Question

derive linear circular polarized light from the equation of an ellipse

Answer 1

Answer on Question #45659, Physics, Optics

Derive linear circular polarized light from the equation of an ellipse.

Solution:

Suppose that a plane polarized light beam of amplitude $A$ is incident on a uniaxial crystal at an angle $\theta$. Let $A\cos\theta$ and $A\sin\theta$ be the amplitudes of $E$-ray and $O$-ray respectively. If $\delta$ be the phase difference between the two emergent beams, then their vibrations can be expressed as

$$\text{For E-ray: } x = A \cos \theta \sin (\omega t + \delta) = a \sin (\omega t + \delta)$$ $$\text{For O-ray: } y = A \sin \theta \sin \omega t = b \sin \omega t$$

where $a = A\cos \theta$ and $b = A\sin \theta$

From second equation we have:

$$\frac{y}{b} = \sin \omega t$$

Hence $\cos \omega t = \sqrt{1 - \sin^2 \omega t} = \sqrt{1 - \frac{y^2}{b^2}}$

From first equation we have:

$$x = a \sin (\omega t + \delta) = a (\sin \omega t \cos \delta + \cos \omega t \sin \delta)$$

or,

$$\frac{x}{a} = \sin \omega t \cos \delta + \cos \omega t \sin \delta = \frac{y}{b} \cos \delta + \sqrt{1 - \frac{y^2}{b^2}} \sin \delta$$

or,

$$\frac{x}{a} - \frac{y}{b} \cos \delta = \sqrt{1 - \frac{y^2}{b^2}} \sin \delta$$

Squaring and rearranging, we get:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} \cos \delta = \sin^2 \delta$$

This is the general equation of an ellipse.

Spesial cases:

1. When $\delta = 0$ sin $\delta = 0$ and cos $\delta = 1$, therefore

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} \cos 0 = \sin^2 0$$ $$\frac {x ^ {2}}{a ^ {2}} + \frac {y ^ {2}}{b ^ {2}} - \frac {2 x y}{a b} = 0$$ $$\left[ \frac {x}{a} - \frac {y}{b} \right] ^ {2} = 0$$

or,

$$y = \frac {b}{a} x$$

This is the equation of a straight line. In this case, the emergent light is plane polarized.

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