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Detection of Plane, Circularly and Elliptically Polarised Light

Consider a plane polarised light of wavelength λ. and amplitude a from Nicol N is incident normally on a doubly refracting calcite crystal cut with its faces parallel to optic axis and is oriented in such a way that vibration in plane polarised light makes an angle El with optic axis of the crystal as shown in Figure 6.13 (b).

 

On entering the crystal the light gets splitted into two components called ordinary and extraordinary and represented as O-ray and E-ray respectively. The O-ray has vibrations perpendicular to optic axis with amplitude a sin θ and E-ray has vibrations along optic axis with amplitude a cos El as shown in Figure 6.13 (c). Both the components travels in same direction but with different velocities. In case of calcite E-ray  travels faster than 0-ray (µO > ) whereas in case of quartz the situation is reversed.

Although before entering the crystal, the two components are in phase but owing to their different velocities within the crystal, the two component are in phase but owing to their different velocities within the crystal, the two component on emergence have different phases.

 

The phase difference or path difference between 0-ray and E-ray depends upon thickness and nature of crystal and wavelength of incident light.

 

Thus incident plane polarised light after passing through the crystal consist of two simple harmonic  vibration, in two mutually perpendicular planes having same period but different amplitude and phases. They superimpose and result into a single motion which can be linear, circular or elliptical polarised depending upon value of El and phase difference.

 

To calculate phase difference betWeen 0-ray and E-ray

 

By definition , µ =λ airmed

So                    λ E =λ /µ E and λ o =λ/µo

 

Number  of waves within  the crystal = thickness /  wavelength

For E –ray ,                 NE =t /λE =tµ E

For E –ray ,                 no =t /λo =tµo

 

As we have considered calcite crystal i.e., negative crystal, so µo > µE Therefore the number of waves .by which E-ray leads 0-ray can be given as

 

1/λ (µOE)

As before entering both the rays were in same phase and each wave corresponds to a phase angle of 2n radians, therefore phase difference between  E -ray and 0-ray after emergences

will be

.

δ =2π t/λ(µoE)

Let displacements of E-ray and 0-ray along and perpendicular to optic axis may be represented as

 

X= a cos θ sin wt                    for E –ray….(1)

Y =a sin (wt +δ)                      for E – ray….(2)

 

Equations (1) and (2) may be written as

X  =A sin wt

Y = a sin (wt-δ)

 

Where A=a cos θ and B a= sin θ

From eduation (4) ,we get

y/B =sin(wt+δ)

y/B =sin wt cos δ + cos wt sin δ

or

from education  (3) ,we get

 

sin wt =x/A therefore ,cos wt =√1-x2/A2

To calculate phase difference between 0-ray and E-ray

By definition,             µ = λ air/ λmed

                                                            λ e =λ / µ E and λo =λ / µo

Number  of waves within the crystal =thickness /  wavelength

 

For E – ray , nE =t / λ E =t µ E / λ

 

For E – ray ,no =t / λo =tµo

 

 

As we have considered calcite crystal i.e., negative crystal, so µo > µE. Therefore the

number of waves .by which E-ray leads 0-ray can be given as

1/λ (µoE)

 

As before entering both the rays were in same phase and each wave corresponds to a phase angle of 2π radians, therefore phase difference between x -ray and 0-ray after emergences will be

δ =2π t / λ (µoE)

 

Let displacements of E-ray and 0-ray along and perpendicular to optic axis may be

represented as

x =a cos θ sin wt

y =a sin θ sin (wt +δ)

 

Equations (1) and (2) may be written as

x =A sin wt

y =B sin (wt +δ)

 

where A =a cos θ  and B  =a sin  θ

 

from equation (4) ,we get

y/B =sin (wt+ δ)

y/B =sin wt cos δ +  cos wt sin δ

 

from equation (3) ,we get

sin wt =x/A there ,cos wt =√ 1-x2/A2

 

Put sin wt and cos wt in equation (5) ,we get

y/B =x/A cos δ  +  1- x2 /A2 sin δ

or                                 y/B – x/A cos δ  =  1-x2 /a2 sin   δ

 

Squaring both sides, we have

Y2/B2 = X2 /A2 cos δ  -2xy/AB cos δ =(1-x2 /A2)  sin 2 δ

Or                                x2/A2 + y2/B2 -2xy /AB cos δ = sin 2 δ

 

This is general equation of ellipse

 

Special case:

 

1 . When  =δ 0, sin  = δ 0 and cos δ = 1

From equation (6),

X2/A2 + y2/B2  -2xy /AB =0

x/A –y / B = 0

y = B/A x

 

This is equation of a straight line. Thus emergent light will be plane polarised.

2. when δ =π/2 ,3π/2 ….. (2n +1)π /2

From  equation (6) ,

 

X2 / A2 +y2/B2 =1

 

This is the equation of symmetrical ellipse. Hence emergent light will be elliptically polarised.

 

3. When   =  δ =π/2 ,3π/2 ….. (2n +1)π /2 and a=b

X2 +y2 =A2

 

This  is the equation of circle of radius a. Hence emergent ray will be circularly polarised.

 

Note. For A= B, the vibration of the incident light makes an angle 45° with the optic axis.

From equation (6),

X2 /A2 +y2 / B2 +2xy /AB =0 or x /A +y/B =0

Y = – B /A x

 

which is again equation of a straight line but with negative slope.

 

5. when δ   =  π/4  , 3 π/4 …. Sin  =cos  =1/  √2

From equation (6)

X2 /A2 +y2/B2 +  2 xy/AB =1/2       – ve sign for δ =π /4

+ ve sign for δ  =3 π /4

 

This equation of ellipse in which major axis .is inclined· to positive and negative x-axis

respectively for δ=π/4and δ =3π /4   respectively