Intensity of a 1 mW Laser versus the Sun

Here is a comparison between the maximum intensity on the retina of the Sun and the beam from a 1 mW HeNe laser. (Adapted from one of Simon Waldman's optics lectures.)

Standard Sun:

• Maximum intensity of sunlight at ground level (directly overhead, no smog, etc.) = 1 kW/m2 or 1 mW/mm2.
• Assuming pupil diameter is 2 mm (i.e., radius of 1 mm), the area is approximately 3 mm2. So, the power of the sunlight through the pupil = 3 mW.
• Focal length of eye's lens = approximately 22 mm. Angular size of Sun from Earth = 0.5 degree = 9 mR. Thus, diameter of image formed = 22 mm x 9 mR = 0.2 mm and the area of image = 0.03 mm2.
• The intensity of the Sun on the retina (Power/Area) = 3 mW/0.03 mm2 = 100 mW/mm2.

Typical 1 mW HeNe laser (or laser pointer):

• Power (P) = 1 mW, wavelength (l) = 633 nm, radius of beam (w) = 1 mm, focal length of eye (f) = 22 mm. So, the diameter of spot = (2 x f x l)/(w x pi) = 9 x 10-3 mm and the area of spot = 6 x 10-5 mm2.
• The intensity of the HeNe laser on the retina is 1 mW/(6 x 10-5 mm2) = 16,667 mW/mm2.

So the 1 mW laser has the potential to produce an intensity on the retina 167 times that of direct sunlight! But there are many more factors to consider in determining the real risk of damage. In addition to those noted below, the actual focal point when looking at a laser at close range will not be at the retina so the spot size will most likely be much larger than the diffraction limit of the calculation. Even if the spot from the laser beam is smaller, natural eye movements or movement of the source (e.g., some moron waving a laser pointer) will result in it hitting any given point for a shorter time than the larger spot from the Sun (which usually doesn't move very quickly).

But, at least, perhaps you'll now have a bit more respect for that little HeNe laser or laser pointer!

The real problem behind this is that it is assumed that the power density is the significant factor in the thermal damage mechanism. The ability of the retina to dissipate heat is not dependent on the area covered, but the periphery (circumference) of the exposed area! The blood vessels are in the retina and not the sclera (the surface under the retina) - it is the blood flow that dissipates the heat and so can only act on the *edge* not the middle of the exposed area. In circumference terms, the ratio drops to 7 times. Furthermore because the larger spot is less efficient at dissipating heat, the effective power delivered by the laser beam is only about 2 times greater than that of the spot formed by the sun.

Related Topics:

• You Only Received One Set of Eyeballs?
• Intensity of a 1 mW Laser versus the Sun
• Why a 1 mW Helium-Neon Laser Still Appears Bright a Mile Away
• Problems With Determining Safe Limits