PH2103: Physics Laboratory III

Polarization of light
Polarisers and waveplates, Malus Law.

Satvik Saha∗
19MS154

Indian Institute of Science Education and Research, Kolkata,

Mohanpur, West Bengal, 741246, India.

November 17, 2020

Abstract
In this experiment, we verify the Malus Law of polarization and study the effect of
polarisers and waveplates on the intensity of light.

1 Theory
We have seen in our discussion of Lissajous figures that the superposition of two orthogonal
vectors, whose amplitudes vary sinusoidally with the same frequency but with some phase
difference, produces a new vector whose tip traces an ellipse. In other words, consider the
parametric curve described by
x(t) = A cos(ωt), y(t) = B cos(ωt + φ).
Eliminating the parameter t gives
x2 y2 2xy
2
+ 2
− cos φ = sin2 φ.
A B AB
Conversely, any light wave polarized as such (elliptically, which includes circular and linear1 ) can
be decomposed into two sinusoidal components along any two perpendicular axes. A polariser
functions by removing the component along one of its axes and allowing the remainder to pass.
A waveplate functions by introducing an additional phase difference between the components
along its axes (informally called the fast and slow axes).

Polarisers Consider a polariser whose transmission axis is inclined at an angle θ with respect
to the oscillation of a linearly polarised wave. If this wave is of the form E(t) = E0 cos ωt, note
that its intensity is given by
1
I0 = 0 chE 2 i = 0 cE02 .
2
At any moment, the electric field vector can be decomposed into components along the trans-
mission axis, and perpendicular to the transmission axis.
Ek (t) = E0 cos ωt cos θ, E⊥ (t) = E0 cos ωt sin θ.
∗
Email: ss19ms154@iiserkol.ac.in
1
A line is simply a degenerate ellipse

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Experiment VI

When passing through the polariser, only Ek survives. Its intensity is simply

1
I = 0 chEk2 i = 0 cE02 cos2 θ.
2
Thus, we have obtained a relation between the initial and final intensities. This is called the
Malus Law.
I = I0 cos2 θ.
Of course, this holds only for linearly polarised light. If instead we start with circularly polarized
light, the components along and perpendicular to the transmission axis will always have the
same amplitude, regardless of the orientation of the polariser2 (this is a simple consequence of
symmetry). Thus, the component passing through must have exactly half the intensity of the
original light, so I = I0 /2.
An analogous argument can be used for unpolarised light. Different waves in the incoming
beam have their electric fields oscillating in different, random orientations. Thus, we can ap-
proximate θ as a random variable which varies uniformly over [0, 2π]. The average of cos2 θ is
thus hcos2 θi = 1/2, so the transmitted intensities of all these waves adds up to I0 /2, half of the
incoming intensity.

Waveplates A waveplate comprises of an anisotropic medium which exhibits different refrac-

tive indices along its fast and slow axes. If light passes perpendicular to both these axes over a
thickness t, the components along the fast and slow axes accumulate a path difference d = t∆n,
where ∆n is the difference in the fast and slow refractive indices. This corresponds to a phase
difference of δ = 2πt∆n/λ.
Suppose a waveplate imparts a phase difference of δ between components along its fast and
slow axis. Thus, a linearly polarised wave striking the waveplate at an angle θ has components
as described earlier. We add a phase δ to one of them to obtain

Ek (t) = E0 cos(ωt) cos θ, E⊥ (t) = E0 cos(ωt + δ) sin θ.

Thus, the waveplate changes the nature of the elliptic polarisation. Note that the intensity
of the outgoing beam, which is equal to the sum of intensities of both outgoing components,
remains unaltered.
1 1
I = Ik + I⊥ = 0 c(E0 cos2 θ + E0 sin2 θ) = 0 cE02 = I0 .
2 2
In the special case that θ = π/4 and δ = π/2, note that
1 1
Ek (t) = √ E0 cos(ωt), E⊥ (t) = − √ E0 sin(ωt).
2 2
Thus, we have converted linearly polarised light into circularly polarised light. We can clearly
see that Ik = I⊥ = I0 /2. The principle of reversibility shows that we can also convert circularly
polarised light into linearly polarised light by running this system backwards. This particular
waveplate is called a quarter waveplate, because it imparts a path difference of λ/4.

