- #1
roam
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We want to examine the thermal effects of irradiating a given material with a laser. The material under consideration can either be homogeneous and isotropic or a diffusive turbid material (e.g. biological tissues). Suppose we decrease the power by a certain factor. Will we still get the same amount of thermal damage if we increase the duration of irradiation by the same factor?
So, the energy density incident on the material (also known as the fluence, ##\psi##) is the energy per unit area. Since optical power (##W##) is energy per unit time, i.e., ##P=E/t \ (J/s)##, therefore, I believe we can write:
$$\psi = \frac{P \ t}{A} \ \text{(typically } J/cm^2) \tag{1}$$
where ##A## is the cross-sectional area of the incident beam.
One reference states that ##\psi## is obtained by integrating the power density ##I## (aka intensity or irradiance, which is the average energy per unit area per unit time, ##W/m^2##) over the irradiation period:
$$\psi = \int I \ dt.$$
where at a radial distance ##r##, ##I## can be expressed in terms of radiance ##L## (power density per unit solid angle)
$$I (r) = \int_{4 \pi} L(r, \hat{s}) \ d \omega.$$
According to this, for a Gaussian beam at normal incidence, the relavant effective beam area is ##A_{\text{eff}}=\pi r^2 /2## and therefore ##I=\frac{2P}{\pi r^{2}}##.
Clearly (1) implies that, for instance, if you let a 2mW laser run for ~28 hours, it is capable of causing the same degree of damage as a 20W laser does in 10 seconds. But is this true?
Is the resulting damage dependant more on the intensity (power density) rather than fluence (energy density)? If so, why? And is it possible to confirm this using equations/analytic expressions?
Any explanation would be greatly appreciated.
So, the energy density incident on the material (also known as the fluence, ##\psi##) is the energy per unit area. Since optical power (##W##) is energy per unit time, i.e., ##P=E/t \ (J/s)##, therefore, I believe we can write:
$$\psi = \frac{P \ t}{A} \ \text{(typically } J/cm^2) \tag{1}$$
where ##A## is the cross-sectional area of the incident beam.
One reference states that ##\psi## is obtained by integrating the power density ##I## (aka intensity or irradiance, which is the average energy per unit area per unit time, ##W/m^2##) over the irradiation period:
$$\psi = \int I \ dt.$$
where at a radial distance ##r##, ##I## can be expressed in terms of radiance ##L## (power density per unit solid angle)
$$I (r) = \int_{4 \pi} L(r, \hat{s}) \ d \omega.$$
According to this, for a Gaussian beam at normal incidence, the relavant effective beam area is ##A_{\text{eff}}=\pi r^2 /2## and therefore ##I=\frac{2P}{\pi r^{2}}##.
Clearly (1) implies that, for instance, if you let a 2mW laser run for ~28 hours, it is capable of causing the same degree of damage as a 20W laser does in 10 seconds. But is this true?
Is the resulting damage dependant more on the intensity (power density) rather than fluence (energy density)? If so, why? And is it possible to confirm this using equations/analytic expressions?
Any explanation would be greatly appreciated.