UV transmission of sunglasses

[Physikfan]

The spectra are photos of Hg spectra with constant exposure time of 1/60 second, light source is a Hg high-pressure lamp, the entrance slit is imaged with a quartz lens via a reflection grating on a screen, to detect the UV lines, the left part of the screen is coated with an ultraviolet luminous color.
The camera is a Canon 10D, f/9.5, f = 105 mm, ISO 3200, for both pictures.

The first picture is the Hg reference spectrum, the lines go deep into the UV.


The second picture shows the Hg spectrum, filtered by cheap sunglasses.
The color of the blue line in the spectrum of the sunglasses also corresponds to the color of the visual impression with unarmed eye.
The color of the violet line in the Hg reflectance spectrum does NOT match the color of the visual impression with unarmed eye, and is an artifact, probably due to overexposure at this point.
To my surprise, the quality of the sunglasses is excellent in terms of UV protection,
I will now use these glasses as goggles for my further UV experiments.
UV experiments with normal glasses are in preparation.

[Paul Moir]

Try a clear piece of polycarbonate 

.

[tggzzz]

So it is reduced. But how much is it reduced, and is the reduction sufficient?

[T3sl4co1l]

Crank up the exposure, how much longer must it be for the same intensity of the UV lines?

Tim

[cdev]

Don't mess with your eyes. Use glasses that you know beyond a shadow of a doubt are adequate.

[Physikfan]

Reference spectrum:
The camera is a Canon 10D, exposure time = 1/10 second, f/13, f = 105 mm, ISO 3200.


Hg spectrum with sunglass absorptance:
The camera is a Canon 10D, exposure time = 1 second, f/13, f = 105 mm, ISO 3200.

[John Heath]

If you note that your arms are a little tanned then maybe some sun protect on the hands and arms well as UV sun glasses.

They make UV LEDs as well as red and blue ones. By putting these at the hg lamp source you will have calibration points for the spectrum.

[Gyro]

Try a clear piece of polycarbonate 

Yes, most prescription glasses, sun or otherwise, are made of polycarbonate.

[T3sl4co1l]

Longer exposure time:
http://galerie.experimentierkasten-board.de/data/media/201/IMG_9677800x195.jpg
http://galerie.experimentierkasten-board.de/data/media/201/IMG_9673800x207.jpg

Yeah but how much longer? 

Tim

[tggzzz]

Longer exposure time:
http://galerie.experimentierkasten-board.de/data/media/201/IMG_9677800x195.jpg
http://galerie.experimentierkasten-board.de/data/media/201/IMG_9673800x207.jpg

Yeah but how much longer? 

I don't think the OP has understood the importance of the word "significant" in "... is the reduction significant".

[Physikfan]

I have added the exposure times of all my pictures in both contributions.
I think I did understand that there is a significant reduction of the UV lines due to the absorptance of the cheap sunglasses.
The exposure time of the reference spectrum is 1/10 of a second, the exposure time of the "sunglasses spectrum" is 1 second.
Still there are no UV lines visible in the "sunglasses spectrum".
At least the reduction of the UV lines seems to be better than 10**-2 and is significant.

The originally used reference spectrum had a smaller entrance slit, therefore a higher resolution and had also an exposure time of 1 second:

[T3sl4co1l]

Thanks!

As for statistics, the dark area looks to be (after turning up brightness an arbitrary amount) roughly 101 +/- 4 counts.  I can't resolve any significant banding in that area, so the attenuation seems to be in the 4-counts-out-of-200 range (counting a nice bright ~full scale signal as ~200), or ~1/50th.  With exposure accounting for a factor of 10, that puts the attenuation around at least 500x or 27dB.  (Note, attenuation factor <1 implies gain.  So, just keep that in mind. )

Tim

[Physikfan]

For a better visual comparison between the intensity of the spectral lines and exposure times five Hg spectra are shown with increasing exposure time from top to bottom, (1/90 s, 1/30s, 1/10 s, 1/2 s, 3 s):

[Physikfan]

Hi Tim

You are right:
If an attenuation is <1, there is a gain!

