1. Laser Safety Calculations
Tom Lister

2. 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

3. 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

4. MPE
AOR Directive
“In carrying out the assessment, measurement or calculation, the employer must follow the [...] standards of the IEC”
: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 nm, the size of the retinal image.”

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

6. 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
MPE = 18·t0.75·C4
Exposure time, t = 20 ms
t0.75 = 5.3·10-2
C4 = ·( ) = 1.3
MPE = 1.24 Jm-2
There was general agreement with this result

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

8. 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

9. 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 ?

10. C6≠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

11. C6≠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

12. 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 ?

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

14. 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

15. 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

16. 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

17. 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 ?

18. 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

19. 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

20. 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 :2014 compared to 2007 version (50 mJm-2) – legally in the UK, the previous limits still apply.

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

22. 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 ?

23. Thank you

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

25. 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 cm.
If you change the exposure time to 100 ms then the NOHD is 47.0 cm
LaserBee calculations courtesy of Geoff Atkins

26. 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

27. BS EN :2014
Page 69: x = 100 mm

28. 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