Answer:
a) 318.2 W/m^2
b) 2.5 x 10^-4 J
c) 1.55 x 10^-8 v/m
Explanation:
Power of laser P = 1 mW = 1 x 10^-3 W
exposure time t = 250 ms = 250 x 10^-3 s
If beam diameter = 2 mm = 2 x 10^-3 m
then
cross-sectional area of beam A = [tex]\pi d^{2} /4[/tex] = (3.142 x [tex](2*10^{-3} )^{2}[/tex])/4
A = 3.142 x 10^-6 m^2
a) Intensity I = P/A
where P is the power of the laser
A is the cros-sectional area of the beam
I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2
b) Total energy delivered E = Pt
where P is the power of the beam
t is the exposure time
E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J
c) The peak electric field is given as
E = [tex]\sqrt{2I/ce_{0} }[/tex]
where I is the intensity of the beam
E is the electric field
c is the speed of light = 3 x 10^8 m/s
[tex]e_{0}[/tex] = 8.85 x 10^9 m kg s^-2 A^-2
E = [tex]\sqrt{2*318.2/3*10^8*8.85*10^9}[/tex] = 1.55 x 10^-8 v/m
(a) The intensity of laser beam is [tex]318.2 \;\rm W/m^{2}[/tex].
(b) The total energy delivered before the blink reflex shuts the eye is [tex]2.5 \times 10^{-4} \;\rm J[/tex].
(c) The required value of peak electric field in the laser beam is [tex]1.55 \times 10^{-8} \;\rm V/m[/tex].
Given data:
The power of laser is, [tex]P=1 \;\rm mW = 1 \times 10^{-3} \;\rm W[/tex].
The exposure time is, [tex]t = 250\;\rm ms = 250 \times 10^{-3} \;\rm s[/tex].
The beam diameter is, [tex]d = 2 \;\rm mm = 2 \times 10^{-3} \;\rm m[/tex].
a)
The standard expression for the intensity of beam is given as,
I = P/A
Here, P is the power of the laser and A is the cross-sectional area of the beam. And its value is,
[tex]A =\pi /4 \times d^{2}\\\\A =\pi /4 \times (2 \times 10^{-3})^{2}\\\\A =3.142 \times 10^{-6} \;\rm m^{2}[/tex]
Then intensity is,
[tex]I = (1 \times 10^{-3})/(3.142 \times 10^{-6})\\\\I =318.2 \;\rm W/m^{2}[/tex]
Thus, the intensity of laser beam is [tex]318.2 \;\rm W/m^{2}[/tex].
(b)
The expression for the total energy delivered is given as,
E = Pt
Solving as,
[tex]E = 1 \times 10^{-3} \times (250 \times 10^{-3})\\\\E = 2.5 \times 10^{-4} \;\rm J[/tex]
Thus, the total energy delivered before the blink reflex shuts the eye is [tex]2.5 \times 10^{-4} \;\rm J[/tex].
(c)
The expression for the peak electric field is given as,
[tex]E = \sqrt{\dfrac{2I}{c \times \epsilon_{0}}}[/tex]
Solving as,
[tex]E = \sqrt{\dfrac{2 \times 318.2}{(3 \times 10^{8}) \times (8.85 \times 10^{9})}}\\\\E =1.55 \times 10^{-8} \;\rm V/m[/tex]
Thus, the required value of peak electric field in the laser beam is [tex]1.55 \times 10^{-8} \;\rm V/m[/tex].
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