If the eye receives an average 1.0 * 10^2 W/m^2 intensity greater than damage to
ID: 1768165 • Letter: I
Question
If the eye receives an average 1.0 * 10^2 W/m^2
intensity greater than damage to the retina can
occur. This quantity is called the damage threshold of the retina.
(a) What is the largest average power (in mW) that a laser beam
1.5 mm in diameter can have and still be considered safe to view
head-on? (b) What are the maximum values of the electric and
magnetic fields for the beam in part (a)? (c) How much energy
would the beam in part (a) deliver per second to the retina?
(d) Express the damage threshold in W cm2
Explanation / Answer
A)
Area
A=pi*d^2/4 =pi*(1.5*10^-3)^2/4
A=1.767*10^-6 m^2
Average power
P=I*A=(1*10^2)*(1.767*10^-6)
P=1.767*10^-4 W
P=0.1767 mW
b)
Maximum Electrci field is
Emax=sqrt[uo*c*I] =sqrt[4pi*10^-7*(3*10^8)*(1*10^2)]
Emax=194.16 V/m
Maximum magnetic field
Bmax=sqrt[uo*I/c]=sqrt[4pi*10^-7*((1*10^2)/(3*10^8)]
Bmax=6.47*10^-7 T
c)
Energy in 1 second
E=P*t =1.767*10^-4*1
E=1.767*10^-4 J or 0.1767 mJ
d)
I=100 W/m^2 =100*10^-4 W/cm^2
I=0.01 W/cm^2