In the special case that δ = π, note that

Ek (t) = E0 cos(ωt) cos θ, E⊥ (t) = −E0 cos(ωt) sin θ.

Thus, the linearly polarized light remains linear, but has been reflected about the parallel
component, i.e. deflected by an angle 2θ. This particular waveplate is called a half waveplate,
2
In all these cases, we of course assume that the polariser/waveplate is normal to the direction of propagation
of the waves.

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Experiment VI

because it imparts a path difference of λ/2.

In any case, note that when the incoming linearly polarised beam is directed along one of
the axes, i.e. one of Ek and E⊥ is zero, the waveplate has no effect. Otherwise, the waveplate
has the effect of distributing the intensity of the beam over both axes. This can be checked by
removing one of the components using a polariser. Additionally, a polariser has no effect on a
linearly polarised beam only when their axes are aligned too (θ= 0 in the Malus Law). Thus,
such a waveplate-polariser system will produce a full intensity outgoing beam only when all of
their axes (oscillation of the electric field, fast/slow axis of the waveplate, transmission axis of
the polariser) are aligned.

Imperfect polarisers Suppose that the polarisers used are imperfect, such that the trans-
mission axis allows a fraction β ≈ 1 of the electric field amplitude to pass and the perpendicular
axis allows a fraction α ≈ 0 of the electric field amplitude. Linearly polarised light striking this
polariser such that its polarisation angle is inclined by θ with the transmission axis will thus
transform into
Ek = βE0 cos(ωt) cos θ, E⊥ = αE0 cos(ωt) sin θ.
The intensity distribution is of the form
1
I(θ) = 0 E02 (β 2 cos2 θ + α2 sin2 θ) = I0 (α2 + (β 2 − α2 ) cos2 θ).
2
Note that Imax = β 2 I0 and Imin = α2 I0 . The quantity

Imax − Imin β 2 − α2
P= = 2
Imax + Imin β + α2

is called the degree of polarisation. Comparing the distribution with the curve a + b cos2 (x − c),
we note that I0 α2 = a and I0 (β 2 − α2 ) = b, so P = b/(2a + b).
With a slightly different definition of the degree of polarisation, we may write

Imin α2 b
P0 = 1 − =1− 2 = .
Imax β a+b
Suppose instead that we start with elliptically polarised light of the form Ex = A cos(ωt),
Ey = B sin(ωt)3 . We see that
1 1
I0 = 0 c(A2 + B 2 ) = 0 cE02 .
2 2
Rotating the frame by θ and scaling by β and α, we see that

Ek = β (A cos θ cos(ωt) − B sin θ sin(ωt)) , E⊥ = α (A sin θ cos(ωt) + B cos θ sin(ωt)) .

The intensity distribution is thus of the form

I(θ) = 0 c β 2 (A cos θ cos(ωt) − B sin θ sin(ωt))2 + α2 (A sin θ cos(ωt) + B cos θ sin(ωt))2 .

Note that
1 1
h(A cos θ cos(ωt) − B sin θ sin(ωt))2 i = A2 cos2 θ + B 2 sin2 θ,
2 2
1 1
h(A sin θ cos(ωt) + B cos θ sin(ωt))2 i = A2 sin2 θ + B 2 cos2 θ.
2 2
3
It is always possible to choose appropriate axes to resolve the components this way. Specifically, any ellipse
has this parametrization along the major and minor axes.

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Experiment VI

This is because the cross terms hcos(ωt) sin(ωt)i = 0. Thus,

1
I(θ) = 0 c (β 2 A2 + α2 B 2 ) cos2 θ + (β 2 B 2 + α2 A2 ) sin2 θ .
 
2
With an appropriate choice of constants, this again simplifies to the form

I(θ) = a + b cos2 θ.

2 Experimental setup
Light from a diode laser is passed through a polariser, which is oriented so as to obtain linearly
polarised light of the maximum possible intensity. Once this adjustment is made, the light
is passed through another polariser (an analyser). The analyser is rotated and the resultant
intensity is measured as a function of this rotation angle.

Now, rotate the analyser so that the transmitted intensity is minimum. This must be the
crossed position, where the transmission axes of the polariser and analyser are perpendicular.
Insert a quarter waveplate between them, and rotate it such that the transmitted intensity is
minimum once again. In this orientation, for the waveplate to have had no effect, we deduce
that the waveplate anisotropy axes are aligned with the transmission axes of the polarisers.
Rotate the waveplate by π/4. It must now produce circularly polarised light, which can be
verified by rotating the analyser; very little variation in intensity should be observed. Any such
variation is the result of the ellipticity of the light, and the ratio of maximum and minimum
observed intensities is a measure of this.

