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1.
You have available a source of
unpolarized light
(intensity I
o
) 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(
θ
)
relative to I
o
transmitted by two polarizers as a function of the angle
θ
between their
transmission axes.
I
0
I
1
I
2
!
I
1
=
1
2
I
0
, and
I
2
=
1
2
I
0
cos
2
θ
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View Full Document In considering the electric field components, we can obtain the relations
between intensities before and after each polarizer.
E
!
"
E
E
!
E
 
I
o
=
1
2
ε
o
c {(E
↑
)
2
+ (E
→
)
2
}
I
1
=
1
2
ε
o
c (E
→
)
2
=
1
2
I
o
I
2
=
1
2
ε
o
c (E
→
cos
θ
)
2
=
1
2
I
o
cos
2
θ
(B) Imagine two polarizers are set with their transmission axes "crossed" at 0° and
90°. What is the transmitted light intensity?
For
θ
= 90°,
I
2
= 0.
(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
θ
o
= 0°,
θ
1
= 60°,
θ
2
=
45°,
θ
3
= 75°,
θ
4
= 15°,
θ
5
= 30°, and
θ
6
= 90°.
I
1
=
1
2
I
o
I
2
= I
1
cos
2
(
θ
1
–
θ
o
) =
1
2
I
o
cos
2
( 60 – 0)
I
3
= I
2
cos
2
(
θ
2
–
θ
1
) =
1
2
I
o
cos
2
(60) cos
2
(15)
I
4
= I
3
cos
2
(
θ
3
–
θ
2
) =
1
2
I
o
cos
2
(60) cos
2
(15) cos
2
(30)
I
5
= I
4
cos
2
(
θ
4
–
θ
3
) =
1
2
I
o
cos
2
(60) cos
2
(15) cos
2
(30) cos
2
(60)
I
6
= I
5
cos
2
(
θ
5
–
θ
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This note was uploaded on 08/06/2008 for the course PHYS 352 taught by Professor Dierolf during the Fall '04 term at Lehigh University .
 Fall '04
 DIEROLF
 Light

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