Filter technical data

[Demonperformer]

Just want to make sure that I am understanding what they are saying.

If I have a filter, the technical data of which states "Max. transmission: 50%", am I right in thinking this means that, for an object that emits the entire spectrum evenly (yeah ... "spherical chicken in a vaccuum"!), I would need an exposure of twice the duration with the filter for the object to appear as the same brightness in the resultant image as an image taken without the filter, in otherwise identical conditions?

Or does it mean that the transmission at peak-transmission is 50% and other bits of the spectrum are a lot less?

Or something else ... ?

Thanks.

[Stu]

Which sort of filter John, an ND? If so then it will be (should be) approximately 50% transmission across the entire spectrum, although not sure what happens at UV and IR!

Found this link which may help....

http://midopt.com/filters/nd030/

[Demonperformer]

Thanks, Stu. It is a filter that blocks part of the spectrum, but I am trying to work out what it is going to do to the rest of the spectrum. The intention is to darken objects which are bright, but want to know what it will do to everything else.

[Merlin66]

We'd need more info.....

I usually end up putting filters on the spectroscope to verify transmission/ bandwidths etc.

 

[Astrobits]

This site has lots of info on optical filters including representative transmission curves if you follow the links:

https://www.edmundoptics.com/resources/application-notes/optics/optical-filters/

Nigel

[Demonperformer]

Of course, I could always try doing an internet search for the specific item I am using rather than a generic version and so find data relating to said specific item - .

So, link to specification page of the filter in question.

Now I know that this filter is primarily used to identify methane on planets such as Jupy and Saturn. But my question refers to what happens in respect of other (non-methane-rich) objects. Clearly, because less light is getting through, it would require a longer exposure time to get the same number of photons on the chip. But how much longer? (don't want the main question to get lost in the rest of this spiel!)

Now the specs on this page completely confuse me. It starts by saying that this filter "absorbs the light at that part of the spectrum" [the methane band ca 890nm], but then the "transmission" graph shows that anything much outside the immediate area of "that part of the spectrum" [say 850nm to 925nm] has a transmission of zero. So, if the light is "at that part of the spectrum" it is absorbed by the filter and if it is outside that part of the spectrum the transmission is zero ... so how does any light at all get through?? Also, I have seen photos on the internet taken with CH4 filters that include the galilean moons and the rings of Saturn, which are not methane objects where the graph says that the transmission should be zero???

My bwain 'urts!

[Merlin66]

OK, softly, softly......

 

The CH4 methane absorbs light at 890nm. To make this "band" easy visible you need to suppress all the surround extraneous light.

The filter is designed to give a transmission of 90% at 890nm with a bandwidth of 20nm. (+/-10nm) this allows 90% of the available Methane light to pass through while blocking all the surrounding light.

So far so good......

Like taking images of Ha gas nebulae the narrower the bandwidth the less background light comes through - but it's always there - so if you image a planet comet etc where there is no active methane to be observed you'll just see the light from the body at 890nm +/-10nm which is basically NIR light.

 

[Demonperformer]

OK, I think it starts to make sense to me.

I was getting confused with the idea that light at 890nm had something to do with methane which, if I understand you, it doesn't.

So, Saturn and the rings both emit some light at 890nm, but the methane in Saturn's atmosphere absorbs that light and no-methane in the rings does not. So with the filter (which only lets light at 890nm through) the light from Saturn has been absorbed by its methane and so it appears dark and the light from its rings hasn't so they appear bright? That would fit with the attached picture (taken from here).

But presumably, only a proportion of the light emitted by the rings is at 890nm (otherwise we would not be able to see them visually). So only a proportion of the light emitted by the rings is reaching the chip through the filter, so they would also be darker on the same chip using the same optics and exposure length. But by how much (for an object with no methane to absorb the light)?

I guess that depends on what exact proportion of the light being emitted from an object is in the critical wavelength. But, as a rule of thumb, if I assume that light is being emitted at uv-ir evenly (more spherical chickens in a vaccuum!), then the visible spectrum is about 300nm and the filter is about 20nm, so I would need an exposure length of about 15x the exposure I would use to capture the object in the visible spectrum?

