polarized light of intensity s0 passes through an ideal polarizer. if the electric vector of the polarized light is horizontal what, in terms of the initial intensity s0, is the intensity of the light that passes through a polarizer if that polarizer is tilted 0.449 rad from the horizontal?



Answer :

80% of the initial intensity s0, is the intensity of the light that passes through a polarizer if that polarizer is tilted 0.449 rad from the horizontal

Polarization could be a property of transversal waves characterizing a particular oscillation direction. Within the case of light, in specific, we are managing a transversal electromagnetic wave where the magnetic and electric fields oscillate in opposite direction among them and perpendicular to the propagation direction. The polarization of light is characterized by the oscillation direction of the electric field within the EM wave. since the electric vector of the polarized light, oscillates horizontally so the light is linearly polarized.when  it  passes through an ideal polarizer, it forms an angle  with the  axis and oscillation direction of the electric field  which  is  θ=0.449rad, the intensity of the light according to Malus's law will be

S(f)=S₀cos2θ

where S(f)=  intensity after crossing the polarize

S₀  = the initial intensity.

S(f)=S₀cos2(0.449)

=>Sf=0.812S0

so the intensity of the polarized  light is about 80% of the initial intensity.

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