Schott DUG11X and Baader-U comparison

[JMC]

Ok, so it's a new day, and new question. The Schott DUG11X filter (http://www.uqgoptics...ting_DUG11.aspx). I'm guessing these are different filters - the charts look different to me with the DUG11X letting in more shorter wavelength UV. Has anyone done a comparison of the DUG11X to the Baader U for IR blocking?

 

Edit - Just found this online (http://www.company7....ilter_bpu2.html). So does the nickname for the Baader U of UG11xx mean that the Baader U is an upgrade to the DUG11X or is that just a coincidence?

[Cadmium]

Your graph doesn't show the whole story.

Both DUG11 and DUG11x rear their ugly infrared heads at about 1030nm. I would guess that at UV exposure times of seconds, that IR will not be good...

http://www.us.schott.com/d/advanced_optics/3e48e184-8016-4760-9d6c-06e5303aee98/1.0/schott-uv-broadband-dug11-may-2013-us.pdf

[JMC]

Interesting thanks. I also suspect it may be a problem, but do you know if anyone has ever looked at it though? My experience with the Baader U is that at longer exposures (seconds) IR does also become a problem.

[Cadmium]

Only if your UV is virtually nonexistent, or purposely suppressed.

 

You could purchase some DUG11, but it might cost you more than a Baader U, and if I am reading the graph correctly, you would own a piece of filter glass that leaks IR stronger than a Baader U.

It would be an interesting but expensive test.

[JMC]

Ok, so hypothetical question. Say it takes 1s to capture an image using a filter designed to let through UV, but which also lets through a very small amount of IR. Now, you add a filter which blocks the UV, and the exposure time is increased to 100s to generate an image. Can it be said that when using that filter the image being captured comprises contributions of 99% UV and 1%IR? Obviously assuming the same reflectivity in the UV and IR.

[Cadmium]

"Can it be said that when using that filter the image being captured comprises contributions of 99% UV..."

You said you added a filter which "blocks the UV", so there isn't "99% UV", there is 0% UV.

The point remains that given the DUG11 graphs, the Baader U would suppress more IR, and have a higher UV to IR ratio than the DUG11 filter glass.

Thus the Baader U out preforms DUG11(x) at any exposure time, and at any UV suppression.

 

Old example.

[JMC]

"Can it be said that when using that filter the image being captured comprises contributions of 99% UV..."

You said you added a filter which "blocks the UV", so there isn't "99% UV", there is 0% UV.

 

Ok, that's me not being clear then in my request then. Filter A, lets through UV and a small amount of IR. With Filter A, exposure is 1s. Now Filter B is added on top of Filter A. Filter B blocks all UV, but lets IR through. Now with Filter A+B, the exposure is 100s.

 

Can the original image with Filter A, be said to be comprised of contributions of 99% UV and 1% IR?

[Andrea B.]

I've never used any specific DUG11X, just for the record.

 

I only know by way of hearsay that the substrate for a BaaderU is DUG11X. The most recent purchases I've made of BaaderU filters only reference a UG11 substrate on the filter box. Which theoretically is correct even if DUG11X was actually used.

 

*************

 

Jonathan, I think you are looking for some relationship explained by the Filter Factor?

The relationship between exposure stops and amount of light transmitted by the filter?

Let me think for a moment.

 

If Filter A (mixed UV + IR) has a filter factor of X,

then exposure should be increased by 2^X stops over a baseline, unfiltered exposure.

 

If Filter B (IR only) has a filter factor of Y,

then exposure should be increased by 2^Y stops over a baseline, unfiltered exposure.

 

Stack A+B (IR only) has a filter factor of X*Y,

so exposure should be increased by 2^X*Y stops over a baseline, unfiltered exposure.

 

If exposure time for Filter A is 1 second and for Stack A+B is 100 secconds,

then the filter factor for Stack A+B is 6.7 stops with respect to a baseline filtered exposure using Filter A.

 

Getting stuck. You multiply factors and add stops......

Uh......

