Polarised monochromatic light falls at normal incidence on a parallel -sided sla

Question

Polarised monochromatic light falls at normal incidence on a parallel -sided slab of a transparent crystal. The plane of polarisation of the incident light makes an angle oof 45 degrees to the optic axis, the latter being parallel to the surface:

b) A second identical slab, oriented in the same way as the first, is inserted into the  transmitted beam beyond the first slab i)what is the state of polarisation  of the light transmitted by the second slab

ii) how is this related to that of the beam incident on the first slab

( you can assume the equation for rlliptically polarised light or otherwise work from  1st principles

Explanation / Answer

we have formula,

x^2/a^2 +y^2/b^2 -(2xy/ab)*cos(theta)=sin^2(theta)

where,

thetea is phase difference angle the ordinary and extraordinary plane polarised lights

a=Asin(theta')

b=Acos(theta') are the amplitude components of the plane polarised light

where,

theta' is angle between amplitude and optical axis is=45 degrees(given)

from the above equation,

then,

if theta= 90 degrees(special case)

x^2/a^2 +y^2/b^2 -(2xy/ab)*cos(90)=sin^2(90)

x^2/a^2 +y^2/b^2 =1 ======> thus the emergent light becomes the elliptically polarised

now for the first slab

theta'=45 degrees

a=Asin(45)

b=Acos(45)

so we get a=b

then above equation can written as,

x^2 +y^2=a^2 =======> thus the emergent light becomes the circularly polarised

now , the second slab is also inserted and oriented as first slab

so,the state of polarisation of the light transmitted by the second slab is also same as the emerging

light from the first slab.which is circularly polarised.

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