Question

Show that the superposition of two linearly polarized light waves having different amplitude and a finite phase difference can be used to produce elliptically plane polarized light waves.

Answer 1

Answer on Question #61897, Physics / Optics

Show that the superposition of two linearly polarized light waves having different amplitude and a finite phase difference can be used to produce elliptically plane polarized light waves.

Answer:

Let $a$ and $b$ be the amplitudes of first linearly polarized light wave and $b$ of second linearly polarized light wave, respectively.

If $\delta$ be the phase difference between the two emergent beams, then their vibrations can be expressed as

For first wave: $x = a \sin(\omega t + \delta)$ (1)

For second wave: $y = b \sin \omega t$ (2)

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.

When $\delta = \pi /2$, then $\sin \delta = 1$ 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.

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