How to setup to get a blazed grating?

Peter · Post by **Peter** » Sun May 30, 2004 8:28 pm

Hi group,

first let me say thanks for the grating formulas. I heard that there is a way to optimize a greating for the 1. diffraction order if it is made as a blazed grating. I like to make such for building a simple spectrograph.

Anybody any idea how to setup the equipment?

Thank you

Peter

Dinesh · Post by **Dinesh** » Sun May 30, 2004 11:27 pm

A blazed grating is a grating that produces a very high diffractive order at a specific angle, ie it "blazes" if hit at just the right angle. It's usually made with sawtooth profile with the angle of the sawtooth at the blaze angle. Holographically it has to be a surface hologram since it's properties are defined by a surface profile. This generally means it has to be made on resist. One way of doing this is to hit the resist from opposite sides with one beam at zero degrees and the other beam at the blaze angle.

If all you want is a bright, direction-dependant grating it may be easier to shoot a grating on Silver Halide with two collimated beams, one at zero and one at the angle you want the maximum diffraction. Such a grating will give a very bright diffractive order at a specific angle for the wavelength at which you shoot with other wavelengths also diffracted at nearby angles. The spread of the wavelngths around the recording wavelength - the dispersion - will depend on whether you shoot with collimated or diverging beams.

Peter · Post by **Peter** » Mon May 31, 2004 5:15 am

Thanks Denish for info.
I calculated that I need a grating having about 200 grooves/mm to get an angle of about 8° for illuminating a small CCD chip with the spectrum produced by the grating.

As you mentioned there is a possibility to produce a very bright blazed grating using two beam from oposite sides. If I do this, what is about the spreading (dispersion) angle? I guess it would produce a grating having 1000 or more lines/mm giving a very large spreading of the spectrum. Am I right or isnt that a problem, how can I possibly solve that problem (the CCD chip is only about 6 x 4mm).

A last question would be, what is the difference in result using collimated or spreaded beams?

Please excuse my bad english, thank you again

Peter

Colin Kaminski · Post by **Colin Kaminski** » Mon May 31, 2004 12:55 pm

Peter, Also remember bright sources of blue and UV light tend to cause Silver Halide holograms to turn black. A process called print-out.

Dinesh · Post by **Dinesh** » Mon May 31, 2004 3:34 pm

If all you want is 200 l/mm, you don't need hoographic film. Photographic film may well be within this limit, especially slow photographic film. I believe Pan X or Tri X can take 200 l/mm. In fact, it may even be possible to rule this density with a sharp tool on a surface. You need to move the tool by 1/200 mm or 5 microns. There are translation tools available at this resolution. There may be however a problem if you want to do this holographically. 200 l/mm is essentially a surface hologram and so has little or no Bragg selectivity. What this means is that the hologram reconstruction (the spectrum) will not be directional at a specific angle, as a volume hologram, but will diffract at any angle of the reconstruction beam. The angle of diffraction (taking orders into acount) is given by applying the grating formula,
lambda = n*d*sin(theta)
backwards to give theta for a specific lambda and d. From this you can calculate the dispersion by calculating the angle for red, theta_red, and for blue, theta_blue, and getting the dispersion as |theta_blue - theta_red|. In addition, you'll probably also get orders, so that the red will repeat, as will the blue etc. To find the dispersion of the orders, use n=1, n=2 etc. How many orders you get will depend on the emulsion thickness, in your case, the emulsion thickness needs to be of the order of a 2 microns or so to supress orders.
If you collimate one beam and diverge the other, it depends on which you use as a reconstruction beam. Assuming you reconstruct with the collimated beam, each component of the grating will also diverge. In white light the diffracted orders will be weak at a small distance from the plate.
If you shoot the hologram from opposite sides in any other medium but photoresist, you'll get a holographic mirror, not a grating. You get a blazed grating with this geometry only with photoresist.
If you need any help, feel free to call or email me. Contact info at the "Dinesh and Joy" link at the bottom of the page.

Dinesh · Post by **Dinesh** » Mon May 31, 2004 3:46 pm

n*lambda = d*sin(theta)
not
lambda = n*d*sin(theta)