Paper-II Chapter-Polarization of light
Nicol Prism:
Construction: A Nicol prism is an optical device, made by a calcite crystal having a length to breath ratio as 3 : 1. The end faces AB and CD a of a calcite crystal ABCD are properly ground to make the angles in the principal section 710 and 1090 . The crystal is then cut into two haves along the plane A0B0, perpendicular to both the principal section and end faces. The two faces are polished optically flat and joined together by a transparent liquid (canada balsam) of refracted index µ = 1.55. The crystal is finally enclosed in a tube blackened inside and formation of Nicol prism is complete.
Use of Nicol prism as polarizer:
Inside the crystal the unpolarise light PM splits into O-ray and E-ray. Again, we have µe < µc < µo, where µc is the r.i. of canada balsam. So when the O-ray reaches the balsam layer, it passes from a denser to a rarer medium. Since the length of the crystal is large, the O-ray is usually incident at the at the calcite balsam surface at an angle greater than its critical angle (690) and gets totally reflected to be finally absorbed by the tube enclosing the crystal.
The E-ray is transmitted by the calcite-balsam surface to emerge from the Nicol as plane polarized light with vibrations parallel to the principal section.
Use of Nicol prism as analyzer:
1 Step I: When an unpolarise ray is incident on a Nicol prism the emergent ray is plane polarised with vibration in the principal section of P. If this emergent ray falls on a second Nicol A whose principal section is parallel to that of P, the vibrations will be in the principal section of A and the ray behaves as E-ray in it and will be transmitted completely. The Intensity will be maximum.
Step II: Now if A is rotated such that its principal section perpendicular to that of P, the ray will behaves as O-ray inside A. So it gets lost by total reflection. No light will emerge from A and the Nicols P and A are said to crossed.
Step III: If A is further rotated till its principal section is again parallel to that of P, the intensity of the emergent light will again be a maximum.
So, a rorating Nicol shows intensity variation with zero minimum.
Limitations: A Nicol prism cannot be used in highly convergent or divergent beams.
Malus law: It states that if a completely plane polarized light is incident on an analyser, the intensity of the emergent light varies as the square of the cosine of the angle between the planes of transmission of the polariser and the analyser.
i.e. 2 Iθ = Icos θ
2 Proof: Let a be the amplitude of light transmitted by polariser and θ the angle between the planes of transmission of the polariser and the analyser. This plane polarised light is incident on the analyger.
The amplitude a be resolved into two components:
(i) asinθ perpendicular to the plane of transmission of the analyser and is elimi- nated. (i) acosθ along to the plane of transmission of the analyser and only is freely transmitted by the analyser. Intensity of light emerging from the analyser is
2 2 2 Iθ = a cos θ = Icos θ
Where I = a2 is the intensity of polarised light incident on analyser.
Case I: If the polariser and analyser are parallel to each other, then θ = 0 or 1800,
Iθ = I Case I: If the polariser and analyser are perpendicular to each other, then θ = 900,
Iθ = 0
Effect of polariser on transmission of polarised light:
(i) If unpolarised light is incident on a polariser, the transmitted light will be linearly polarised light. The intensity of the transmitted polarised light will be half the intensity of unpolarised light incident on the polarizer.
(ii) The intensity of the transmitted light does not change on rotation of the polariser.
(iii) If partially polarised light is incident on a polarizer, the intensity of the transmitted light will vary from Imax to Imin in one full rotation of the polarizer.
3 (iv) If plane polarised light is incident on a polarizer, the intensity of the transmit- ted light will vary from zero to a maximum value in one full rotation of the polarizer.
(v) If circularly polarised light is incident on a polarizer, the intensity of the transmitted light stays constant in any position of the polariser. hence the intensity is same for all position of the polariser.
(vi) In case of elliptically polarised light, the intensity of the light transmitted through the polariser varies with the rotation of the polariser from Imax to Imin.
Imax is found when the polariser axis coincides with the semi-major axis of the ellipse.
Imin is found when the polariser axis coincides with the semi-minor axis of the ellipse.
