1. A Nd:YAG laser has an output power of 10mW at 1064nm. If the laser is not con
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Question
1. A Nd:YAG laser has an output power of 10mW at 1064nm.
If the laser is not continuous but operates in pulsed mode with the same output power and same size, and the pulse duration is 10 ps every 10 ns, what is ratio of the peak electric field of the pulsed laser to that of the continuous laser?
Hint: The intensity of a laser is the optical power per unit area. Intensity is related to the electric field CE E through the equation I For simplicity, you can consider that the pulse is rectangular 2cu in shape.Explanation / Answer
As intensity=power/area
So,you need to calculate the spot diameter of the laser falling on target surface
Diameter of laser beam at the focal point=1064 nm
Radius=diameter/2=1064/2=532 nm
Area=pi*(radius)^2=3.14*(532nm)^2=888695.36 nm^2
intensity=I2=10 mW/888695.36 nm^2=10*10^-3 W/888.695.36*10^-18 m^2=[1.125*10^-6*10^-2]/[(10^-18 m^2)*(10^4 cm^2/m^)]=1.125*10^-8/10^-14=1.125*10^6 W/cm^2
I2=1.125*10^6 W/cm^2
Also for a pulse laser,energy=power*time=10mW*10ns=(10 *10^-3J/s)*(10*10^-9 s)=10^-10J
Power=energy/pulse width=10^-10J/10ps=10^-10J/10^-12s=100 W
So intensity=100W/888695.36 nm^2=1.125*10^-4W/(10^-18 m^2*10^4 cm2/m2)=1.125*10^10W/m^2
Intensity=I1=E^2/2cmo
So for the ratio of I1/I2=E1^2/E2^2
or E1/E2=(I1/I2)^1/2=(1.125*10^10/1.125*10^6)^1/2=100
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