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With your gelatin and dichromate, you should be dissolving in about 35 mls of water. However, most dcg holographers us ammonium dichromate, not potassium. Potassium dichromate is less sensitive than ammonium dichromate and you need a longer exposure than you'd normally use. As a rough estimate, I'd ...

I'm sorry I can't make out the image you uploaded. As I understand, you placed a lighter underneath the plate, and a coin next to it. If the lighter was about 5 mm thick, the coin was 5 mm below the lighter? If this is all correct, the coherence of your laser is very probably more than 5 mm, and the...

BobH wrote: ↑Sat Nov 25, 2023 10:57 am give you the last word.

Thanks!
By the way, does Hans Bjelkhagen know that you're accusing him of spinning "unrelated mathematical gobbledygook"

Essentially it says to disregard the boundary going from air to emulsion containing fringes. Assume the whole world is the index of the emulsion, and apply coupling theory to the fringes themselves during replay. Not true. In equation 2 of the Kogelnik paper: k² = (ω²/c²)ε - jωμσ the relative diele...

Sorry, I forgot to post the picture supporting 488nm. The laser is a parameter of a 488nm wavelength laser. The reason why I want to ask whether this device supports shooting holograms is that I previously purchased a meteorological ion laser to shoot holograms. The model is jdsu 2211D. The depth o...

I noticed that there are many waveband products to choose from in the obis series, and I want to use the 488nm band laser as the light source for making holograms. Here are some parameters, I want to know how this is different from 315m, what are the advantages and disadvantages. I only know that t...

So, do you actually expect ALL of us to buy this book to follow your arguments? Are you then incapable of actually giving just a single equation, quoting a single paragraph?. Perhaps, since your author references Kogelnik, you can give us the equation that your author quotes? We ALL need to buy this...

Oh, wait! I think I see your point. If an edge-lit hologram is recorded with TM polarisation, then r.s = 0, where r and s are the vector representations of the reference and signal waves, because r and s have an angular separation of π/2. Since we have M = k[ r.s (cos(phase))], the modulation vanish...

Then quote those equations. How about you get the book? Normally, in an environment of physics/ mathematics and engineering, one does not quote a reference by insisting that one buys a book just for one section. Normally, in an environment of physics/mathematics and engineering, it should be possib...

Then quote those equations. Also, interpret those equations. If you look at Hans Bjelkhagen's expression for the diffraction efficiency for a reflection hologram, you'll see it has no variable for polarisation. I can't attach the appropriate page because I'm told: "Sorry, the board attachment q...