Huygens’ theory of double refraction:
Huygens’ explained the phenomenon of double refraction by his theory of sec- ondary wavelets. Here we give below his extended theory.
(a) When a wavefront is incident on a doubly refracting crystal, every point on the crystal becomes the origin of two wavefronts,one the ordinary (O) and the other extraordinary (E).
(b) Ordinary wavefront corresponds to ordinary rays which obey the laws of re- fraction and have the same velocity in all direction. The ordinary wavefront is thus spherical.
(c) Extraordinary wavefront corresponds to extraordinary rays which do not obey the laws of refraction and have different velocities in different directions. The ex-
4 traordinary wavefront is thus an ellipsoid of revolution, with the optical axis as the axis of revolution.
(d) The velocity of the O-ray and E-ray is the same along the optic axis. For this the sphere and the ellipsoid touch each other at points that lie on the optic axis of the crystal.
(e) In negative crystal the ellipsoid lies outside the sphere, S being the source implying that in such crystal the E-ray wavefront travels faster than the ordinary wavefront, except along the optic axis.
In positive crystal the sphere is outside the ellipsoid implying that in such crystal the O-ray wavefront travels faster than the extraordinary wavefront, except along optic axis.
Refractive indices of a crystal:
Refractive index of a substance is defined as c n = v
5 where c is the velocity of light in vacuum v is the velocity of light in the substance Refractive index of the crystal for O-ray is c nO = = constant vo because, the wave velocity of O-ray is a constant being the same in all directions and for all orientation of the optic axis. Refractive index of the crystal for E-ray is c ne = = not constant ve because, the wave velocity of E-ray is not a constant but varies with directions. So r.i. of the crystal for the E-ray will be different for different directions.
The principal r.i. for E-ray in positive crystal is c ne = (ve)min The principal r.i. for E-ray in negative crystal is c ne = (ve)max
In positive crystal, ne > n0
In negative crystal ne < no
Birefringence or amount of double refraction of a crystal is defined as
4n = ne − no
For +ve crystal, ne > n0, hence 4n = +ve
For -ve crystal, ne < n0, hence 4n = −ve
Huygens’ construction of wave surfaces on the plane of incidence in Uniaxial crys- tal:
6 Case I: Optic axis inclined to upper face but lying in the plane of inci- dence: Let a plane wavefront AB be incident obliquely on the upper face of a uniaxial -Ve crystal (e.g. Calcite) cut such that the optic axis lies in the plane of incidence along AX, but inclined to the upper face at some angle.
On the crystal surface, the point A of the wavefront AB is the centre of both O-ray and E-ray. At the same time, the disturbance from B reaches C, O-ray travels a distance AD and e-ray travels AF, so BC AD AF t = = = c vo ve Hence, v BC AD = BC o = c no With A as centre and BC as radius, we draw a circle cutting the optic axis at X. The no circle gives the position of O-wave surface.
v BC AF = BC e = c no With A as centre and BC as radius, we draw a circle cutting the optic axis at X. In (ne)min -Ve crysta ne < no, and (ne)min = no along the optic axis. So AX is the semi-minor axis of ellipse. The semi-major axes AF. If an ellipse is drawn with given minor and major axes touching the circle at X, it gives the position of E-wave surface.
Tangents CD and CF drawn from C on O- and E-wave surfaces gives the O- and E-wave refracted wavefronts.
7 Case II: Optic axis parallel to upper face but lying on the plane of inci- dence:
(a) Oblique incidence: As the optic axis is parallel to the crystal face and lies in the plane of incidence, we get the O-wave and E-wave surface as shown in fig.
The circle and ellipse will touch at X along the optic axis AX, CD and CF are O-ray and E-ray refracted wavefronts corresponding to the incident wavefront AB.
(b) (i)Normal incidence: This is illustrated in fig.