3 Experimental data and analysis

3.1 Processing and plotting
All data has been gathered into an Excel spreadsheet, read using pandas and processed using
numpy. The code used has been listed below.

# !/ usr / bin / env python3

import pandas as pd
import matplotlib . pyplot as plt
import numpy as np
from scipy import optimize

cos2 = lambda x , a , b , c : a + b * np . cos ( x - c ) **2

cos4 = lambda x , a , b , c : a + b * np . cos ( x - c ) **4
coscos = lambda x , a , b , c , d , e , f : ( a + b * np . cos ( x - c ) ) * ( d + e * np . cos (
x - f))

sheets = pd . read_excel ( ’ data . xlsx ’ , sheet_name = None )

polarisers = [ ’ Polariser ␣ I ’ , ’ Polariser ␣ II ’]
waveplates = [ ’ Waveplate ␣ I ’ , ’ Waveplate ␣ II ’]

for name in polarisers :

sheet = sheets [ name ]
angle , intensity = sheet [ ’ Angle ’] , sheet [ ’ Intensity ’]
coeff2 , cov2 = optimize . curve_fit ( cos2 , angle , intensity )
coeff4 , cov4 = optimize . curve_fit ( cos4 , angle , intensity )
plt . scatter ( angle , intensity , label = name , s =12)
plt . plot ( angle , cos2 ( angle , * coeff2 ) , ’ -r ’ , label = " Fit ␣ to ␣ $a ␣ + ␣ b \ cos ^2( x ␣ -␣
c)$")

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Experiment VI

plt . plot ( angle , cos4 ( angle , * coeff4 ) , ’ --g ’ , label = " Fit ␣ to ␣ $a ␣ + ␣ b \ cos ^4( x ␣ -
␣c)$")
print ( coeff2 , np . sqrt ( np . diag ( cov2 ) ) )
plt . xlabel ( " Angle ␣ ( rad ) " )
plt . ylabel ( " Intensity ␣ ( percentage ␣ of ␣ total ) " )
plt . legend ( loc = " lower ␣ right " )
plt . show ()

a , b , c = coeff2
x = np . linspace ( -1 , 1 , 100)
normalized = ( intensity - a ) / b
plt . scatter ( np . cos ( angle - c ) , normalized , label = ’ Normalized ␣ intensity ␣ vs ␣ $
\ cos ( x ␣ -␣ c ) $ ’ , s =12)
plt . plot (x , x **2 , ’ -r ’)
plt . scatter ( np . cos ( angle - c ) ** 2 , normalized , label = ’ Normalized ␣ intensity
␣ vs ␣ $ \ cos ^2( x ␣ -␣ c ) $ ’ , s =12)
plt . plot ( x [ x > 0] , x [ x > 0] , ’ -g ’)
plt . xlabel ( " $ \ cos \\ theta$ ␣ and ␣ $ \ cos ^2\\ theta$ " )
plt . ylabel ( " Normalized ␣ intensity " )
plt . legend ()
plt . show ()

for name in waveplates :

sheet = sheets [ name ]
angle , intensity = sheet [ ’ Angle ’] , sheet [ ’ Intensity ’]
coeff , cov = optimize . curve_fit ( coscos , angle , intensity )
plt . scatter ( angle , intensity , label = name , s =12)
plt . plot ( angle , coscos ( angle , * coeff ) , ’ -r ’ , label = " Fit ␣ to ␣ $ ( a ␣ + ␣ b \ cos ( x ␣ -␣
c ) ) ( d ␣ + ␣ e \ cos ( x ␣ -␣ f ) ) $ " )
print ( coeff , np . sqrt ( np . diag ( cov ) ) )
plt . xlabel ( " Angle ␣ ( rad ) " )
plt . ylabel ( " Intensity ␣ ( percentage ␣ of ␣ total ) " )
plt . legend ( loc = " lower ␣ right " )
plt . show ()

(a) Polariser I (b) Polariser II

Figure 1: Intensities from the polariser-analyser setup as a function of the rotation angle. Note
that the fit against a + b cos2 (x − c) is almost perfect, while the a + b cos4 (x − c) curve does not
fit at all.

