Welcome to Laser Pointer Forums - discuss green laser pointers, blue laser pointers, and all types of lasers

LPF Donation via Stripe | LPF Donation - Other Methods

Links below open in new window

ArcticMyst Security by Avery

has anyone ever thought of this?

laserhead2112

laserhead2112

0
Joined
Feb 26, 2011
Messages
191
Points
0
Hey all

I see that there is a lot of talk of the safe viewing of the dot of a 445nm laser at 10 feet or so. Now idont know the focal length or lens diameter of the human eye but maybe someone with an lpm can try this.
Get a small lens shine the laser at the ceiling and focus the dot though the lens on the lpm and see what the mw reading is.

Thanks for reading.
And have fun trying this:D
 





JLSE

JLSE

1
Joined
Dec 13, 2007
Messages
3,580
Points
0
LOL, tried this a while back with my coherent lasercheck and a 1W+ 445... I held a lens
MUCH larger than my pupil, the results were interesting in that the do that
it made a dot on the wall...

Cant remember the reading, but it was from approx 10ft and wasnt that high...

Still wouldnt risk the same scenario with my own peeps.
 
Last edited:
laserhead2112

laserhead2112

0
Joined
Feb 26, 2011
Messages
191
Points
0
Lol. Yeah I wouldn't stare at it that's for sure. I used a 25mm lens on a piece of white paper and that little
Tiny dot was still pretty bright.
 
Cyparagon

Cyparagon

0
Joined
Sep 12, 2007
Messages
9,399
Points
113
The lens will just serve to focus the light on the power sensor. Whether the beam diameter is 7mm or .07mm, the power meter will read the same. It would be better to place an aperture over the power meter with a diameter of 7mm. 7mm because that's about as big as a pupil gets. In fact, I think this is how exposure is measured when doing audience scanning, too.

You can predict this using equations. I would use the following:

I=P/(4πr²)

Is the intensity of a spherical wave in W/m² at distance r with point source power P.
I would introduce a reflectivity constant k (from 0 to 1) denoting the reflectivity of the target, as a coefficient to power. Also, if the target is a flat surface, it will be a half sphere and the intensity will be double; we will add a 2 in the denominator Intensity times area equals resultant power, and we can find the area via πr² = πd²/4 where d is the diameter of the aperture (or theoretical pupil):

Pr=P/(4πr²)×k×2×πd²/4
Pr=Pkd²/(8r²)

10 feet is about 3 meters. Quantitatively, I have no idea how reflective a white ceiling would be, so I'll assume a really white 90%. For a 2W laser at 3 meters with diffuse reflection of 90% into a 7mm pupil, I predict:

Pr=2×.9×.007²/(8×3²)
Pr=1.8*.007²/72
Pr=1.2µW

So it won't even register. Of course, the reflection isn't uniform, but this should be the average value.
Feel free to correct me if I missed something.

Edit: Mr. Wannaburn, the lasercheck's aperture is 8mm, strangely enough. So if you just aim the lasercheck at the dot, you can predict a worst-case scenario of exposure for yourself.

Edit again: Apologies if it looks like a rainbow threw up in here. I'm just trying to label variables ;)
 
Last edited:
JLSE

JLSE

1
Joined
Dec 13, 2007
Messages
3,580
Points
0
My lasercheck will measure down to .01 µW and thats with my hand covering the sensor.
and the attenuator removed.

The problem is finding the proper wavelength and dialing it in.

These meters are a pain if you dont know the exact nm. and even
then I dont trust it...

For measuring He/Ne, they are nice and sensitive.
 
InfinitusEquitas

InfinitusEquitas

0
Joined
Mar 27, 2011
Messages
14,125
Points
113
Hi Guys,

I've been thinking about this a lot actually. There is A LOT of conflicting information regarding the subject of safely viewing the dots.

Goes without saying that with 1W+ lasers safety goggles are an absolute must, but for the purposes of this question lets pretend safety goggles don't exist.

Let's say the wall in question is a random white/semi white wall: http://www.apartmenttherapy.com/uimages/ny/11-18-white-wall-1.jpg

At what distance would it be safe to view a dot at various power levels, 50mw, 100mw, 200mw, 400mw, 1W? 1.2W etc..?

Is there any kind of general rule to this?
 
Meatball

Meatball

0
Joined
Feb 1, 2008
Messages
2,894
Points
0
Determine a reflectivity constant for the wall, and then plug everything into the intensity equation posted above by Cyparagon. Look for outputs that are <250uW or so.

Even such low power levels can be uncomfortably bright to look at if its all directed down your peepers.
 
InfinitusEquitas

InfinitusEquitas

0
Joined
Mar 27, 2011
Messages
14,125
Points
113
It's been 6 years since I've dealt with anything but excel formulas :(

Time for a refresher I guess.
 
snoutke

snoutke

0
Joined
Nov 1, 2010
Messages
479
Points
18
25 feet distance will be far enough to view 1W 445nm, 400mw 532nm and 250mw 660nm without goggles and with no discomfort.

Talking from experience.
 
BShanahan14rulz

BShanahan14rulz

0
Joined
Dec 29, 2009
Messages
3,136
Points
63
somebody +rep cyparagon for me. It's a pain in the ass to color-code your posts.

At least the math is reassuring ;-)
 
Haloid95

Haloid95

0
Joined
Apr 5, 2011
Messages
909
Points
0
Cyparagon said:
The lens will just serve to focus the light on the power sensor. Whether the beam diameter is 7mm or .07mm, the power meter will read the same. It would be better to place an aperture over the power meter with a diameter of 7mm. 7mm because that's about as big as a pupil gets. In fact, I think this is how exposure is measured when doing audience scanning, too.

You can predict this using equations. I would use the following:

I=P/(4πr²)

Is the intensity of a spherical wave in W/m² at distance r with point source power P.
I would introduce a reflectivity constant k (from 0 to 1) denoting the reflectivity of the target, as a coefficient to power. Also, if the target is a flat surface, it will be a half sphere and the intensity will be double; we will add a 2 in the denominator Intensity times area equals resultant power, and we can find the area via πr² = πd²/4 where d is the diameter of the aperture (or theoretical pupil):

Pr=P/(4πr²)×k×2×πd²/4
Pr=Pkd²/(8r²)

10 feet is about 3 meters. Quantitatively, I have no idea how reflective a white ceiling would be, so I'll assume a really white 90%. For a 2W laser at 3 meters with diffuse reflection of 90% into a 7mm pupil, I predict:

Pr=2×.9×.007²/(8×3²)
Pr=1.8*.007²/72
Pr=1.2µW

So it won't even register. Of course, the reflection isn't uniform, but this should be the average value.
Feel free to correct me if I missed something.

Edit: Mr. Wannaburn, the lasercheck's aperture is 8mm, strangely enough. So if you just aim the lasercheck at the dot, you can predict a worst-case scenario of exposure for yourself.

Edit again: Apologies if it looks like a rainbow threw up in here. I'm just trying to label variables ;)
Click to expand...

Thanks for taking time to write that :)
 
You must log in or register to reply here.




Top