Question #65282

Obtain an expression for elliptically polarised light resulting due to superposition of two orthogonal linearly polarised light waves

Expert's answer

Answer on Question #65282, Physics / Optics

Obtain an expression for elliptically polarised light resulting due to superposition of two orthogonal linearly polarised light waves.

Solution:

Consider two coherent and plane polarised light waves. The plane of oscillation of first wave is perpendicular to the plane of oscillation of the second wave. Fluctuations in first wave are located along the axis xx. Fluctuations in second wave are located along the axis yy. Light vector E\vec{E} is a result of the addition of these two oscillations.



Projections of light vectors of these waves:


Ex=A1cosωt(1)E_x = A_1 \cos \omega t \quad (1)Ey=A2cos(ωtφ)(2),E_y = A_2 \cos (\omega t - \varphi) \quad (2),


where A1A_1 and A2A_2 are amplitudes, ω\omega is cyclic frequency, φ\varphi is phase difference.


Of (1)cosωt=ExA1(3)\text{Of (1)} \Rightarrow \cos \omega t = \frac{E_x}{A_1} \quad (3)Of (2)Ey=A2(cosωtcosφ+sinωtsinφ)(4)\text{Of (2)} \Rightarrow E_y = A_2 (\cos \omega t \cos \varphi + \sin \omega t \sin \varphi) \quad (4)Of (4)sinωtsinφ=EyA2cosωtcosφ(5)\text{Of (4)} \Rightarrow \sin \omega t \sin \varphi = \frac{E_y}{A_2} - \cos \omega t \cos \varphi \quad (5)(3) in (5):sinωtsinφ=EyA2ExA1cosφ(6)\text{(3) in (5):} \sin \omega t \sin \varphi = \frac{E_y}{A_2} - \frac{E_x}{A_1} \cos \varphi \quad (6)Of (6)(sinωtsinφ)2=(EyA2ExA1cosφ)2(7)\text{Of (6)} \Rightarrow (\sin \omega t \sin \varphi)^2 = \left(\frac{E_y}{A_2} - \frac{E_x}{A_1} \cos \varphi\right)^2 \quad (7)Of (7)(sinωtsinφ)2=(EyA2)22×EyA2×ExA1cosφ+(ExA1cosφ)2(8)\text{Of (7)} \Rightarrow (\sin \omega t \sin \varphi)^2 = \left(\frac{E_y}{A_2}\right)^2 - 2 \times \frac{E_y}{A_2} \times \frac{E_x}{A_1} \cos \varphi + \left(\frac{E_x}{A_1} \cos \varphi\right)^2 \quad (8)Of (3)cosωtsinφ=ExA1sinφ(9)\text{Of (3)} \Rightarrow \cos \omega t \sin \varphi = \frac{E_x}{A_1} \sin \varphi \quad (9)Of (9)(cosωtsinφ)2=(ExA1sinφ)2(10)\text{Of (9)} \Rightarrow (\cos \omega t \sin \varphi)^2 = \left(\frac{E_x}{A_1} \sin \varphi\right)^2 \quad (10)(8)+(10):sin2φ=(ExA1)2+(EyA2)22×EyA2×ExA1cosφ(11)(8) + (10): \sin^2 \varphi = \left(\frac{E_x}{A_1}\right)^2 + \left(\frac{E_y}{A_2}\right)^2 - 2 \times \frac{E_y}{A_2} \times \frac{E_x}{A_1} \cos \varphi \quad (11)


Expression (11) is a equation of ellipse.

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