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Experiment VI

(a) Polariser I (b) Polariser II

Figure 2: The intensities from the polariser-analyser setup has been normalised as Inormal =
(I − a)/b. These have been plotted against cos(x − c) and cos2 (x − c). The curves y = x2 and
y = x have been drawn in red and green respectively to emphasize the cos2 relationship.

The fit parameters and uncertainties from Fig. 1 are as follows.

a b c
Set I 21.95 ± 0.01 −20.69 ± 0.02 −7.80 ± 0.01
Set II 21.34 ± 0.01 −20.13 ± 0.01 −7.83 ± 0.01

Because of the manner in which the fit has been made (b < 0), we have Imax = a and
Imin = a + b. Thus,

Imax − Imin a − (a + b) −b Imin a+b −b

P= = = , P0 = 1 − =1− = .
Imax + Imin a + (a + b) 2a + b Imax a a

We calculate these for the two sets of data as follows.

PI = 0.891, PI0 = 0.943,

0
PII = 0.893, PII = 0.943.

We estimate the ellipticity of the light from the waveplate using data from Fig. 3.
Imin Imax η = Imin /Imax
Set I 6.05 ± 0.1 10.75 ± 0.1 0.563
Set II 6.80 ± 0.1 9.05 ± 0.1 0.751

Thus, neither set exhibits circularly polarised light, since Imin /Imax is not close to unity. Indeed,
the particular pattern of alternating magnitudes of the maxima and minima cannot be explained
by supposing elliptically polarised light. The fitting function I = (a+b cos(x−c))(d+e cos(x−f ))
has been conjectured without a theoretical basis.

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Experiment VI

(a) Waveplate I (b) Waveplate II

Figure 3: Intensities from the waveplate setup as a function of the rotation angle. Note that
they do not fit an elliptically polarised light distribution, even allowing for imperfect polarisers.
Thus, the light from the waveplate is not circularly polarised. The ellipticity is estimated as
η = Imin /Imax .

3.2 Error Analysis

We write
2 2  2  2
∂P ∂P 2b 2a
(δP)2 = (δa)2 + (δb)2 = (δa)2
+ (δb)2 ,
∂a ∂b (2a + b)2 (2a + b)2
∂P 0 2 ∂P 0 2
 2  2
b 1
(δP 0 )2 = (δa)2 + (δb)2 = (δa) 2
+ (δb)2 ,
∂a ∂b a2 a
 2 
δImin 2 δImax 2
  
δη
= + .
η Imin Imax

Thus, we calculate
δPI = 0.002, δPI0 = 0.001,
0
δPII = 0.001, δPII = 0.001.
δηI = 0.011, δηII = 0.013.

3.3 Reported Values

We see that the degree of polarisation of light produced is P = 89.2±0.2% (or P 0 = 94.3±0.1%,
depending on which definition is preferred).
We also note that the light produced by the waveplates is not circularly polarised. The
measure of ellipticity η is not close to unity, instead it is 0.563 ± 0.011 in the first set and
0.751 ± 0.013 in the second.

4 Discussion
We see that the Malus Law holds very well, and any deviations can be explained by the imper-
fection of the polariser. The degree of polarisation of these polarisers is quite good.
The behaviour of the polarised light from the waveplate is unexpected, and requires deeper
analysis to justify the form (a + b cos(x − c))(d + e cos(x − f )). This is not consistent with

7
Experiment VI

elliptically polarised light, and is certainly not with a circular polarisation. We may attempt to
consider the effects of an imperfect first polariser, or an imperfect waveplate to try and explain
the form of the intensity curve.
One hypothesis for the four extrema in the waveplate curve is that the polarisation of the
light resembles an off-centre or squashed ellipse. The magnitude of a vector from the centre
to the edge of such a shape goes through a periodic cycle with two different maxima and two
minima, much like our intensity distribution. Such a polarisation may have arisen from an
imperfect waveplate, perhaps tilted, or some external electric field which biases the oscillating
vector in some particular direction.

4.1 Sources of error

The imperfections in the polariser comprise a source of systematic error. The least count of the
rotary motion sensor introduces random error, as does any external lighting.

5 Conclusion
In conclusion, we have verified the Malus Law and have analysed the effect of waveplates on
lieanrly polarised light.