Let’s say it is midday on the summer solstice in New York City. I have a magnifying glass with a circular lens and am frying ants on the pavement.
What is the largest magnifying glass diameter that I can use before the brightness of the focused spot becomes acutely damaging to my eyes?
By acutely damaging, I mean producing obvious damage from a single session (e.g. playing around for one hour), not “If you make a habit of this it will increase your lifetime risk of cataracts by X%” or something like that.
What is the reflectivity of the pavement? How clear is the air? Is the sky completely cloudless? How long do you spend staring directly at the bright spot? Without such additional information (including, probably, other factors I have not thought of) your question is unanswerable.
A bright enough light focused for long enough on the same point on the retina can cause a scotoma, an area where the light sensitive cells have been killed off. However, you have not given us nearly enough information for anyone to be able to tell how long it would take for this to happen in the scenario you envisage. Furthermore, a small retinal scotoma, unless, perhaps, it is right in the fovea, at the center, of the visual field, probably will not have a very large detrimental effect on vision, and may well go unnoticed. After all, we all have a “blind spot” in each of our eyes, where the optic nerve connects to the retina, and do not notice that except under specially contrived circumstances.
The only source I could find on danger per watt were laser safety guides - a Class IV laser can cause eye damage by diffuse reflection alone instantly.
A class 4 laser is classified as a laser more powerful than a class 3B laser(Class 3B laser’s diffuse reflections off paper are not harmful). Here is a graph showing the power limits, and it shows that any laser with more than 500mW of power is Class IV.
If we ignore the thickness of the lens eating up your power (e.g. you’re using a Fresnel lens) and if we prented that the lens perfectly focuses the light onto an infinitesimal spot then a lens with a radius of 39.9cm is the maximum limit - there is 1W/m[SUP]2[/SUP] of sunlight hitting the ground so a lens can only have an area half that before melting your eyes ((0.5m[sup]2[/sup]/pi)[sup]0.5[/sup]=0.3989422804m)
This is probably a lowball estimate, since the lens will eat up some of the sun’s power and since the laser’s diffuse reflection is dangerous off paper, which has a higher albedo (diffuse reflectivity) than the pavement.
This is wrong. The Y-axis in that plot has units of W/m[sup]2[/sup]/nm, and describes the solar energy contained over a 1-nm bandwidth. You need to integrate that curve over the entire electromagnetic spectrum (i.e. find the area under that curve) in order to find out the totalinsolation. When you do, you come up with about 1000 W/m[sup]2[/sup] on a cloudless day.