I guess that would be as good a starting-point as any. Alternatively, it could be that I am making so many approximations that that 15x could be anywhere from 1.5x to 150x. Or, as my father would have put it: "suck it and see"!

Thanks.

[Merlin66]

Hmmm, yes but no.....

The Methane band is found at 890nm, so if Methane is present that's the most dominant source......

Your rough maths is generally correct, but the results obviously will vary.......

 

[Demonperformer]

21 hours ago, Merlin66 said:

I usually end up putting filters on the spectroscope to verify transmission/ bandwidths etc.

This is an interesting thought in relation to another project in which I am engaged (I guess that I am allowed to drag my own thread off-topic!) in which it would be useful to get a "transmission graph" for specific individual filters.

Looking at the price of spectroscopes online is not a lot of help. It seems they come in varieties that cost from a few tens of pounds to a few thousand pounds. Where in that spectrum (if you will excuse the pun) I would need to be I have no idea.

But another thought occurs to me: I have a ASI224MC camera (which is reasonably sensitive across the wavelengths of the filters in which I am interested) and I have a white-light LED panel (which I use to take flats). If I were to get something like this, I could screw my filter-to-test onto it and put them both into the nosepiece of the 224. If I then stand that (nose-down) on the LED panel, I would get white light going from the panel, through the test filter, through the staranalyzer, creating a spectrum that is recorded on the chip. This could then be converted into a graph using visual spec software (or similar).

As a person who has some expertise in spectroscopy, how does that sound?

Doable? Completely ridiculous? Doable with some modification? My gut-feeling is that, if doable (with or without modification), I would probably be hard-pushed to find a ready-made spectroscope that would do the job as well for under the ~£100 needed for the staranalyzer, but again ... what would I know?!

Thanks.

[Merlin66]

I think the Alpy from Shelyak is probably the cheapest slit spectrograph at the moment.

The grating idea lack the refinement of being able to calibrate the spectrum you would obtain. It's designed to work with stellar, small point of light images. Flooding it with light would not give any usable results. You really need a narrow slit and some reference light source - neon or fluoro lamp.

Building your own spectrograph is doable. There's an interesting 3D printed version being trialled at the moment......

(Have a look at Samir's shoebox solution...may give you some ideas)

http://www.samirkharusi.net/spectrograph.html

[Demonperformer]

Thanks for the feedback and advice. 

I will have a look through it all when I have a bit more time to examine it in detail. 

Thanks.

[Astrobits]

Now we know which filter you are talking about, have you looked here:

http://www.schursastrophotography.com/ccdmoonplanets/JupiterFiltertests011616.html

there is some interesting/useful info.

Nigel

[Demonperformer]

On 20/02/2018 at 06:31, Demonperformer said:

so I would need an exposure length of about 15x the exposure I would use to capture the object in the visible spectrum?

Found another comparison image (from here), but this one was taken with a 50nm filter rather than the ZWO 20nm.

The arguement I have used above would give 50nm compared to 300nm means 6x the exposure time. The exposure time used on these goes from 1/32s to 1/4s, which is 8x. The pixel values (seems to be fairly constant from various points in the rings) in the CH4 picture are roughly 0.7x those in the other, which would suggest the exposure time needs to be increased more than 8x to get to the same brightness. Not totally sure what these pixel values mean in 'real terms', but if I divide the 8x used by 0.7, I get 11.4x, which is close to twice the 6x my theory would suggest. So maybe I would need to double my original estimate of 15x to 30x to get the desired result.

[Demonperformer]

I've just had another thought about imaging with this filter. My local streetlights have been switched to LEDs and I wonder if this filter will effectively remove this LP and so darken the sky-background for my images. Will have to go see if I can find a graph of wavelengths produced by these lights, but I would not have thought the 890nm waveband would feature particularly strongly ...