A 20 MW laser is outfitted with a polarizer. The laser passes through a second p

Question

A 20 MW laser is outfitted with a polarizer. The laser passes through a second polarizer that is oriented such that it's axis of polarization is rotated 60o from the axis of polarization of the first polarizer. The laser has a beam divergence, which is the angle that the beam sweeps, of 0.002 radians, and is incident on a screen located a distance of 1000 m away.

Part A - Assuming that the first polarizer cuts the laser's intensity in half, what is the magnitude of the electric field at the screen?

  E=(15/40c) ×10^3Vm     E=(5/20c) ×10^6Vm     E=(5/20c) ×10^3Vm     E=(15/40c) ×10^6Vm     E=(15/0c) ×10^3Vm     E=(5/0c) ×10^6Vm     E=(150c) ×10^6Vm     E=(30c) ×10^3Vm  

Explanation / Answer

power transmitted through first polarizer, P1 = Po/2

= 20*10^6 W/2

= 10*10^6 W/2

power trnasmitted throiugh second polarizer, P2 = P1*cos^2(60))

= 10*10^6*(0.5)^2

= 10*10^6*0.25

= 2.5*10^6 W

diameter of the beam on the screen, d = r*theta

= 1000*0.002

= 2 m

Area of the beam on screen, A = pi*d^2/4

= pi*2^2/4

= pi m^2

I = power/Area

= 2.5*10^6/pi

Intensity of light wave, I = Emax^2/(2*c*mue)

Emax = sqrt(2*I*c*mue)

= sqrt(2*I*c/(C^2*epsilon)) (since, mue = 1/(epsilon*C^2))

= sqrt(2*I/(c*epsilon))

= sqrt(2*2.5*10^6/(pi*epsion*C) )

= sqrt(5/(pi*epsilon*C))*10^3 V/m

or

= sqrt( (5/(pi*epsilon*C)*10^6 ) V/m

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