I(UV,transmitted) =  I(UV, Incident) x 10**-3.

Please, how could you find these numbers:

the dark area looks to be (after turning up brightness an arbitrary amount) "roughly 101 +/- 4 counts."  I can't resolve any significant banding in that area, so the attenuation seems to be in the "4-counts-out-of-200 range (counting a nice bright ~full scale signal as ~200), or ~1/50th"

I have also another question:

Do you think there is a software available so that you could calculate the intensity distribution of the spectral line directly from such a picture spectrum?

Regards

Physikfan

[John Heath]

There is one way to find the brightness numbers but it is not easy Change the picture file format to .bmp . This in a microsoft format that is raw and simple with no compression. From here you look at the .bmp format file with an editor in hex format . It gets worse. You then look for a pattern where the hex numbers are high. These will be you spectrum lines. 1 horizontal line of spectrum will be 3 times the the width of your picture , 3000 for 1000 wide something like that.

A is easier going if the .bmp file is changed to 256 gray scale black and white 1 byte per pixel. We do not care what the color is and 1000 pixels  spectrum horizontal line is now 1000 bytes.

I have a feeling this is not that helpful but if desperate it is an option.

Also if you take the spectrum while the bulb is cold and staring to harm up the spectrum will be closer to mercury. After it harms up there is a bunch of unrelated lines from metal hydrates.

I have a picture of this from a spectrum taken by Mr Manfred. 

[T3sl4co1l]

I just opened it in an image editor and looked at the colors of various pixels...

Tim

[Physikfan]

Hi Tim and John Heath

Please, could you give us a link to such an image editor?
Is there any chance to get a software for calculating the intensity distribution of the spectral line directly from such a picture spectrum?

Regrads

Phyikfan

[T3sl4co1l]

I mean you can open it in Paint, pick color, edit color and read the number.  Does Irfanview have a color pick tool?  I don't use it, I don't remember...

Otherwise, any fully featured editor (PS, GIMP..) will do for sure.

As for what that number means: depends on the camera response.  Should be proportional to power, but if there's any log or gamma correction going on, it'll be different.  Could set up a test with fixed stop and exposure, and different intensity of source (suitably calibrated).  With the grating, you can then also see if it's nonlinear or proportional or whatever at different wavelengths.

Tim

[John Heath]

Hi Tim and John Heath

Please, could you give us a link to such an image editor?
Is there any chance to get a software for calculating the intensity distribution of the spectral line directly from such a picture spectrum?

Regrads

Phyikfan

Your spectrum , your spectrum grey scale with histogram equalized for better dynamic range. Note the little lines that were hiding in the background. Thirdly a histogram chart. with separated spectrum lines , sort of. Unfortunately they are distorted amplitude wise with equalization and there order for left to right is not the same as the spectrum picture On the positive side we are getting warmer. 

[Physikfan]

The following graphical representation of the spectra corresponds to the pictures shown above:

The first spectrum is the Hg reference spectrum, the lines go into deep UV.
The camera is a Canon 10D, exposure time = 1/10 second, f/13, f = 105 mm, ISO 3200.


The second picture shows the Hg spectrum, filtered by cheap sunglasses.
The camera is a Canon 10D, exposure time = 1 second, f/13, f = 105 mm, ISO 3200.

[John Heath]

This is super. I had not noted that you have already accomplished this. Nicely done. Is there any way to sharpen up the focus. You could use log 10 on the graph but that would just be sweeping the problem under the carpet. I assume you tried all the easy fixes.

[helius]

UV protection on sunglasses and regular safety goggles is OK for use with blacklights, but for lasers, arc lamps, or welding, you really do need specialized equipment.

[Physikfan]

Hi helius

I agree completely with your statement.

My measurements should more demonstrate the practical value of such a simple spectroscopic apparatus in the UV.

Here are the glasses:

[Physikfan]

The software used for calculating the spectra from the pictures of the UV screen:

https://www.rspec-astro.com/

[texaspyro]

AvE did a video on auto-darkening welding helmets.  By removing the darkening shutter, he found that the darkening shutter only reduced visible light.  The UV was removed passively by the front lens material.