Answer on Question #45658, Physics, Optics
Derive equation of an ellipse for polarized light.
Solution:
Suppose that a plane polarized light beam of amplitude A A A θ \theta θ A cos θ A\cos\theta A cos θ A sin θ A\sin\theta A sin θ E E E O O O δ \delta δ
For E-ray: x = A cos θ sin ( ω t + δ ) = a sin ( ω t + δ ) \text{For E-ray: } x = A \cos \theta \sin (\omega t + \delta) = a \sin (\omega t + \delta) For E-ray: x = A cos θ sin ( ω t + δ ) = a sin ( ω t + δ ) For O-ray: y = A sin θ sin ω t = b sin ω t \text{For O-ray: } y = A \sin \theta \sin \omega t = b \sin \omega t For O-ray: y = A sin θ sin ω t = b sin ω t
where a = A cos θ a = A\cos \theta a = A cos θ b = A sin θ b = A\sin \theta b = A sin θ
From second equation we have:
y b = sin ω t \frac{y}{b} = \sin \omega t b y = sin ω t
Hence cos ω t = 1 − sin 2 ω t = 1 − y 2 b 2 \cos \omega t = \sqrt{1 - \sin^2 \omega t} = \sqrt{1 - \frac{y^2}{b^2}} cos ω t = 1 − sin 2 ω t = 1 − b 2 y 2
From first equation we have:
x = a sin ( ω t + δ ) = a ( sin ω t cos δ + cos ω t sin δ ) x = a \sin (\omega t + \delta) = a (\sin \omega t \cos \delta + \cos \omega t \sin \delta) x = a sin ( ω t + δ ) = a ( sin ω t cos δ + cos ω t sin δ )
or,
x a = sin ω t cos δ + cos ω t sin δ = y b cos δ + 1 − y 2 b 2 sin δ \frac{x}{a} = \sin \omega t \cos \delta + \cos \omega t \sin \delta = \frac{y}{b} \cos \delta + \sqrt{1 - \frac{y^2}{b^2}} \sin \delta a x = sin ω t cos δ + cos ω t sin δ = b y cos δ + 1 − b 2 y 2 sin δ
or,
x a − y b cos δ = 1 − y 2 b 2 sin δ \frac{x}{a} - \frac{y}{b} \cos \delta = \sqrt{1 - \frac{y^2}{b^2}} \sin \delta a x − b y cos δ = 1 − b 2 y 2 sin δ
Squaring and rearranging, we get:
x 2 a 2 + y 2 b 2 − 2 x y a b cos δ = sin 2 δ \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} \cos \delta = \sin^2 \delta a 2 x 2 + b 2 y 2 − ab 2 x y cos δ = sin 2 δ
This is the general equation of an ellipse.
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