Laser Safety Calculations Tom Lister

NOTES The following three calculations were given as ‘homework’ for delegates of the IPEM LPA Update in Bath, 28th Nov 2014 and worked through as an exercise during the afternoon. The answers given are not necessarily ‘correct’. There was a consensus agreement that the consideration of pulsed sources and α< αmin are appropriate conservative approaches for these particular calculations

Calculation 1 Alexandrite epilatory laser λ = 755 nm Aperture = 15 mm (‘focal spot’ 12 mm at 40 mm) Beam half angle = 2.7° Max Energy Density = 35 Jcm-2 Pulse Duration = 20 ms Operating frequency = 1.5 Hz

MPE AOR Directive “In carrying out the assessment, measurement or calculation, the employer must follow the [...] standards of the IEC” 60825-1:2014 “The MPE levels represent the maximum level to which the eye or skin can be exposed without consequential injury immediately or after a long time and are related to the wavelength of the laser radiation, the pulse duration or exposure duration, the tissue at risk and, for visible and near infra-red laser radiation in the range 400 nm to 1 400 nm, the size of the retinal image.”

Maximum Permissible Exposure 755 nm, 20 ms pulse Thermal effects on the retina 2.7° x? 12 mm 40 mm

C6=1? 60825-1 page 27 If we assume α < αmin, C6=1 “Most laser sources have an angular subtense α less than αmin, and appear as an apparent “point source” (small source) when viewed from within the beam (intra-beam viewing).” If we assume α < αmin, C6=1 Table A.1 in 60825-1 MPE = 18·t0.75·C4 Exposure time, t = 20 ms t0.75 = 5.3·10-2 C4 = 100.002·(755-700) = 1.3 MPE = 1.24 Jm-2 There was general agreement with this result

NOHD 60825-1:2014 “distance from the output aperture beyond which the beam irradiance or radiant exposure remains below the appropriate corneal maximum permissible exposure (MPE)”

NOHD C6=1 Maximum energy output Spot size required 35 Jcm-2 over a 12 mm diameter spot 35·π·(0.62) = 39.6 J Spot size required 39.6 J / 1.24 Jm-2 = 32 m2 Radius = √(32/π) = 3.2 m

NOHD C6=1 68 m ? Distance required Manufacturer: NOHD = 100 m 3.2 m /47·10-3 = 68 m Manufacturer: NOHD = 100 m 3.2 m There was greater variation in the NOHDs calculated by the delegates, although in all cases the values were far in excess of the dimensions of any clinical rooms likely to be used. 47 mrad 68 m ?

C6≠1? 60825-1 p13: “The angular subtense of the laser source should not be confused with the divergence of the beam. The angular subtense of the laser source cannot be larger than the divergence of the beam but it is usually smaller than the divergence of the beam.” 2.7° smallest spot size α 12 mm 40 mm 100 mm 100 mm 17 mm

C6≠1? 60825-1 p13: “The angular subtense of the laser source should not be confused with the divergence of the beam. The angular subtense of the laser source cannot be larger than the divergence of the beam but it is usually smaller than the divergence of the beam.” If we assume α = beam divergence, C6≠1 C6 = α/αmin = 47/1.5 = 31⅓ MPE = 18·t0.75·C4·C6 (Table A.2) MPE = 1.24 x 31⅓ MPE = 39.0 Jm-2 This approach is incorrect and overestimates MPE

NOHD C6≠1 12 m ? Spot size required Distance required Radius = √(1.0/π) = 0.57 m Distance required 0.57 m /47·10-3 = 12 m 0.57 m ? 47 mrad 12 m ?

Calculation 2 EVLT laser λ = 1470 nm Aperture = 400 μm Numerical Aperture = 0.37 Max Power = 12 W (±15%) Pulse Duration = 100-9900 ms Pulse Interval = 100-1000 ms Max pulse rate = 5 Hz

Maximum Permissible Exposure 1470 nm, >100 ms pulse length Thermal effects on the cornea If we assume α < αmin, C6=1 MPE = 5600·t0.25 Exposure time, t = 100 ms t0.25 = 0.56 MPE = 3150 Jm-2 Exposure time, t = 9900 ms t0.25 = 1.8 MPE = 9933 Jm-2

NOHD 9900 ms Maximum energy output Spot size required 12 W over 10 s; 9900 ms on, 100 ms off Approx. 108 J Spot size required 108 / 9933 = 1.09·10-2 m2 Radius = √(1.09·10-2 /π) = 5.9 cm

Beam Divergence NA = 0.37 NA = n·sin Refractive index of fibre = 1.62 (Wikipedia)  = sin-1(NA/n) = 230 mrad This value can be calculated also by using an equation within IEC standards

NOHD 9900 ms 26 cm ? Distance required 0.059 m /0.23 rad = 26 cm There are rounding errors in these calcuations 230 mrad 26 cm ?

NOHD 100 ms Maximum energy output Spot size required Distance required 12 W over 10 s; 100 ms on, 100 ms off Approx 6 J Spot size required 6 / 3150 = 1.90·10-3 m2 Radius = √(1.90·10-3 /π) = 2.5 cm Distance required 0.025 m /0.23 = 10.8 cm Other MPE/NOHD calculations are more restrictive and therefore should be followed

Calculation 3 Ophthalmic YAG Laser λ = 1064 nm Aperture = 50 mm (8 μm spot size) Focal length 107 mm Cone Angle = 16° Max Energy = 10 mJ Pulse Duration = 4 ns Single pulse

Maximum Permissible Exposure 1064 nm, 4 ns pulse length Thermal effects on the retina (any pulse length) Assume α < αmin, C6=1 MPE = 2·10-2·C7 C7 = 1 MPE = 20 mJm-2 This is more restrictive in 60825-1:2014 compared to 2007 version (50 mJm-2) – legally in the UK, the previous limits still apply.

Divergence Cone angle (16°)? Divergence 8° (140 mrad)?

NOHD 5.7 m ? Spot size required Distance required 20 mJm-2 / 10 mJ = 2 m2 Radius = √(2 /π) = 80 cm Distance required 0.80 /0.14 = 5.7 m Effect of ionisation? 80 cm ? 140 mrad 5.7 m ?

Thank you

Beam diameter d63 definition? If d86 is 12 mm, then d63 = 8.5 mm

LaserBee NOHD Assume full power of 12 W for different times. Set power profile to continuous and enter 12 W. This gives the default 100 s NOHD of 25.8 cm. If you change the exposure time to 100 ms then the NOHD is 47.0 cm LaserBee calculations courtesy of Geoff Atkins

LaserBee NOHD Assume that the cone angle is the required full angle d63 divergence source diameter of 8 m (we are measuring the NOHD from the focal spot). Enter a pulse energy of 10 mJ and lookup the NOHD from the tab to get 2.84 m. Increasing the energy by 20% to allow for fluctuations makes the NOHD 3.1 m

BS EN 50825-1:2014 Page 69: x = 100 mm

Maximum Exposure Assume 7 mm pupil 18.8 Jcm-2 x π·(0.352) = 7.25 J “The pupil is a variable aperture but the diameter may be as large as 7 mm when maximally dilated in the young eye.” 18.8 Jcm-2 x π·(0.352) = 7.25 J