1Waves and .nb 505- James C. Wyant (2000 Modified) 1

Waves and Polarization

WP - 1

A Show that E r, t = ÅÅÅÅÅr Cos kr-wt is a solution to the wave equation.

@ D @ D Solution 3-D Wave Equation:

1 ∑2 E “2 E - ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = 0 (1) c2 ∑t2 In polar coordinates:

1 ∑ ∑E 1 ∑ ∑E 1 ∑2 E “2 E = ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ r2 ÅÅÅÅÅÅÅÅÅÅÅ + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ Sin q ÅÅÅÅÅÅÅÅÅÅÅ + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ (2) r2 ∑r ∑r r2 Sin q ∑q ∑q r2 Sin q 2 ∑f2 For spherical waves there is no dependence on q and f so we have i y i y j z j @ D z k { @ D k { @ D 1 ∂ ∂E r, t ∇2 E r, t = r2 . Thus (3) r2 ∂r ∂r 1 ∑ ∑ 1 ∑2 ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ r2 ÅÅÅÅÅÅÅÅÅ E r, t - ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅ E@r, t =D0 (4) r2 ∑@r ∑Dr cJ2 ∑t2 N The problem is to show that i y j z @ D @ D k { A E r, t = Cos kr −ωt (5) r is a solution to the above equation if certain conditions concerning k, w, and c are satisfied. I will use@ MathematicaD to do@ the algebra.D Let

a e r_, t_ = Cos kr −ωt ; r Then, substituting Eq. (5) into Eq. (4) yields @ D @ D

1 2 1 FullSimplify ∂r r ∂r e r, t − ∂t,t e r, t r2 c2 a −c2 k2 +ω2 Cos kr− t ω c2 r A H @ DL @ DE H L @ D 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 2

A 2 2 2 Thus, E[r,t]= ÅÅÅÅÅr Cos kr -wt is a solution if −c k +ω =0, or

2 w2 k = ÅÅÅÅÅÅÅc2 . @ D H L

WP - 2

Âd Suppose that a reflecting surface has a complex reflection coefficient r = ro e . This means that if a normal incident electric field is given by

Ei (x,t)=A cos(kx - wt + f)

the reflected wave is given by

Er x, t = ro A cos(-kx - wt + f + d)

Let a plane wave be incident at an angle to the plane reflecting surface. Determine Ei (x,y,t) and Er (x,y,t). Show thatH theL total field can be written as

Etot =A 1 - ro ) cos[k (xcosq +y sinq) - wt + f] + 2 ro A cos[ky sinq - wt + f + d/2 ] cos[kx cosq - d/2 ].

Interpret this result for the following four cases and find the spatial frequency of the standing wave component: i) H q = 0, ro = 1, ii) q = 0, ro ∫ 1, iii) q ∫ 0, ro = 1, iv) q ∫ 0, ro ∫ 1.

θθ

Y

X

Solution

Ei = a Cos k x Cos θ + y Sin θ −ωt +φ ;

Er =ρo a Cos k −x Cos θ + y Sin θ −ωt +φ+δ ; @ H @ D @ DL D

@ H @ D @ DL D 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 3

Etot = Ei + Er a Cos φ−t ω+k x Cos θ + y Sin θ + a Cos δ+φ−t ω+k −x Cos θ + y Sin θ ρo

While this is the correct answer, it is not in the form we want. @ H @ D @ DLD @ H @ D @ DLD If we let a = k y Sin[q] - w t + f + d/2 and b = k x Cos[q] - d/2 then

Etot = a Cos α+β +ρo a Cos α−β a Cos α+β + a Cos α−β ρo

@ D @ D Subtracting and adding (a ρo Cos α+β yields @ D @ D

Etot = Factor a Cos α+β −ρo Cos α+β + TrigExpand ρo a Cos α−β + Cos α+β −a Cos α+β −1 +ρo + 2 a Cos@ α CosDL β ρo

Etot = Etot . @α→H k y@ Sin Dθ −ωt +φ+δ@ DLD2, β→ k x Cos θ@ −δ H2 @ D @ DLD −a Cos@φ−tDHω+k x CosL θ + k y Sin@ D θ @ −D1 +ρo + δ δ 2 a Cos − k x Cos θ Cos +φ−t ω+k y Sin θ ρ 2ê 8 @ D 2 ê o@ D ê < @ @ D @ DD H L which is the result we wanted. A @ DE A @ DE

ü i) q = 0, ro = 1

Etot . θ−>0, ρo −> 1 δ δ 2 a Cos kx− Cos +φ−t ω 2 2 ê 8 < NormalA standingE waveA having zerosE spaced l/2.

