Response:
a) 318.2 W/m^2
b) 2.5 x 10^-4 J
c) 1.55 x 10^-8 v/m
Reasoning:
The laser power P = 1 mW = 1 x 10^-3 W
duration t = 250 ms = 250 x 10^-3 s
Taking a beam diameter of 2 mm = 2 x 10^-3 m
therefore
the beam's cross-sectional area A = = (3.142 x
)/4
A = 3.142 x 10^-6 m^2
a) The intensity I = P/A
where P refers to the laser's power
and A represents the beam's cross-sectional area
I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2
b) The total energy delivered E =Pt
where P is the beam's power
and t is the exposure duration
E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J
c) The peak electric field can be computed as
E =
where I signifies the beam's intensity
and E is the electric field
c is the speed of light = 3 x 10^8 m/s
= 8.85 x 10^9 m kg s^-2 A^-2
E = = 1.55 x 10^-8 v/m