polarization intensity

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1.
You have available a source of unpolarized light (intensity Io) and a number of
"perfect" linear polarizers (each of them transmitting without losses all the light
polarized parallel to its "transmission axis" and blocking totally all light polarized
perpendicular to this axis.)
(A) Using a sketch of the light polarization and the polarizer orientation, derive
and show the simple relation for the light intensity I() relative to Io
transmitted by two polarizers as a function of the angle  between their
transmission axes.

I0
I1
I1 =
1
2 I0 ,
and
I2
I2 =
1
2
2 I0 cos 
In considering the electric field components, we can obtain the relations
between intensities before and after each polarizer.


E
E
E
Io =
1
2
2
2 o c {(E ) + (E ) }
I1 =
1
2
2 o c (E ) =
I2 =
1

2
2 o c (E cos  ) =
1
2 Io
1
I cos2 
2 o
(B) Imagine two polarizers are set with their transmission axes "crossed" at 0° and
90°. What is the transmitted light intensity?
For  = 90°, I2 = 0.
E

(C) Five additional polarizers are placed between the two "crossed" polarizers
with their transmission axes set at 60°, 45°, 75°, 15°, and 30° in that order,
with the 60° polarizer adjacent to the 0° polarizer and so on (all angles are
measured from the axis of the first 0° polarizer). What is the intensity relative
to the incident one, transmitted by these seven polarizers?
The orientation angles of a series of polarizers are o = 0°, 1 = 60°, 2 =
45°, 3 = 75°, 4 = 15°, 5 = 30°, and 6 = 90°.
I1 =
1
2 Io
1
2
1
cos2 (2 – 1) =
2
1
cos2 (3 – 2) =
2
1
cos2 (4 – 3) = 2
1
cos2 (5 – 4) = 2
I2 = I1 cos2 (1 – o) =
Io cos2 ( 60 – 0)
I 3 = I2
Io cos2 (60) cos2 (15)
I 4 = I3
I 5 = I4
I6 = I 5
Io cos2 (60) cos2 (15) cos2 (30)
Io cos2 (60) cos2 (15) cos2 (30) cos2 (60)
Io cos2 (60) cos2 (15) cos2 (30) cos2 (60)
cos2 (15)
I7= I6 cos2(6 – 5)=
cos2(60) =
1
2
2
2
2
2
2 Io cos (60) cos (15) cos (30) cos (60) cos (15)
1
6(60) cos4(15) cos2(30) = 0.0051 I
I
cos
o
o
2
(D) Can you rearrange the sequence of the five polarizers between the crossed
ones such that the total transmitted intensity is as high as possible? With the
arrangement what is the total transmitted intensity?
For the maximum transmission, we need the equi-angular difference of two
adjacent polarizers. That is, the orientation angles of a series of
polarizers are
o = 0°, 1 = 15°, 2 = 30°, 3 = 45°, 4 = 60°, 5 = 75°, and 6 = 90°.
Then,
I7 =
1
12(15) = 0.3298 I
I
cos
o
o
2
[~ 65 times stronger than (C)!!! ]
2. A single slit of width b, illuminated with He-Ne laser light ( = 6328Å) produces a
Fraunhofer diffraction pattern with first order minima appearing at angular distance 
= 3° from the optical axis.
(A) How wide is the slit?
For a single slit diffraction,
I() = I(0) {
sin (/2)
(/2)
}2
where  = kb sin  = (2b/) sin 
The first minimum occurs at /2 = ±, where


b=
sin 1
b
sin 1 = ±.

6.328  10-7 m
=
= 1.21  10-5 m = 12.1 m
sin 3°
(B) If the laser light could be chopped into pulses of 210-14 sec duration, what would
change in the observed diffraction pattern? Describe in words and with a sketch.
Coherence time t = 2  10-14 sec