ü ii) q = 0, ro π 1

Etot . θ−>0 δ δ −a Cos kx+φ−t ω −1 +ρ + 2 a Cos kx− Cos +φ−t ω ρ o 2 2 o ê 8 < Same @standing waveDH pattern asL in i), but reducedA amplitude.E A Also, haveE travelling wave in +x direction.

ü iii) q π0, ro = 1

Etot . ρo → 1 δ δ 2 a Cos − k x Cos θ Cos +φ−t ω+k y Sin θ 2 2 ê 8 < A @ DE A @ DE 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 4

We have a standing wave in the x direction. The phase of the standing wave varies as a sinusoid in the y direction.

ü iv) q π 0, ro π 1

Etot −a Cos φ−t ω+k x Cos θ + k y Sin θ −1 +ρo + δ δ 2 a Cos − k x Cos θ Cos +φ−t ω+k y Sin θ ρ 2 2 o @ @ D @ DD H L We haveA a superposition@ DE of caseA iii) with a plane wave@ travellingDE along the incident direction with reduced amplitude.

WP - 3

A given argon laser has a cw output of 1 watt. The output beam has a Gaussian amplitude distribution which is truncated at the point where the amplitude falls to 1/e its axial value. The resulting beam diameter is 2 mm.

a) What is the on-axis electric field?

b) What would the on-axis electric field be if we had a uniform amplitude distribution across the beam? (1 watt total power)

Solution

ü a) The Poynting vector is given by

1 2 2 < S >= ε cE2 ; E = E −r ro 2 o o

where r = 1 mm. ê ” o

The total power is given by

ro 2 π 2 2 1 2 −2 r ro P = < S > dA = εo c Eo r r θ 2 0 0

ro ê 2 2 1 ” 2 −2 r ro 2 −2 P =πεo cEo r r =πεo cEo 2 − 1 ‡ 0 ‡ ‡ −4 ro ê J N H L ‡ ê 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 5

Solving for the square of the on-axis electric field yields

2 4 P Eo = 2 −2 πεo cro 1 −

2 P 2 1 volt Eo = −2 = −3 −12 8 ro πεo cH 1 −L 10 π 8.85 10 310 1 −−2 m

The on-axis electric field is then $%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% $%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% H L H L H L volt E = 23550.4 o m

ü b) 1 ε cE 2 π r2 = 1 watt 2 o o

2 1 2 1 volt volt Eo = −3 = −12 8 −3 = 15485 εo c π 10 8.85 10 310 π 10 m m

$%%%%%%%%%%%%%%%% $%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% H L WP - 4

I have two electric fields ` Âf  kz-wt E1 = E1 a1 e 1 e and ` Âf ÂHkz-wtL ÷E÷÷÷2” = E2 a2 e 2 e ` ` where a1 and a2 are unit vectors in the direction of oscillation of the electric field. By breaking the electric field H L into components÷÷÷÷” in the x and y directions show that the time average irradiance is proportional to

2 2 ` ` E1 + E2 + 2 E1 E2 a1 ÿ a2 Cos f2 -f1 .

H L @ D Solution First we need to break the electric field into x and y components. Let one electric field be at an

angle q1 with respect to the x-axis and the second be at an angle q2 with respect to the x-axis. Then.

