A quarter-wave plate is made from a piece of calcite with its optic axis oriente

Question

A quarter-wave plate is made from a piece of calcite with its optic axis oriented parallel to

its surface.

a. Determine the thickness of the plate in terms of the wavelength of the radiation and

the two refractive indices.

b. For incident light which is linearly polarised, determine the angle between the plane

of polarisation and the optic axis of the crystal in order that the transmitted light be

circularly polarised.

(you can assume the equation for elliptically polarised light in the notes but

otherwise work from 1st principles)

show working please

Explanation / Answer

By taking Z axis is along optic axis(e-wave) and y axis is perpendicular to optic axis(o wave). If the incident wave of E0 amplitude makes an angle (phi) with z-axis , the output components are<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

Ey=E0 sin(phi) cos(no kx-wt) and Ez=E0cos(phi) cos(ne kx-wt).

the phase difference = (n0 kd- ne kd)=kd(n0-ne)=pi/2 ( QWP gives pi/2 phase diff)

k=2*pi/lambda,

Therefore d=lamba/4*(n0-ne)

b) When the angle is 45 degree the y and z components of the incident wave have equal amplitudes .Thus equal amplitude and a phase difference of pi/2 represents a left circularly polarized wave

From equations Ey=E0 sin45 cos(wt-pi/2) and Ez=E0cos 45 cos(wt)

Ey=E0/sqrt(2) sin(wt) and Ez=E0/sqrt(2) cos(wt)

Ey^2+Ez^2=E0^2 represents a circle

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