. The article I read didn't give any details as to what it was made from. (It also mentions that the signal electrode was transparent, but doesn't give any details about that, either.) This wasn't the only difference, however: for one, it only produced a 120-line raster without interlacing, and the secondary anode voltage was only about 600V, so perhaps that has something to do with it?
Could you send me the publishing details of the textbook you have quoted? I'll try to see if I can get a copy from Bookfinder.com.
As always, thanks for your help.
PS: Oh hang on, there's one more I have thought of: it seems that in the early days of TV, to create the effect of a dark space like an office in the middle of the night, they used the normal blinding light but simply reduced the gain on the camera signal. I suppose that it's the equivalent of Day For Night shooting in film. Does anyone know how convincing this was? If it's anything like its cinema counterpart it would have been dreadful, but you never can tell.