quarter wave plate

birefringence

1. Birefringence of mica using quaterwaveplate
PRESENTATION BY :
MUKESH.U
M.Sc Ed Physics

2. * Full wave plate
* Half wave plate
* Quarter wave plate

7. Theory
Huygen explanation 0f double refraction in uniaxial crystal.
●Consider a point source in calacite crystal (-ve crystal).
●In negative can ystal the ellipsoid of e-ray is outside and e- Ray travel with
maximum velocity in direction perpendicular to the optic axis.
●We know
Pathi in air =path in medium× refractive index.
For negative crystal
n
o
greater than n
e.
Path difference=[n
o
-n
e
] t
t- thickness of the plate
For positive crystal n
e
greater than n
o
so path difference = {n
e
-n
o
} t

8. ●Quaterwaveplate is the double refracting uniaxial crystal of calacite or quartz of
suitable thickness whose refracting are cut parallel to the direction of optic axis .
●The incident plane polarized light is perpendicular to it's surface and o-ray and
e-ray travel along the same direction with different velocities.
●To produce path difference of quaterwaveplate in calacite
{n
o
-n
e
} t =quaterwaveplate
t=λ/4({n
o
-n
e
}
●If the plane polarised light whose plane of vibration is inclined at of angle 45° to
the optic axus is incident on quarter waveplate the emergent light is circularly
polarised.
● Ouaterwaveplate do not polarizer the unpolarized light and ideally they not
reduce the intensity of incident light beam.

11. Theory of quaterwaveplate as function of angle.
Let Eo is the amplitude of electric vector emerging from polarizer. Let Φ is the
angle b/w the direction of polarizer and optic axis of doubly refracting crystal.
●The amplitude of o-ray and e- ray
E
1
= E
0
sinΦ
E
2
= E
0
cosΦ

At time t, the state of vibration of two ray at crystal surface ,is
described
E
1
(t)=E
1
sinwt
E
1
(t)=E
o
sinΦ sinwt
E
2
(t)=E
2
sinwt
E
2
(t)=E
o
cosΦ sinwt.

12. Wkt for quaterwaveplate the thickness is given by
t=quaterwaveplate/4{n
o
-n
e
}
Quaterwaveplate=path difference introduces phase difference
Delta=2π\λ(λ/4)=π/2
When the o-Ray and e- ray combine to give resultant on emerging from the
plate then
E
x
=E
o
sinΦ sinwt
E
y
=E
o
cosΦ sin wt+π/2
E
y
=E
0
cosΦ coswt
Vector rotation if E-vector indirection of propagation perpendicular to the x-
and y-ax sbabout a fixed axis if Φ = o then E
x
=o
E
y
=E
0
coswt and resultant
E= √{E
x
2
+E
y
2
}=√{0+E
0
coswt
2
}
E= E
0
coswt
I=E
2
=E
0
cos
2
wt = E
o
coswt or I α cos
2
wt
Max I
O
= E
O
2

13. : We thus obtain plane polarised light of intensity I
O
= E
O
2
Intensity analysis with the position of an analyser without quaterwaveplate and
with quaterwaveplate
Φ=0 or Φ=90°
If Φ=45°
SinΦ= cosΦ=√1/2
Then E
X
=E
O
√2 sinwt
E
y
=E
O
√2 coswt
and E=√Ex
2
+E
y
2

=Eo√2
The light is circularly polarised and have intensity
I=Io√2=E
O
2
Now light is transmitted wolithout loss of intensity whatever may be the
position of analyser.

14. Procedure

Without quaterwaveplate in path adjust the path of rays from
halogen lamp so that it fall on light sensor connected to
computer through go link.
set the polarizer on zero position rotate the analyser so that light
in minimum.then quaterwaveplate clamped into its holder rotated
so that light passing through the analyser is again minimum
intensity .
Now the plane of polarisation of light from polarizer made angle
zero or 90 with the logic axis of quaterwaveplate.
The intensity is measured by keeping the analyser at 0
0
30
0
45
0

over the range -90
0
to 90
0
over .
The intensity of current is directly proportional to the intensity of
the incident light.

17. RESULTS
It is observed that the variation of intensity of
plane polarised light with the rotation of position
of analyser

18. THANK U