e Cos θ φ1 kz−ωt e1 = 1 1 ; φ kz−ωt e1 Sin θ1 1 H L and i @ D H L y j z k @ D { e Cos θ φ2 kz−ωt e2 = 2 2 ; φ kz−ωt e2 Sin θ2 2 H L Next wei need to@ addD the twoH x-componentsL y and the two y-components. j z k @ D { 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 6

e3 = e1 + e2; MatrixForm e3 kz−t ω + φ kz−t ω + φ 1 Cos θ1 e1 + 2 Cos θ2 e2 kz−t ω + φ kz−t ω + φ 1 Sin θ1 e1 + 2 Sin θ2 e2 H L @ D H L i H L @ D H L @ D y Thej irradiance is obtained by multiplying each component byz its complex conjugate. k @ D @ D { e4 = ComplexExpand e3 Conjugate e3 ; MatrixForm Simplify e4 2 2 2 2 Cos θ1 e1 + 2 Cos θ1 Cos θ2 Cos φ1 −φ2 e1 e2 + Cos θ2 e2 2 2 @ @ DD 2 2 Sin θ1 e1 + 2 Cos φ1 −φ2 Sin θ1 Sin θ2 e1 e2 + Sin θ2 e2 @ @ DD ji @ D @ D @ D @ D @ D zy Thej total irradiance is given by the sum of the irradiance for the x-componentz and the irradiance fork the@ y-component.D @ D @ D @ D @ D {

Simplify Plus @@ e4 2 2 e1 + 2 Cos θ1 −θ2 Cos φ1 −φ2 e1 e2 + e2

@ D ` ` Since Cos q1 -q2 is equal to the dot product of the two unit vectors a1 and a2, we8 have the desired@ result.D @ D <

@ D WP - 5

Prove using Jones calculus that a retarder at 45o between vertical and horizontal quarter-wave plates is equivalent to a rotator. (Orientation of sandwich not important.)

Solution First we need to define rotation matrices and the Jones matrices for quarter-wave plates and retarders.

ü Quarter-wave plate with fast axis vertical 10 qfavJones =π 4 ; 0 −

ê J N ü Quarter-wave plate with fast axis horizontal 10 qfahJones =− π 4 ; 0

ê J N ü Retarder with fast axis vertical 10 rfavJones φ_ :=φ 2 0 − φ

ê i y @ D j z k { 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 7

ü Rotation Matrix Cos θ Sin θ rotJones θ_ := −Sin θ Cos θ

The retarder must be rotated so it is at 45 degrees. @ D @ D @ D J N @ D @ D ret45Jones = rotJones −45 Degree .rfavJones φ .rotJones 45 Degree ; Put the retarder between the vertical and horizontal quarter-wave plates to show that we end up with a rotation matrix. @ D @ D @ D

rotation = FullSimplify qfahJones.ret45Jones.qfavJones ; MatrixForm rotation φ φ Cos 2 Sin 2 @ D −Sin φ Cos φ 2 @ 2 D i @ D @ D y j z Thus,j we have rotation matrixz of angle f/2. k @ D @ D { Now we want to rotate the quarter-wave plate, retarder, quarter-wave plate sandwich, to show that it's orientation does not change it's effect. We will rotate it an angle a.

MatrixForm FullSimplify rotJones −α .rotation.rotJones α φ φ Cos 2 Sin 2 −Sin φ Cos φ 2 @ 2 @ @ D @ DDD i @ D @ D y j z Thus,j the orientation of thez sandwich is not important. k @ D @ D {

WP - 6

Use Jones calculus to show that a half-wave plate converts right handed circularly polarized light into left handed circularly polarized light and the phase of the light can be changed by rotating the half-wave plate.

Solution Right handed can be represented by

1 1 rcJones = ; − 2 A half wave plate with the fast axis horizontal can be written as è!!!! J N 10 rfahJones =− ; 0 −1 The rotation matrix can be written as J N 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 8

Cos θ Sin θ rotJones θ_ := −Sin θ Cos θ = −θ θ outputPolarization @rotJonesD @ D .rfahJones.rotJones .rcJones; @ D J N TrigToExp TrigFactor @outputPolarizationD @ D MatrixForm − θ − 2 2 @ D @ D −2 θ @ @ DD êê è!!!!2 ji zy j z j è!!!! z Butj this is equalz to k { −2 θ 1 − 2

Thus, right handed circular has been changed to left handed circular and the phase is changed at twice the rotation J N rate ofè!!! the half-wave plate.

WP - 7

Derive a Mueller matrix for a half-wave plate having a vertical fast axis. Utilize your result to convert a right-handed state into a left-handed state. Verify that the same wave plate will convert a left-handed state to a right-handed state. Advancing or retarding the relative phase by p should have the same effect. Check this by deriving the matrix for a horizontal fast axis half-wave plate.