 

There are 2^X stops increase from a baseline, unfiltered exposure to Filter A exposure

Then there are 6.7 stops increase from Filter A exposure to Stack A+B exposure.

Thus we have 2^X + 6.7 stops from a baseline, unfiltered exposure to Stack A+B exposure.

 

Therefore, 2^X*Y = 2^X + 6.7

 

But is that a filter factor for IR only? Or what?

I'm getting stuck on that part now.

 

Or I could have the whole danged thing wrong. :D

Sorry, I'll go think some more.

[Cadmium]

The imaginary number i, as in, "i am not a mathematician"...

All I am doing is pointing out the obvious conclusion based on the DUG11 graphs, that the Baader U will (should) suppress IR better than the DUG11.

To know for sure, you can purchase some DUG11 and compare it to a Baader U.

[Andrea B.]

Well, yes, that is all true about DUG11 and BaaderU. :)

 

But --->>> the general question remains an excellent one.

Can one apportion a UV filter factor based on certain empirical observations about exposure times? Not being much of an optics engineer myself (my engineering years were in telecom), I lost my way trying to be formal with filter factors. So I'm hoping someone stops by and points us in the right direction. Or in some direction. B)

[Cadmium]

Well, that is 'a' general question, more like your general question, not my general answer to the original general question.

The original general question was how does DUG11 compare with the Baader U.

Shane's Baader U graph only shows detailed transmission up to about 950nm:

http://www.beyondvisible.com/BV3-filter.html

 

As far as all the other general questions, I would just say one word, 'ratio', which we have discussed before, and most agreed about.

Ratio:

http://www.ultravioletphotography.com/content/index.php/topic/1913-investigating-ir-leak-from-the-baader-u-filter/page__view__findpost__p__13342

[Andrea B.]

Oh, sorry, my fault, we are talking past one another! I was trying to play with the filter factors thing to figure out the later hypothetical question not the general question. Ok, enough confusion for one night.

[Cadmium]

You are a mathematician, a real one! :-) Math is your expertise, only natural for you.

[Andrea B.]

The mathematics is very, very much past tense, however. One really does forget languages or math or chemistry -- or whatever -- when the particular language is not spoken for years and years. We just donated boxes and boxes of mathematical texts to the Fordham University library, which is Michael's alma mater for undergraduate work. I truly do not mind forgetting math because I've gotten to learn all sorts of new, more interesting things in recent years.

 

********

 

Back to Jonathan's hypothetical........

 

If you look at the transmission chart of a dual bandpass filter and estimate the areas under the UV curve peak and the IR curve peak, then you can reasonably estimate the ratio of UV-to-IR quantities. (All this as duly noted and mentioned above!!) But -- that assumes that the actual illumination is somehow "perfect", doesn't it? In practice, you are not going to get equal amounts of UV or IR in the illumination being used to make the photo. And, we have observed in practice, that even with a UV-pass filter having a tiny, tiny 1/2% leakage of IR as shown on a chart, we can and do see our UV photo ruined by IR contamination because the actual sunlight present at the time contains a ratio like 3-45-52% of UV-Visible-IR light. Even though a UV-pass filter might pass all the UV light it can get, there might not be enough of it.

 

Sooo, if we were to try to quantify the UV by using exposure times as Jonathan was thinking about, we would still need to untangle the illumination quantities from the transmission quantities.

 

Does this make sense to anyone??

 

(Believe me, there are days when nothing makes sense. :D B) )

[JMC]

There's also the question of sensor sensitivity in the different regions Andrea - much higher sensitivity in the IR than UV, and magnifying the effects of even small amounts of IR transmission. I fully acknowledge this is very much a hypothetical question needing a number of assumptions. But even keeping those in mind, I'm still interested in knowing whether from exposure times alone using different filters is it possible to infer relative contribution of UV and IR to an image.

[Andrea B.]

Well, I started the calculation. Is the reasoning set up correctly?

Can anyone get past that last 2^X*Y - 2^X = 6.7 or is that complete gar-bage?

If you get the X filter factor there, then flip it for the % transmission.

