effect of colour filters

[Demonperformer]

I am getting a bit confused by the different information provided by different websites on the effect of filter colours, so I have boiled-down my (astro) problem to the following (non-astro*) hypothetical:

I have two LEDs, one red and one white, the red one being 10x brighter than the white one. I wish to take a monochrome photograph of them together so that they appear to be the same brightness. Would a dark-blue filter achieve this objective, or do I need another colour?

(Don't get too hung-up on the '10x' bit - this is just a sample figure - it is the 'direction of travel' that is important to me.)

Thanks.

*MODS: if you think this should be in the 'non-astro' lounge (I've umed & ahed for ages without reaching a definitive decision) please move it there. Thx

[Gina]

The opposite colour to red is green.

[Demonperformer]

So a green filter would make them appear to be (more) the same brightness in a monochrome photo?

[Ricochet]

You would need to know the transmission curve of the filter in question to determine how much of the red spectrum it passes. It might be true that "a" green filter will do what you want but that doesn't mean that "any" green filter will do the job. It is probably best to state your exact astro problem because the answer may very well be different to the answer to your hypothetical situation. See how much the transmission curves differ on three different "green" filters.

[Demonperformer]

Now that has focussed my thinking ... find an appropriate image to explain what is going on.

If I want to dim the red light in comparison to the white light, I need a filter that is going to block red differentially. Looking at the attached, it would seem that the 38A would do that.

So, if I take an exposure of Mars (primarily red) through a blue filter, it should darken it compared to the surrounding (primarily white) stars [I know not all stars are white, but this is just my first approximation]. Yes, I will need longer exposures to get the same faintness of stars [introducing its own problems], but as Mars is red, a greater proportion of the light coming from Mars will be blocked, making it appear fainter in comparison to the surrounding stars than it would without the filter. And therefore, I will be able to take even longer exposures before the glare of Mars overpowers the image, resulting in an overall improvement in the faintness of stars I can capture with Mars in the picture [hence the red & white LED example].

I was getting confused by comments on websites that seemed to be implying I needed a filter of the colour of the object I wanted to make fainter. Probably my reading of them rather than what they were actually saying!

Thanks.

[ronin]

By rights if you put a blue filter in front of the red led nothing should get through, the filter should be blocking red and green and passing only blue. Same if you put a green filter in front as it should pass only green and block red and blue.

It all depends on how good or narrow the filter is, the speactra passed tends to spread out a bit. Also depends on how pure the red wavelength is but an led tends to be narrow/pure.

A red filter would pass the red led output and also pass the red component of the white led, so with a red filter you should get red light coming through on the eyeball side of the filter.

At a guess/estimate you would need say a white led 10x brighter then the red led to "see" about the same at the eye. As the red filter should pass all the red led output but block the blue and green and assorted other bits of the output from the white led, so reducing the final apparent "red" brightness.

[Mr Spock]

In photographic terms red and blue are opposide. In B&W photography you would use a red filter to darken a blue sky, with orange for less dark and yellow even less dark. So, a blue filter will turn the red LED dark. But, that is in B&W

As an aside, if this is B&W photography, can't you use an ND filter?

[Demonperformer]

Thanks for the responses.

Yes, it is going to be B&W, but I don't see how an ND filter would differentially decrease the brightness of red mars more than that of white stars. Wouldn't it just make them all dimmer?

Thanks.

[symmetal]

If you want to reduce the amount of red getting to your sensor and pass everything else you need a minus red filter, which is a cyan filter (it passes green and blue). Using Wratten filters these would be colour compensating filters CC05C to CC50C

The CC50C has the strongest red rejection, passing approx 30% red, 70% green and 85% blue. If you used 2 CC50C filters to give even more rejection of red you'd get approx 9% red, 50% green and 70% blue.

For a cheaper alternative and to do some testing get some cyan lighting gel to fit over the front of the scope. Lee Filters 116 or 354  could be suitable. You can get cheaper alternatives on ebay. 

Alan

[Demonperformer]

Thank you for that, Alan. It is certainly an approach I had not thought about.

I have a spare 2">1.25" filter adapter. Purchase something like this, cut out a small square and stick it to the adapter (should even be enough to put one square on each side to "double stack" it) and fit into the nosepiece of the camera. At under £10 It has to be worth a go, especially as a quick search for 1.25" mounted CC50C filters suggests that they could rarer than hens' teeth.

I guess the quality would be lower than a proper mounted filter, but more subs and good flats might compensate for that.

Thanks.

 

[symmetal]

Demonperformer, yes that sounds like a good method to try out. I didn't think genuine wratten filters would be that cheap. Being gel, they aren't coated filters of course so will cause some image degradation due to surface reflections. I suppose 1.25" size filters are only used for astronomy so there wouldn't be much call for complementary colour filters, just the RGB primaries and narrowband.

As you say for the price it's worth giving it a go. 

Thinking about it, why don't we do CMY imaging instead of RGB. You would get twice the signal through each filter so better signal to noise. But to derive the RGB from CMY requires adding and subtracting the CMY results and the resulting RGB has no improvement in noise.

Alan