Disc Photographic
A look at what is currently available on eBay
 Photographic Lighting Reflector Disc holder Arm US $35.95 
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 New 42"/110 CM 5-in-1 Disc Photographic Reflector US $16.99
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 107cm White/Silver Zip Disc Photographic Reflector US $32.42
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 100cm x 150cm Photographic Reflector Disc Silver/White US $32.42
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 FREE SHIP 32" 80cm 5in1 New Disc Photographic Reflector US $27.01
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 32" 5 in 1 Multi Disc Photographic Lighting Reflector US $38.95
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![[Fomex] Light Disc Ø 80cm, Multi (5 in 1) Video Light Reflector Photographic](http://www.jisc.net/images/e/120841140898_0.jpg)
 82cm White/Silver Zip Disc Photographic Reflector US $27.02
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 55cm Photographic Grey / Focusing Panel Disc US $27.02
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 FREE SHIP 22" 56cm 5in1 New Disc Photographic Reflector US $17.28
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Optics (diffraction) question?
An opaque ball of diameter D = 40 mm is placed between
a source of light with wavelength l = 0.55 um and a photographic
plate. The distance between the source and the ball is equal to
a = 12 m and that between the ball and the photographic plate
is equal to b = 18 m.
(a) Explain why does it work?
(b) Find the image dimension y' on the plate if the transverse dimension of the source is y = 6.0 mm;
(c) Find the minimum height of irregularities, covering the surface
of the ball at random, at which the ball obstructs light.
(d) Can we put the disk with the same diameter instead of the sphere and get the same effect?
Sounds like the undescribed "it" that "works" refers to Poisson's spot, also known as Arago's spot, a bright spot that is seen at the center of the shadow of a perfectly circular object. In this case we have an image of the source because it's diffuse, not a point source.
A. The spot is the result of constructive interference from light sources (per the Huygens principle) at all points on the periphery of the object. The image is made of all the point sources making up the diffuse source.
I'd suppose simple geometry answers B. y' = 6 mm * 18/12 = 9 mm. But there'll be path length differences delta-L away from the center which will create null rings at radius r and successively odd-integer multiples of r, where delta-L = (odd-integer *) lambda/2, and you should calculate this radius and see if it's less than 9 mm. r ~= sqrt((18+0.275E-6)^2-18^2). I get ~ 3.15 mm.
C. Now we have to find the difference between the radius and the point at which the two tangent rays intersect, which should be simple, involving trig and Pythagorus. This tells how high the irregularities should be. Since the source is diffuse, it might get tricky, but I think it's valid to consider it point-by-point. However, the irregularities should also be fine-grained enough in areal distribution not to introduce half-wavelength path errors.
D. Should get the same effect from a disk according to the ref.