Solution

ü Mueller matrix for a half-wave plate having a vertical fast axis The Mueller matrix for a quarter-wave plate having the fast axis vertical is

100 0 010 0 qfavMueller = ; 000−1 ji001 0zy j z j z j z Thus, the Mueller jmatrix for a half-wavez plate having the fast axis vertical is j z k { hfavMueller = qfavMueller. qfavMueller; hfavMueller MatrixForm 10 0 0 01 0 0 00−10 êê ji 00 0 −1 zy j z j z j z j z Thej Stokes vectorsz for right and left handed circular polarization are given by k { 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 9

1 1 0 0 rcStokes = ; lcStokes = ; 0 0 ji1zy ji−1zy j z j z j z j z j z j z Convert right jto leftz j z j z j z k { k { hfavMueller . rcStokes MatrixForm 1 0 0 êê ji −1 zy j z j z j z j z Convertj z left to right k { hfavMueller . lcStokes MatrixForm 1 0 0 êê ji 1 zy j z j z j z j z j z k { ü Mueller matrix for a half-wave plate having a horizontal fast axis The Mueller matrix for a quarter-wave plate having the fast axis horizontal is

1000 0100 qfahMueller = ; 0001 ji00−10zy j z j z j z Thus, the Mueller jmatrix for a half-wavez plate having the fast axis horizontal is j z k { hfahMueller = qfahMueller. qfahMueller;

hfahMueller MatrixForm 10 0 0 01 0 0 00−10êê − ji 00 0 1 zy j z j z j z j z Thusj the Mueller zmatrix for a half-wave plate having the fast axis horizontal is the same as the Mueller matrix for ak half-wave plate {having the fast axis vertical.

WP - 8

a) Assume I want to rotate the polarization of a plane polarized wavefront 90o by using only "perfect" linear .What is the maximum flux transmission I can obtain using only two polarizers? What is the minimum number of perfect polarizers required to have a flux transmission greater than 95%?

b) The indices of refraction for for the sodium yellow line are no = 1.544 and ne = 1.553. Calculate the thickness of a quarter-wave plate made from quartz. 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 10

Solution

ü a) From the Malus Law

90 ° 2 n trans n_ := Cos n If there are two polarizers, @ D A E 1 trans 2 = 4 Determining n for a flux transmission greater than 95% @ D For n = 1, trans n ≤ 0.95, n++ ; Print n, " ", N trans n 49 0.950883 @ @ D D @ @ @ DDD Thus, we need 49 perfect polarizers.

ü b) 0.589 µm Solve 1.553 − 1.544 t == ,t 4 t → 16.3611 µm AH L E Can also have a thickness of (4 m+1) 16.3611 mm, where m is an integer since (4) 16.3611 mm is a wave plate. 88 <<

WP - 9

Use Jones calculus to show that two half-wave plates at angle q between them are equivalent to a rotator through angle 2q.

Solution The matrix for a retarder with the fast axis vertical is

10 rfavJones φ_ :=φ 2 0 − φ

ê The rotation matrix is given iby y @ D j z k { 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 11

Cos θ Sin θ rotJones θ_ := −Sin θ Cos θ

The matrix for two half-wave@ Dplates at@ anD angle q between them is given by @ D J N @ D @ D MatrixForm TrigReduce rotJones −θ .rfavJones π .rotJones θ .rfavJones π −Cos 2 θ Sin 2 θ −Sin 2 θ −Cos 2 θ @ @ @ D @ D @ D @ DDD @ D @ D SinceJ this is minus the rotationN matrix for a rotation of -2 q we have the result we want. @ D @ D

WP - 10

Let Ex = Ax cos(kz-wt) and Ey = Ay cos(kz-wt+f). Show that

E 2 E 2 E E x + y − 2 x y Cos φ = Sin φ 2 Ax Ay Ax Ay

J N J N J N J N @ D @ D Solution

ex = ax Cos kz−ωt ;ey = ay Cos kz−ωt +φ ;

2 ex 2 ey ex ey FullSimplify + − 2 Cos φ @ ax D ay @ ax ay D Sin φ 2 i y i y AJ N j z J N j z @ DE Therefore, k { k { @ D e 2 e 2 e e x + y − 2 x y Cos φ = Sin φ 2 ax ay ax ay

