2.710 Optics
Quiz 2 Solutions
Spring ‘09
1. Solution:
π ��
π z
g1 (x, z) = a1 exp i2π
sin + cos
,
λ
6 λ
6
�
� x
��
π
π z
g2 (x, z) = a2 exp i2π − sin + cos
,
λ
3 λ
3
�
At z = 0,
�x
a1 = 1
a2 =
1
2
√ �
�
x 1
x
3 2
I(x) = |
g1 (x, 0) + g2 (x, 0)|
2 = �a1 ei2π λ · 2 + a2 e−i2π λ · 2 �
�
�
�
√ ���
2a1 a2
x 1
3
= (a21 + a22 ) 1 + 2
cos 2π
+
2
a1 + a2
λ 2
2
�
√ ��
�
��
1
2
×
1
×
1
x 1+ 3
2
= 1+
1+
cos 2π ·
1
4
λ
2
1+ 4
�
�
�
�
5
4
x
2λ
√
=
1 + cos 2π
where Λ =
4
5
Λ
1+ 3
4
2λ
√ → period
→ contrast
Λ=
5
1+ 3
�
�
5
4
5 9
9
1
I(0) = × 1 +
= × = =2
4
4 5
5
4
4
m=
After phase-shifting g1 by π/2, we get
�
� x π ��
5
4
5
1
⇒ I(0) = = 1
I(x) =
1 + cos 2π +
4
5
Λ 2
4
4
2. Solution:
λ = 1.0µm,
f = 5cm,
1
Λ = 10µm
According to Lecture 19, slide 12, with duty cycle α = 12 ,
∞
�q �
x
1 �
gt (x) =
sinc
ei2πq Λ
2 q=−∞
2
∞
�q � �
1 �
q�
Gt (u) =
sinc
δ u−
2 q=−∞
2
Λ
⎧
q−1
⎪
2(−1) 2
⎪
if q = odd
� q � sin ( πq ) ⎨ πq
2
sinc
=
=
1
if q = 0
πq
⎪
2
⎪
2
⎩0
� 0, even
if q =
�
�
�
�
1
3
1
1
Gt (u) = · · · −
δ u+
+ δ u+
10µm
π
10µm
3π
�
��
� �
��
�
-3rd order
-1st order
�
�
�
�
1
1
1
1
3
+ δ(u) + δ u −
− δ u−
+···
10µm
10µm
3π
�2 �� � �π
��
� �
��
�
0th
order
(DC)
+1st order
+3rd order
We see that the even orders are missing.
� �� �
�
�
�
�
x
1
3λf
1
λf
1
��
��
��
gpp− (x ) ∝ Gt
= ··· −
δ x +
+ δ x +
+ δ(x�� )
λf
Λ
π
Λ
2
3π
�
�
�
�
1
λf
1
3λf
+ δ x�� −
−
δ x�� −
+ ···
π
Λ
3π
Λ
λf
10−6 × 5 × 10−2
=
= 5mm,
Λ
10 × 10−6
1
1
1
1
1
gpp− (x�� ) ∝ · · ·− δ(x�� +15mm)+ δ(x�� +5mm)+ δ(x�� )+ δ(x�� −5mm)− δ(x�� −15mm)+· · ·
π
2
π
3π
3π
Since
2
1
1
gpp+ (x�� ) = gpp− (x�� )gpm (x�� ) = δ(x�� + 5mm) −
δ(x�� − 15mm)
π
3π
� ��
x�
x
1
1 +i2π 103xµ�m
�
gout (x ) ∝ Gpp+
= e−i2π 10µm −
e
λf
π
3π
� � 2 � �2
� ��
�
�
�
1
1
1
1
2x�
�
� 2
Iout (x ) = |gout (x )| =
+
+2
−
cos 2π
π
3π
π
3π
10µm
�
�
�
10
2
2x
= 2 − 2 cos 2π
9π
3π
10µm
Reminder:
|a1 eiφ1 + a2 eiφ2 |2 = (a1 eiφ1 + a2 eiφ2 )(a1 e−iφ1 + a2 e−iφ2 )
= a21 + a22 + a1 a2 ei(φ1 −φ2 ) + a1 a2 ei(φ2 −φ1 )
= a21 + a22 + 2a1 a2 cos(φ2 − φ1 )
10
( 9π
Imax − Imin
2 +
V =
= 10
Imax + Imin
( 9π2 +
2
)
3π 2
2
)
3π 2
10
− ( 9π
2 −
10
+ ( 9π2 −
2
)
3π 2
2
)
3π 2
2
2� × 3π�
3
2
�
=
10 =
5
2� × 9π�2
�
Now modify the pupil mask so that its complex transmissivity is:
This can be done by inserting a piece of glass of optical thickness equal to π in the
upper hole. The physical thickness, if n = 1.5, is:
2π
(2k + 1)λ
(2k + 1) × 1µm
(n−1)t = (2k+1)π ⇒ t =
=
= (2k+1)µm, where k is any integer
λ
2(n − 1)
2 × (1.5 − 1)
With this modification,
gout (x� ) =
1 −i2π 10xµ�m
1 +i2π 103xµ�m
e
+
���� 3π e
π
result of π−phase shift in the +3rd order
⇒ Iout
�
�
10
2
2x�
= 2 + 2 cos 2π
9π
3π
10µm
Now the intensity is maximum at x� = 0.
3
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Spring 2009
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