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Unruh, Jay on
URL: http://confocal-microscopy-list.275.s1.nabble.com/How-to-calculate-an-estimate-of-power-density-for-laser-illumination-tp7580574p7580577.html
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I agree about measuring the power after the objective.
As far as illumination size is concerned: many high power experiments are done without full field illumination (lower NA going into the back aperture). In that case, you can see the illumination profile and measure it. If this is not possible, then I would use the objective manufacturer's field of view specifications. Alternatively, you could use a field diaphragm to crop the illumination until the edges are visible and then measure the power and illuminated area. That same power density should then apply to the illumination without the diaphragm.
Jay
-----Original Message-----
From: Confocal Microscopy List [mailto:
[hidden email]] On Behalf Of Craig Brideau
Sent: Friday, June 28, 2013 12:19 PM
To:
[hidden email]
Subject: Re: How to calculate an estimate of power density for laser illumination ?
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There will be quite a bit of power loss through the objective, depending on your magnification and numerical aperture. You really should measure the power after the objective, if possible.
Craig
On Fri, Jun 28, 2013 at 6:07 AM, Christophe Leterrier <
[hidden email]> wrote:
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http://lists.umn.edu/cgi-bin/wa?A0=confocalmicroscopy> *****
>
> Dear microscopists,
>
> I'm sure this is quite a stupid question. I keep seeing laser power
> expressed as W/cm2 or kW/cm2 in super-resolution articles. I have a
> laser for which I know the power out of the coupling fiber (let's say
> 50 mW), that I use to illuminate the full field of a 100X, NA 1.49
> objective (similar to a TIRF laser setup, but not inclined). How can I
> estimate the power density on the sample, assuming that there is no
> power loss in the objective? It looks like I need to know the field of
> view area, but I'm not sure what surface is actually illuminated : it
> is at least as large as the ocular field of view, but could it be larger?
>
> Thanks for your help,
>
> Christophe
>