[Cadmium]

Jonathan? OK, well I am quite interested in the original question, even though I have expressed by doubt in the comparison between the Baader U graphs I have seen and the Schott DUG11(x) graphs I have seen.

So which one of us is going to get some DUG11 and compare? :-)

I just guess it is going to have to be me.

I have never done this yet because... here is why:

If the Baader U were DUG11, or even using the DUG11 substrate and then further coated to suppress the 1030nm+ leak, then why is there still a 900nm+ leak form the Baader U (as seen in Shane's detail - RED LINE):

http://www.beyondvis...BV3-filter.html

Even Shane's linear graph that goes up to 1100 doesn't show anything in the 1000nm+ range. Why?

That is why I don't think the Baader U uses DUG11(x) as its substrate.

However, I would like to try the DUG11(x), but it is more expensive than the Baader U, and hard to cut for the same reason the Baader U is hard to cut (without chipping the coatings on the edge).

Also, DUG11X is designated "on demand', which means they don't make it unless some multinational conglomerate wants a train load of it...

Of course, if you can find some that a conglomerate has sitting around, then you might be able to get some, but at a premium.

DUG11 is a little more attainable, but not much, and is still as or more expensive than the Baader U.

[JMC]

Hi Cadmium, thanks for the note. I have reached out to Schott as well, and asked them about it (and whether they have run a comparison with the Baader U, or whether they would like to). I don't always get an answer from these big guys when I ask, but you never know. I will ask around here to try and get some of the DUG11x, but my connections with the industry are purely as a buyer, so it's just a matter of trying to track some down. If you can find a way of getting DUG11x and maybe DUG11, perhaps we can have a chat about sharing costs etc to get some in?

[Andrea B.]

Oh yes, me too. Cadmium if you can source some DUG11 or DUG11x then I'm happy to join a group buying pool. Please PM us if this is a possibility.

[JCDowdy]

I corresponded with Schott back in 2013 about DUG11 & DUG11X. As I recall what they told me, the UG11 substrate must be cut before coating in a plasma deposition chamber at their facility in Yverdon, Switzerland. The cost of the coating run was separate from the cost of the substrate and they could mix different sizes.

 

The quote I received at the time, 07Oct13, was a minimum of 1000 euro and a quantity of 8, Ø50±0.2 X 2±0.2mm, was $5544. For a quantity of 2 the price was $1298 each and a quantity of 11 1inch rounds was $2671. All I really wanted was a single 1in sample, but as you already know this is not a stocked item, so I did not place an order.

 

The eBay seller bjomejag sells Omega's version of a red rejection coated UG11 as 330WB70 which some of us have already tried. I have tried the Ø25mm X 3mm 330WB70 (see PN "DUG 11" on 2nd photo) in a rear mounted arrangement and can report that it white balances differently from the BaaderU.

[ulf]

John,

 

Do you have any images you'd like to show using the the Ø25mm X 3mm 330WB70?

I found and ordered this filter recently.

It is on its way and I wonder if it is usable for photo, compared to my Baader U.

[JCDowdy]

I don't have any at hand at the moment. I will see if I can dig something up for you this evening.

[Cadmium]

https://www.itos.de/uploads/pics/dug_graph_01.jpg

They have DUG11X in 12mm circles.

But frankly, the graph doesn't get me excited.

[Andrea B.]

This linked PDF states that the BaaderU is made using UG11X.

 

http://www.baader-planetarium.com/de/downloads/dl/file/id/316/product/2952/baader_u_filter_in_der_terrestrischen_fotografie.pdf

[Cadmium]

Quite interesting! An excerpt for those of us who don't read German:

 

The "Baader U-filter ZWL 350 nm" Is made of UG-11x-Glass.

In this filter, the UV-A transmission is retained, the NIR However, permeability is virtually completely eliminated.

 

So all you have to do is get Schott to make you one (or more actually) of these at a price that is less than what you pay for a Baader U.

Not an easy task I would think.

 

Of course the PDF info could be slightly erroneous, we don't know for sure, and thus it would be best to test this with a smaller sample that is in stock, such as the source I listed above.