6. The light comes through the polarizer, then the birefringent uniaxial plate w

Question

6. The light comes through the polarizer, then the birefringent uniaxial plate with optical axis at the angle of 45° with respect to the polarizer and then analyser. Calculate the relative transmission of the light passed through the system, when analyser is (a) p (b) crossed with a polarizer. The refractive indexes of the retarder plate in ne and no, the thickness of the plate is d, the wavelength of the light M. Describe, how you can observe birefringent colours, using this configuration. arallel;

Explanation / Answer

A polarizing sheet has a characteristic polarizing direction along its transmission axis. The sheet transmits only the component of the electric field vector parallel to its transmission axis .The component perpendicular to the transmission axis is completely absorbed , this is called selective absorption. Therefore, light emerging from the polarizer is linearly polarized in the direction parallel to the transmission axis .The linear polarizer is oriented such that its transmission axis is horizontal. Light incident on the polarizer is unpolarized .

Angle of 45° : I = Io/2  

i.e., the intensity transmitted by the second polarizer is independent of the angle .

Above equation is the transmission function of the second linear polarizer if circularly polarized light is incident on it. The above analysis verifies that circular polarization should be obtained by placing the optic axis of a quarter-wave plate at 45 degrees with respect to an incident linearly polarized light. We also conclude that rotating the second linear polarizer does not change the transmission of the circularly polarized light .

E = Eoe iwt(e i cos . cos( ) sin .sin( ))

Keep in mind that a half wave plate introduces a phase shift of radians between the two components of the E vector. The intensity I of the transmitted light being proportional to |E| ,

I |E| 2 = E.E = Eo 2 [cos2 . cos2 ( ) + sin2 .sin2 ( ) cos . cos( ).sin .sin( )(e i + e i)

= Eo 2 [cos. cos( ) + sin .sin( )]2

Since (e i + e i)= 2 cos = 2,

I = Io[cos. cos( ) + sin .sin( )]2 .  

consider the case where the optic axis of the /2 plate is at 45 degrees with respect to the transmission axis of the first linear polarizer.

Substituting = /4 radians in equation 6 we get, I = 1 2 Io[1 + 2 cos( 4 ).sin( 4 )]  

Using sin(2a) = 2sin(a)cos(a),

I = 1 2 Io[1 + sin(2 2 )]

Since sin(a 2 ) = cos(a),

I = 1 2 Io[1 cos(2)]

Using cos(2a) = 2 cos2 (a) 1,

I = 1 2 Io[2 2 cos2 ()] = Io[1 cos2 ()]  

I = Io sin2 ()

A retarder (or waveplate) resolves a light wave into two orthogonal linear polarization components by producing a phase shift between them. Depending on the induced phase difference, the transmitted light may have a different type of polarization than the incident beam  

I = A cos2 + B  

The intensity I of the transmitted light is proportional to |E| 2 , I |E| 2 = E.E

Equation is valid for light which is completely linearly polarized (B = 0 in equation 3). If unpolarized or partially polarized light is used, B will not be zero. For non-zero B, rotating the polarizer cannot completely extinguish the light  

Birefringent materials are the basis of the construction of retarders because they can introduce a phase difference between the two components of the electric field incident on the retarder.  

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