Optical density (0.0.) is defined as the logarithm (power of ten) of the attenuation of light transmitted through protective eye wear at a particular wavelength. For example, if glasses attenuate light at 1,064 nm. to 1 / 100,000 of its original intensity, the optical density is 5 (the attenuation is 100,000 = 105 ). The A.N.S.I. standards Z136.3 of 1996 ~te that the maximum permissible power density on the cornea for continuous viewing (30,000 seconds) at 514 ~. is 1x 10-6 watts/cm.2 If the intensity of reflected argon-ion-Iaser light at the surgeon's face is 1.0 watt / cm.2 during removal of a port-wine stain, what must be the minimum 0.0. of the surgeon's protective glasses to comply with this requirement? Dot the box adjacent to the correct answer below:
D3.0 D4.0 D5.0 D6.0 D 7.0.
To determine the minimum optical density required for the surgeon's protective glasses, we can use the equation:
Optical Density = log10 (1 / Transmission Factor)
The transmission factor represents the fraction of light that is transmitted through the protective glasses, which is equal to the attenuation.
Given that the attenuation is 1/100,000 (which is 105), we can calculate the transmission factor:
Transmission Factor = 1 / 100,000 = 0.00001
Now, we can find the optical density:
Optical Density = log10 (1 / 0.00001) = log10 (100,000) = 5
So, the minimum optical density required for the surgeon's protective glasses to comply with the requirement is 5.
Therefore, the correct answer is D5.0.