Answer to Question #103190 in Optics for Mahesh devkota

Question

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

Answer 1

In general, projections of light vectors of polarized light on the axes X and Y

$E_x=E_0\cos(\omega t-kz)$

$E_y=E_0\cos(\omega t-kz+\phi)$

they satisfy the equation

$\frac{E^2_x}{E^2_0}-2\frac{E_xE_y}{E^2_0}\cos\phi+\frac{E^2_y}{E^2_0}=\sin^2\phi$

This equation is an equation of an ellipse whose axes are oriented relative to the coordinate axes X and Y arbitrarily. The orientation of the ellipse and magnitude of its semiaxes depends only on the angle $\phi$ (phase difference).

if $\phi=\pi$

$E_x+E_y=0$

we have the equation of the line. That is, light is linearly or plane polarized.

If $\phi=0$

$E^2_x+E^2_y=E^2_0$

we have the equation of the circle. That is, light is circularly polarized.