Question #182698

The electric field vectors of two light waves propagating along the positive zdirection

are given as

E1(z,t) = xˆ E01cos(kz - (omega)t)

E2(z,t) = yˆ E02cos(kz-(omega)t + (si))

where xˆ and yˆ are unit vectors along x and y-axes, respectively. Show that when

these two waves superpose, we obtain elliptically polarised light. Also show that

the linear and circular polarisations are special cases of elliptical polarisation


1
Expert's answer
2021-05-05T14:50:45-0400

We have given electric field vectors of two light waves propagating along the positive z-direction.


E1(z,t)=xˆE01cos(kzωt)(1)E_1(z,t) = xˆ E_{01}cos(kz - \omega t)-----(1)

E2(z,t)=yˆE02cos(kzωt+ϕ)(2)E_2(z,t) = yˆ E_{02}cos(kz-\omega t + \phi)-----(2)


Where, E_{01} and E_{02} are amplitudes, \omega is cyclic frequency ,

From equation (1),

cos(kzωt)=E1E01(3)cos(kz- \omega t) = \dfrac{E_1}{E_{01}}-----(3)

From equation (2),


E2=E02[cos(kzωt)cosϕsin(kzωt)sinϕ](4)E_2 = E_{02}[cos(kz- \omega t)cos \phi - sin(kz- \omega t)sin \phi]-----(4)


From equation (4),


sin(kzωt)sinϕ=cos(kzωt)cosϕE2E02(5)sin(kz-\omega t)sin \phi = cos(kz - \omega t)cos\phi - \dfrac{E_2}{E_{02}}-----(5)


Putting (3) equation in (5), We get:


sin(kzωt)sinϕ=E1E01cosϕE2E02(6)sin(kz- \omega t)sin \phi = \dfrac{E_1}{E_{01}} cos\phi - \dfrac{E_2}{E_{02}}----(6)


Squaring both sides,


(sin(kzωt)sinϕ)2=(E1E01cosϕE2E02)2(7)(sin(kz- \omega t)sin \phi)^2 = (\dfrac{E_1}{E_{01}} cos\phi - \dfrac{E_2}{E_{02}})^2---(7)


(sin(kzωt)sinϕ)2=(E1cosϕE01)2+(E2E02)22×E1cosϕE01×E2E02(8)(sin(kz- \omega t)sin \phi)^2 = (\dfrac{E_1cos \phi}{E_{01}})^2+ (\dfrac{E_2}{E_{02}})^2 - 2\times \dfrac{E_1 cos \phi}{E_{01}} \times \dfrac{E_2}{E_{02}}---(8)


From equation (3),


Multiplying both sides by sinϕsin \phi we get,

cos(kzωt)sinϕ=E1E01sinϕ(9)cos(kz- \omega t)sin\phi = \dfrac{E_1}{E_{01}} sin\phi---(9)

Squaring both sides,


(cos(kzωt)sinϕ)2=(E1E01sinϕ)2(10)(cos(kz- \omega t)sin\phi)^2 = (\dfrac{E_1}{E_{01}} sin\phi)^2---(10)


Adding Equation (8) and (10)


We get,


sin2ϕ=(E1E01)2+(E2E02)22×E1cosϕE01×E2E02sin^2 \phi = (\dfrac{E_1}{E_{01}})^2 + (\dfrac{E_2}{E_{02}})^2 - 2\times \dfrac{E_1 cos \phi}{E_{01}} \times \dfrac{E_2}{E_{02}}


which is the equation of Ellipse.


In general, projections of light vectors of polarized light on the axes X and Y


Ex=E0cos(ωtkz)E_x = E_0cos(\omega t- kz)


Ey=E0cos(ωtkz+ϕ)E_y = E_0cos(\omega t - kz+ \phi)


they satisfy the equation


(ExE0)22ExEyE02cosϕ+(EyE0)2=sin2ϕ(\dfrac{E_x}{E_0})^2 - 2\dfrac{E_xE_y}{E_0^2}cos\phi + (\dfrac{E_y}{E_0})^2 = sin^2 \phi


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\phi (phase difference).


If ϕ=π\phi = \pi


Ex+Ey=0E_x + E_y = 0


we have the equation of the line. That is, light is linearly or plane polarized.


If ϕ=0\phi = 0


Ex2+Ey2=E02E_x^2+ E_y^2 = E_0^2


We have the equation of the circle. That is, light is circularly polarized.








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