Here CD and FG are the O- and E-refracted wavefronts corresponding to the incident wavefront AB.The refracted wavefronts (CD for O-rays and FG for E-ray) are parallel to each other and also to the crystal surface. Although the O- and E-ray travel in the same direction, yet there will be double refraction as they travel with λ λ different velocities. This is used in the construction of 4 and 2 .
(ii)Normal incidence, optic axis perpendicular to the upper face: This is illustrated in fig.
8 In this case AB is the incident wavefront. As the incidence is normal, there is no double refraction as O-ray and E-ray travel with the same velocity along the optic axis, the direction of travel. Here O- and E-rays coincide in crystal OO or EE.
Retardation Plates: There are two types of retardation plates. (i) Quarter wave plate (ii) Half wave plate
(i) Quarter wave plate: A quarter wave plate is a thin plate birefringent crystal having the optic axis parallel to its refracting faces and the plate of a double refracting crystal having a thickness λ so as to produce a path difference or retardation of 4 or a phase difference or phase π retardation 2 between the O- and E-waves propagating through it.
A quater wave plate introduces between E-ray and O-ray a phase difference 2π δ = × 4 λ 2π λ δ = × λ 4 π δ = 2 Computation of minimum thickness: Let a beam of light of wavelength λ be incident normally on a plane parallel plate of a doubly refracting crystal with faces parallel to optic axis. The beam breaks up into O- and E-waves, both traveling along the same path perpendicular to the faces, but with different velocities.
For -Ve crystal (e.g. calcite), E-waves move faster the O-waves. let t be the thickness of the plate.
The optical path for O-ray within the crystal= not The optical path for E-ray within the crystal= net For -ve crystal, ne < n0, So the path difference between two waves,
9 on emergence, is
4n = (no − ne)t
λ Again, for quarter wave plate 4 = 4 So, we get λ (n − n )t = o e 4 Minimum thickness of Quarter wave plate for -Ve native crystal λ t = (1) 4(no − ne)
For +Ve crystal (e.g. quartz), ne < n0, we get λ (n − n )t = o e 4 Minimum thickness of Quarter wave plate for +Ve native crystal λ t = (2) 4(no − ne) So, a plate of thickness given by (1) and (2) will serve as a quarter wave plate for a given wavelength λ.
Application: Quater wave plate are used for the production of circularly and elliptically polarised light.
(ii) Half wave plate: A half wave plate is a thin plate birefringent crystal having the optic axis parallel to its refracting faces and the plate of a double refracting crystal λ having a thickness so as to produce a path difference or retardation of 2 or a phase difference or phase retardation π between the O- and E-waves propagating through it.
A half wave plate introduces between E-ray and O-ray a phase difference 2π δ = × 4 λ
10 2π λ δ = × λ 2 δ = π
Computation of minimum thickness: Let a beam of light of wavelength λ be incident normally on a plane parallel plate of a doubly refracting crystal with faces parallel to optic axis. The beam breaks up into O- and E-waves, both traveling along the same path perpendicular to the faces, but with different velocities.
For -Ve crystal (e.g. calcite), E-waves move faster the O-waves. let t be the thickness of the plate.
The optical path for O-ray within the crystal= not The optical path for E-ray within the crystal= net For -ve crystal, ne < n0, So the path difference between two waves, on emergence, is
4n = (no − ne)t
λ Again, for quarter wave plate 4 = 2 So, we get λ (n − n )t = o e 2 Minimum thickness of Quarter wave plate for -Ve negative crystal λ t = (1) 2(no − ne)
For +Ve crystal (e.g. quartz), ne < n0, we get λ (n − n )t = o e 2 Minimum thickness of Quarter wave plate for +Ve negative crystal λ t = (2) 2(no − ne) So, a plate of thickness given by (1) and (2) will serve as a half wave plate for a given wavelength λ.
11 Babinate compensator: A Babinate compensator is an optical device whose function is to compenstate a path difference. It is used in conjunction with a polariser and analyser combination to investigate elliptically polarised light.
The compensator helps in determining the axis of the ellipse and the ratio of their lengths.
