A green beam of light (wavelength 5435 A) produced by a He-ne laser, is unpolari

Question

A green beam of light (wavelength 5435 A) produced by a He-ne laser, is unpolarized and has a total power of 3.00 mW.  The beam passes through  4  linear polarizers.  The angles of the transmission axes of the polorizers are all relative to the vertical axis of the first polorizer.

given::  h = 6.62607 x 10^(-34)  J-s

              c=  2.99792  x 10^ 8  m/s

P=  10.00 mW randomly polarized

1st

2nd = 45.0 degrees

3rd = 70.0 degrees

4th(LAST polarizer) = 90.0 degrees

a.)  if each polarizer has a maximum transmission of 95.0%, how much optical power emerges from the last polarizer?  

HINT:  An ideal linear polarizer will transmit exactly 50% of a randomly polarized beam.

b.)  in the situation above, how many photons per second emerge from the last polarizer on the right??

Explanation / Answer

a)

Through 1st polarizer,

P1=0.95*Po/2=0.95* 5 =4.75 mW

Through second polarizer,

P2=0.95*P1(cos x)^2=0.95*4.75*(cos 45 degrees)^2=2.2565 mW

Through third polarizer,

P3=0.95*P2(cos x)^2=0.95*2.2565*(cos 25 degrees)^2=1.76 mW

Through fourth polarizer,

P4=0.95*P3(cos x)^2=0.95*1.76*(cos 20 degrees)^2=1.476 mW

*********************************************************

Energy of 1 photon=hc/lembda= (6.62607 * 10^(-34) *2.99792 * 10^ 8)/(5435 * 10^-10)

No. of photos per second = P4/Energy of 1 photon=(1.476*10^6)/(3.65*10^-19)=4.04 * 10^24

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