J N J N J N J N @ D @ D WP - 11

Show that in terms of the the degree of polarization can be written as

1 2 s 2 + s 2 + s 2 V = 1 2 3 so ê H L Solution Polarized Light Degree of Polarization = V = Total Amount of Light

Let the total amount of light be so

For the unpolarized portion of the light 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 12

2 2 2 so − s1 + s2 + s3 0 0 i è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! y j 0 z j z j z j z Forj the polarized portion of zthe light j z k { 2 2 2 s1 + s2 + s3 s1 s2 è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ji s3 zy j z j z j z Therefore,j z j z k { s 2 + s 2 + s 2 V = 1 2 3 so è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! WP - 12

A linearly polarized laser source is used with a Twyman-Green interferometer. A 50-50 non-polarization sensitive beamsplitter is used in the interferometer. The flat mirror in the reference arm of the interferometer is tilted slightly to give straight equi-spaced fringes in the output. A quarter-wave plate is placed in the test arm of the interferometer. The fast axis of the quarter-wave plate makes an angle q with respect to the direction of polarization of the light incident upon the quarter-wave plate. A perfect is placed in the output of the interferometer. The transmission axis of the polarizer is in the direction of polarization of the light coming from the reference arm.

a) Calculate the irradiance of the interference pattern as a function of q. What is the fringe visibility as a function of q?

b) Assume the polarizer is non-perfect in that it transmits 95% of the light having a polarization along the transmission axis of the polarizer and 5% of the light having the orthogonal polarization. What is the observed fringe visibility as a function of q?

Solution Since the light goes through the quarter-wave plate twice it will act as a half-wave plate and it will rotate the direction of polarization by an angle 2q.

ü a) The irradiance of the interference pattern is given by

2 i = i1 + i2 + 2 i1 i2 Cos φ , where i2 = i1 Cos 2 θ

è!!!!!!!!!!! @ D @ D 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 13

2 i = i1 1 + Cos 2 θ + 2 Cos 2 θ Cos φ

imax − imin 2 Cos 2 θ visibility = = 2 imax + imin 1 + Cos 2 θ H @ D @ D @ DL @ D ü b) @ D 2 2 i = i1 1 + Cos 2 θ + 2 Cos 2 θ Cos φ 0.95 + Sin 2 θ 0.05 0.95 2 Cos 2 θ visibility = 0.95 1 + Cos 2 θ 2 + 0.05 Sin 2 θ 2 HH @ D @ D @ DL H L @ D H LL H @ DL WP - 13 H @ D L H L @ D

a) What are the four filters we associate with the four Stokes parameters?

b) Suppose that an ideal polarizer is rotated at a rate w between a similar pair of stationary crossed polarizers. Give the ratio of the modulation frequency of the emergent flux density to the rotation frequency of the polarizer.

c) I am working on a problem involving partially polarized light. Which should I use, Jones or Mueller matrices? Explain (briefly).

d) I am working on a problem using completely polarized light where I want to keep track of the phase of the beam. Should I use Jones or Mueller matrices? Explain (briefly).

Solution

ü a) All filters transmit 50% of natural radiation.

i) isotropic filter

ii) linear horizontal polarization filter

i) linear polarization at 45° filter

i) right-handed circular polarization filter

ü b) As the polarizer rotates 360° no light will be transmitted at 4 orientations. Therefore, the ratio is 4. 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 14

ü c) Mueller matrices should be used since Jones matrices can be used only with completely polarized light.

ü d) Jones matrices should be used since they keep track of the phase.

WP - 14

A linearly polarized source is used with a Young's two-pinhole interferometer. The same amount of light is transmitted through each pinhole. A half-wave plate is placed over one pinhole. The fast axis of the half-wave plate makes an angle q with respect to the direction of polarization of the light incident upon the half-wave plate. A perfect polarizer is placed in the output of the interferometer. The transmission axis of the polarizer is in the direction of polarization of the light coming from the pinhole without the half-wave plate.

a) What is the fringe visibility as a function of q?

b) Assume the polarizer is non-perfect in that it transmits 95% of the light having a polarization along the transmission axis of the polarizer and 5% of the light having the orthogonal polarization. What is the observed fringe visibility as a function of q?