Construction: Babinate compensator consists of two wedges A and B of quartz crystal, having equal small acute angles. They are placed with their hypotenuse- planes in contact so as to form a rectangular block as shown in fig.
Of the wedge, the left one is cut with its optic axis perpendicular to the refracting edge, while the right one with its optic axis parallel to it. Thus the O-ray in one will behave as E-ray in the other and vice versa. One of the wedge (A) is fixed in position and the other (B) slidable in its own plane by a micrometer screw F as shown in fig.
Theory: When plane polarised light falls normally on the first wedge A, with its plane of vibration inclined at θ with optic axis, it breaks up into O-ray and E-ray. E-ray, Parallel to the optic axis, moves slower than O-ray, perpendicular to the Optic axis. As θ is so small that the separation between E-ray and O-ray is ignored. On entry into B, the O-ray becomes E-ray and E-ray becomes O-ray. So they interchange the velocities in transit from A To B and cancel each other effect.
Let t1 is the thickness of wedge A and t2 is the thickness of wedge B ne is the r.i. of quartz for E-ray no is the r.i. of quartz for O-ray
12 So, Path difference between E-ray and O-ray in wedge A
41n = (ne − no)t1
Path difference between E-ray and O-ray in wedge A
42n = (no − ne)t2
Total path difference
4 = 41 + 42 = (ne − no)(t1 − t2) and Phase difference 2π δ = 4 λ 2π δ = (n − n )(t − t ) λ e o 1 2
Case I: At the centre of the compensator t1 = t2, phase difference
δ = 0
The emergent light is plane polarised in the original plane.
Case II: If the compensator be placed between two crossed Nicols, a central dark band is obtained in this position.
Case III: Let the micrometer screw is rotated, the phase difference δ changes. When δ = 2π, a dark band appears. Let for this dark band , the shift of micrometer screw is 2b. So, for a shift x, a phase difference will be 2π π δ = x = x 2b b
advantage:
(i) On sliding themovable wedge relative to the fixed one, any value of (t1 − t2) can be made at the centre. In this way and at that position, the compensator serves as a quater and half wave plate by introducing the required phase difference between
13 E-ray and O-ray. (ii) If any phase difference exists between E-rays and O-rays, it can be neutralized or compensated by it. For this it is called a compensator.
Uses of Babinet’s compensator:
(i) Any desired value of path-difference or phase difference can be obtained at the centre of the compensator, all types of polarised (circularly, plane or elliptical) light can be obtained with the help of it. (ii) It can be used for a complete analysis of elliptically polarised light.
Application Of Babinet’s compensator:
Case I: Production of elliptical and circularly polarised light:
Bebinet’s compensator introduces a phase difference for a shift x πx δ (1) b The prism B is shifted by x = b/2 from its position of coincidence with prism A, then phase difference from (1) π δ = 2 If the direction of vibration of the incident plane polarised light makes an angle α(6= 450) with the optic axis of A, then the emergent light will consists of two rectangular vibrations of unequal amplitude and differening in phase by π/2. This is elliptically polarised light.
If α = 450, the rectangular vibrations will have equal amplitude and the emergent light will be circularly polarised light.
Case II: Analysis of elliptically polarised light:
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14 Production of circularly polarised light:
A circularly polarised light can be produced by superposing two coherent linear vibrations, of equal qmplitudes, at right angles to each other with phase difference of π 2 . This is done with the help of a quartz wave plate. Here
x = acos(ωt + π/2) = asinωt and y = acosωt gives on superposition a circular vibration
x2 + y2 = a2
Production of elliptically polarised light:
An elliptically polarised light can be produced by superposing two coherent linear vibrations of unequal qmplitudes, at right angles to each other with phase difference π of 2 . T Here x = acos(ωt + π/2) = asinωt and y = bcosωt gives on superposition elliptical vibration x2 y2 + = 1 a2 b2
Problem: Write an expression for a linearly polarised wave of angular frequency ω moving along +z direction with plane of vibration making 600 to the xz-plane.
