We have given electric field vectors of two light waves propagating along the positive z-direction.
E1(z,t)=xˆE01cos(kz−ωt)−−−−−(1)
E2(z,t)=yˆE02cos(kz−ωt+ϕ)−−−−−(2)
Where, E_{01} and E_{02} are amplitudes, \omega is cyclic frequency ,
From equation (1),
cos(kz−ωt)=E01E1−−−−−(3)
From equation (2),
E2=E02[cos(kz−ωt)cosϕ−sin(kz−ωt)sinϕ]−−−−−(4)
From equation (4),
sin(kz−ωt)sinϕ=cos(kz−ωt)cosϕ−E02E2−−−−−(5)
Putting (3) equation in (5), We get:
sin(kz−ωt)sinϕ=E01E1cosϕ−E02E2−−−−(6)
Squaring both sides,
(sin(kz−ωt)sinϕ)2=(E01E1cosϕ−E02E2)2−−−(7)
(sin(kz−ωt)sinϕ)2=(E01E1cosϕ)2+(E02E2)2−2×E01E1cosϕ×E02E2−−−(8)
From equation (3),
Multiplying both sides by sinϕ we get,
cos(kz−ωt)sinϕ=E01E1sinϕ−−−(9)
Squaring both sides,
(cos(kz−ωt)sinϕ)2=(E01E1sinϕ)2−−−(10)
Adding Equation (8) and (10)
We get,
sin2ϕ=(E01E1)2+(E02E2)2−2×E01E1cosϕ×E02E2
which is the equation of Ellipse.
In general, projections of light vectors of polarized light on the axes X and Y
Ex=E0cos(ωt−kz)
Ey=E0cos(ωt−kz+ϕ)
they satisfy the equation
(E0Ex)2−2E02ExEycosϕ+(E0Ey)2=sin2ϕ
This equation is an equation of an ellipse whose axes are oriented relative to the coordinate
axes X and Y arbitrarily. The orientation of the ellipse and magnitude of its semiaxes depends only on the angle ϕ (phase difference).
If ϕ=π
Ex+Ey=0
we have the equation of the line. That is, light is linearly or plane polarized.
If ϕ=0
Ex2+Ey2=E02
We have the equation of the circle. That is, light is circularly polarized.
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