Solution The irradiance of the two-beam interference pattern is given by

i = i1 + i2 + 2 i1 i2 Cos φ ;

The half-wave plate rotates the direction of polarization by an angle 2 q. The irradiance of the beam passing è!!!!!!!!!!! through the half-wave plate and@ theD polarizer is given by

2 i2 = i1 Cos 2 θ ;

The irradiance of the two-beam interference pattern is then given by @ D 2 i = i1 1 + Cos 2 θ + 2 Cos 2 θ Cos φ

ü a) H @ D @ D @ DL The visibility is given by

imax − imin 4 Cos 2 θ 2 Cos 2 θ v = = 2 = 2 imax + imin 2 1 + Cos 2 θ 1 + Cos 2 θ

@ D @ D H @ D L @ D 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 15

ü b) The irradiance of the main interference fringe pattern is reduced by a factor of 0.95 and we have to add in 0.05 times the irradiance of the orthogonally polarized beam.

2 2 i = i1 1 + Cos 2 θ + 2 Cos 2 θ Cos φ 0.95 + Sin 2 θ 0.05

imax − imin 0.95 2 Cos 2 θ v = = 2 2 imax + imin 0.95 1 + Cos 2 θ + 0.05 Sin 2 θ HH @ D @ D @ DL @ D H LL H @ DL WP - 15 H @ D L H L @ D

a) Using less than 25 words define polarization angle. b) Using less than 25 words define critical angle. c) What is the approximate power reflectance at normal incidence for common glass? d) What is the approximate power reflectance at a nearly 90 degree angle of incidence for common glass?

Solution

ü a) Angle of incidence for which only the polarization component polarized normal to the incident plane, and therefore parallel to the surface, is reflected

ü b) When light passes from a medium of lower index to a medium of higher index the angle of refraction is always less than the angle of incidence. When the incident rays approach an angle of 90° the refracted rays approach a fixed angle called the critical angle.

ü c) 4%

ü d) 100% 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 16

WP - 16

A quartz plate in the setup shown below has its axes at 45 degrees relative to the transmission axis of the linear

polarizer. The indices of refraction for quartz are no = 1.544 and ne = 1.553. The laser wavelength is 633 nm. a) What is the minimum quartz thickness for having no light reflected back to the laser? b) What is the minimum quartz thickness (other than zero) for having the maximum amount of light reflected back to the laser?

Laser

Polarizer Quartz

Solution

ü a) The quartz plate needs to rotate the direction of polarization by 90 degrees. Thus, we want the quartz plate to act as a half-wave plate. Since the light is going through the quartz plate twice the quartz plate must be a quarter-wave plate.

0.633 µm Solve 1.553 − 1.544 t == ,t 4 t → 17.5833 µm AH L E 88 << ü b) We want the quartz plate to act as a wave-plate. Since the light is going through the quartz plate twice the quartz plate must be a half-wave plate.

0.633 µm Solve 1.553 − 1.544 t == ,t 2 t → 35.1667 µm AH L E 88 << 1Waves and Polarization.nb Optics 505- James C. Wyant (2000 Modified) 17

WP - 17

A 633 nm wavelength linearly polarized source is used with a Young’s two-pinhole interferometer. The same amount of light is transmitted through each pinhole. A 17.58 micron thick quartz plate is placed over one pinhole where the axes of the quartz plate are at 45 degrees relative to the direction of polarization of the incident light. The indices of refraction for quartz are no=1.544 and ne=1.553. If when the quartz plate is removed the fringe contrast is unity, what is the fringe contrast with the quartz plate in place?

Solution Calculation of retardation of quartz plate in units of the wavelength

∆n t 1.553 − 1.544 17.58 µm 1 = = λ 0.633 µm 4

Therefore, oneH beam has circularL polarization. The circularly polarized beam can be thought of as two linearly polarized beams with the amplitude of each beam is ÅÅÅÅÅÅÅÅÅÅ1 that of the incident beam. 2

ω t+φ 1 ω t ˆ eo ω t ˆè!!!! e = eo + i + j 2 2 H L H L H L i y j 1 2 z io i = io j 1 + + è!!!! Cos φ z+ è!!!! k 2 2 { 2 2 ji 2 zy j imax − imin @ Dz 2 1 contrastj = è!!!! = z = k imax + imin {4 è!!!! 2

è!!!!