Question

Show that plane polarized light and circularly polarized light are special cases of
elliptically polarized light.

Answer 1

Answer on Question#39072, Physics, Optics

Show that plane polarized light and circularly polarized light are special cases of elliptically polarized light.

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

For E-ray: $x = A \cos \theta \sin(\omega t + \delta) = a \sin(\omega t + \delta)$ (1)

For O-ray: $y = A \sin \theta \sin \omega t = b \sin \omega t$ (2)

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:

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

This is the general equation of an ellipse.

Spesial cases:

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

$$\begin{array}{l} \frac {x ^ {2}}{a ^ {2}} + \frac {y ^ {2}}{b ^ {2}} - \frac {2 x y}{a b} \cos 0 = \sin^ {2} 0 \\ \frac {x ^ {2}}{a ^ {2}} + \frac {y ^ {2}}{b ^ {2}} - \frac {2 x y}{a b} = 0 \\ \left[ \frac {x}{a} - \frac {y}{b} \right] ^ {2} = 0 \\ \end{array}$$

or,

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

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

2. When $\delta = \pi / 2 \sin \delta = 1$ and $\cos \delta = 0$, therefore

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

or,

$$\frac {x ^ {2}}{a ^ {2}} + \frac {y ^ {2}}{b ^ {2}} = 1$$

This is the equation of an ellipse. In this case, the emergent light is elliptically polarized.

When $\delta = \pi /4\sin \delta = 1 / \sqrt{2}$ and $\cos \delta = 1 / \sqrt{2}$, therefore

$$\frac {x ^ {2}}{a ^ {2}} + \frac {y ^ {2}}{b ^ {2}} - \sqrt {2} \frac {x y}{a b} = \frac {1}{2}$$

which is again an equation of ellipse.

3. When $\delta = \pi /2$ and $\theta = 45{}^{\circ}$, therefore $a = b$, and

$$\frac {x ^ {2}}{a ^ {2}} + \frac {y ^ {2}}{a ^ {2}} = 1$$

This is the equation of an circle. In this case, the emergent light is circularly polarized.

In general, the resultant of two plane polarized beams is an elliptically polarized light. Under certain conditions ($\delta = 0$ or, $\delta = \pi /2$ and $\theta = 45{}^{\circ}$), however, the resultant light is plane or circularly polarized.

Thus, the plane polarized light and circularly polarized light are the special cases of elliptically polarized light.