Coherence length x = c· t = 6  10-6 m = 6 m
Therefore, the diffraction pattern will disappear if the path length difference
x
b·sin is larger than the x. That is, sin max = b ≈ 0.5

max = 30°.
The diffraction pattern disappears for  > 30°.
No
Diffraction
Pattern
30°
3.
A telescope of 10 magnification (constructed by two positive lenses) is pointed at
the moon and focused to give an image at infinity. [Besides giving answers, make
schematic sketches with proper rays for all questions A-C.]
(A) If the focal length of the objective is 60 cm, what are the focal length of the
eyepiece and the distance between two lenses?
fo
Telescope Magnification M = f
e
= 10
fo = + 60 cm, then fe = + 6 cm
The distance d between the objective and the eyepiece lenses in a
telescope is
d = fo + fe = 66 cm.
soobj = + ∞ and siobj = fo  soeye = fe and sieye = + ∞ .
Objective
Eyepiece
Moon
fo
fe
(B) How far and in which direction must the eyepiece be moved to project an
image of the moon on a screen 30 cm away from the eyepiece?
Let the moving distance of the eyepiece be + x ( + is the right side).
Still soobj = + ∞ and siobj = fo, but we want soeye = fe + x and sieye
= + 30 cm.
By the lens equation for the eyepiece,
1
1
=
fe
soeye
+
1
sieye
 soeye = + 7.5 cm
 x = + 1.5
cm
The eyepiece must be moved 1.5 cm to the right side.
Objective
Screen
Eyepiece
x
Moon
fe
fo
30cm
(C) If the eyepiece is exchanged to a negative lens of the same focal length, what
will be changed in (A) and (B)?
If fe = – 6 cm, then the distance d between the objective and eyepiece
lenses is
d = fo + fe = 60 cm – 6 cm = 54 cm.
Objective
Eyepiece
Moon
fe
fo
To make a real image on the screen at 30 cm away,
Still soobj = + ∞ and siobj = fo, but soeye = fe + x < 0 (virtual object)
and sieye = + 30 cm.
Then, soeye = – 5 cm  x = + 1 cm.
The eyepiece must be moved 1 cm to the right side.
Eyepiece
Screen
30cm
Objective
fe
Moon
fo
x
7. What is the size of the diffraction-limited image point which a perfect eye produces
on the retina for light of  = 500 nm and for an object point at infinite distance (focal
distance of the eye lens 25 mm), with a pupil diameter of 2 mm (at daylight vision)
and 8 mm (adopted to vision at night)? If you would have to "design" the eye as
economically as possible (but using its full resolution potential), how would you
choose the distance between the photo-receptor cells on the retina to work in an
optimum way for the whole visible spectral range (390 ≤  ≤ 780 nm) ?
The radius q of the image for an object point at infinite distance is

q = 1.22 2a feye where 2a is the diameter of eye pupil.
500nm
For daytime, 2a = 2 mm .  qday = 1.22· 2 mm · 25mm = 7625 nm
= 7.6m
For nighttime, 2a = 8mm  qnight = 1.9 m
For the visible spectral range (390 nm <  < 780 nm), the size of image q:
q
Daytime (2a = 2mm)
Nighttime (2a = 8mm)
Short Wavelength (390nm)
5.9 m
1.5 m
Long Wavelength (780nm)
11.8 m
3.0 m
Without losing the resolution, the distance between the photoreceptor cells
must be smaller than 1.5 m.
(5)
We need to use the highest EO-coefficent. For that reason the field needs to be applied in
z-direction. For propagation we use the y-direction. This leads to the following ellipse
x 2   z2  2
 2   2  z r33 E z  1
no  ne 
we get a change in refractive index for light polarized in z-direction:
r33 E zn 3e
nz  ne 
2
If we align the incoming polarization along the 45 between x and z we get a phase
difference of
n 33

  z  x 
r33 El  ( n e  n o )l
2c
c
In order to compensate for the natural birefringence we need to run our modulator on an
offset of
n  n 
Voffset  ( e 3 o )2(l)
ne r33
This formula holds if we make the phase shift to zero, but a phase shift of m2 is OK as
well.
n 3

m2  e r33Voffset  ( ne  no )l
2c
c

c  2
Voffset  ne  n o l  m2

 ne3 r33
We need then to apply a voltage of
2c
V 
 Voffset
n33 r33
+
propagation
z
y
Polarization
x
-
6. The plate will just exchange S2 by –S2 and and S3 by –S3 leaves S1 alone. This
can be represented by a 180o rotation around the axis X. for linear polarized light
S3=0. for that reason we get a rotation of the polarization direction of 2 with (
being the angle between fast axis and original linear polarization). The resulting
polarization will be at -.
9. Light is scattered in all direction at the first interface it hits the second interface at an
angle . If  is larger then the angle for total internal reflection then we won’t see any of
the scattered light
R  d tan 
.
sin  
nair
nglass
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