Let E0 be the amplitude of the wave. So x-components of wave 1 E = E cos600cos(kz − ωt) = E cos(kz − ωt) 0x 0 2 0
15 So y-components of wave √ 3 E = E sin600cos(kz − ωt) = E cos(kz − ωt) 0y 0 2 0 For linearly polarised light √ 1 3 E~ (z, t) = ˆiE + ˆjE = ˆi E cos(kz − ωt) + ˆj E cos(kz − ωt) 0x 0y 2 0 2 0
Problem: Find the state of polarisation when the x and y components of the electric field are
(i)Ex = E0sin(ωt + kz),Ey = E0cos(ωt + kz)
We have
2 2 2 2 2 2 Ex + Ey = Eo (sin (ωt + kz) + cos (ωt + kz)) = E0
This represents a circular vibration. At any position for z = 0
~ E(0, t) = ˆiE0x + ˆjE0y = ˆiE0sin(ωt) + ˆjE0cos(ωt) rotates clockwise and the light is circularly polarised.
Problem: Find the state of polarisation when the x and y components of the electric field are
E0 (ii)Ex = E0cos(ωt + kz),Ey = √ cos(ωt + kz + π) 2
Here E0 E0 Ex Ey = √ cos(ωt + kz + π) = −√ cos(ωt + kz) = −√ 2 2 2 This represents a linearly polarised light making and angle 1 θ = tan− − √ 2
16 Problem: Two linearly polarised light waves are in phase but have different am- plitudes. They are represented by
~ E1(z, t) = ˆiA1cos(kz − ωt) + ˆjB1cos(kz − ωt) ~ E2(z, t) = ˆiA2cos(kz − ωt) + ˆjB2cos(kz − ωt) ~ ~ ~ Show that E1 + E2 = E is also linearly polarised. Find its polarisation direction.
Here ~ ~ ~ E(z, t) = E1(z, t) + E2(z, t) ! ~ E(z, t) = ˆi(A1 + A2) + ˆj(B1 + B2) cos(kz − ωt)
E~ (z, t) = Ccos(kz − ωt) where amplitude C = ˆi(A1 + A2) + ˆj(B1 + B2) is a constant, so the resultant wave is ~ ˆ linearly polarised. Let E = ˆiEx + JEy So, we can write ! ~ ˆ E = ˆiEx + JEy = ˆi(A1 + A2) + ˆj(B1 + B2) cos(kz − ωt)
Hence,
Ex = (A1 + A2)cos(kz − ωt)
Ey = (B1 + B2)cos(kz − ωt) So, we get E A + A ! x = 1 2 Ey B1 + B2 ! B1 + B2 Ey = Ex A1 + A2
Ey = mEx (y = mx) So, it is a linearly polarised vibration in the XY-plane hving angle B + B ! θ = tan− 1 2 A1 + A2
17 Problem: If n0 is r.i. for O-rays, ne the principal r.i. for E-rays, then show taht the r.i. nθ for E-rays in a direction θ with optic axis is given by 1 cos2θ sin2θ 2 = 2 + 2 nθ n0 ne
The equation of the wave surface is given by
x2 y2 + = 1 (1) a2 b2
Here x = rcosθ, y = rsinθ Velocity of E-ray along OY axis = ve, hence b = vet Velocity of O-ray along OX axis = vo, hence a = vot Velocity of E-ray along r = OP = vθ, hence r = vθt So, we get from (1)
x2 y2 + = 1 (1) a2 b2 r2cos2θ r2sin2θ + = 1 a2 b2 2 2 2 2 (vθt) cos θ (vθt) sin θ 2 + 2 = 1 (vot) (vet) 2 2 2 2 vθ cos θ vθ sin θ 2 + 2 = 1 vo ve cos2θ sin2θ 1 2 + 2 = 2 vo ve vθ multiplying by c2 on both sides, we get
c2cos2θ c2sin2θ c2 2 + 2 = 2 vo ve vθ c2cos2θ c2sin2θ c2 2 + 2 = 2 no ne nθ cos2θ sin2θ 1 2 + 2 = 2 no ne nθ
18