2.71 Optics
Final Exam Solutions
Spring ‘09
1. Consider the following system.
(a) If we position an on-axis point source at the center of the object plane (front
focal plane of L1), a collimated ray bundle will emerge to the right of L1 and
its diameter is set by S1; therefore, S1 is the aperture stop (A.S.). Similarly, S2
limits the lateral extent of an imaged object (consider an off-axis point source)
and thus, it’s our field stop (F.S.).
(b) The entrance pupil is the image of the A.S. by the preceding optical components.
To find its location we use the imaging condition,
1
1
1
+
=
So Si f1
2f1
So =
3
So f1
2f12 /3
=
So − f1
f1 ( 23 − 1)
⇒
Si =
⇒
Si = −2f1 (virtual)
So the entrance pupil is located at 2f1 to the right of L1. To find its radius, we
compute the lateral magnification,
ML = −
Si
= 3 → rEnP = 3a1
So
For the exit pupil (Ex.P.),
f1
So =
+ f2 ,
3
ML =
Si =
− 3ff12 ( f31 + f2 )
( f31
+ f2 )
( f31 + f2 )f2
f1
3
+�
f�2 − �
f�2
= −3
= f2 + 3
f2
(inverted)
f1
f22
(to the right of L2)
f1
→
rExP = 3
f2
a1
f1
• The exit window is the same as S2.
• The entrance window is the image of S2 through the preceding optical ele­
ments (i.e. combination of L1 and L2). It is f1 to the left of L1.
1
(c) Solution:
The numerical aperture is:
tan α ≈ α ≈ sin α ≈ NA ≈
a1
f1
The field of view is: FOV = 2β =
2Xs
2 f1 a2
2 a2
=
=
3f1
3 f 1 f2
3 f2
(d) The location of S1 limits the FOV because of the requirement for the C.R. to
go through the center of the aperture stop (A.S.). It can be seen that the least
restrictive A.S. location is at the Fourier plane (f1 to the right of L1 ⇐⇒ f2 to
the left of L2).
2. Solution:
(a) Focal length f should be satisfied with
1
1
1
+
=
So S1
f
1
1
5
24
+ =
→ f = cm
12 8
24
5
(b) Lateral magnification MT = −
Si
8
2
=− =−
So
12
3
2
(c) Solution:
R
So
R = 1.2 cm → Diameter of the lens: 2.4 cm
NA ≈ sin θ ≈ tan θ = 0.1 =
(d) λ = 1µm
1µm
λ
Rayleigh resolution = 1.22 (NA)
= (1.22) (0.1)
= 12.2µm
(e) To achieve |MT | = 1, So = Si = 2f → So = Si =
48
cm
5
3. Michelson Interferometer
When the mirror #2 is tilted by θ,
the reflected light is rotated by 2θ.
Unfolding the optical paths, we have this situation:
At the observation plane,
2π
E1 (x) = ei λ (z0 +2z1 +zc )
2π
2π
E2 (x) = ei λ (z0 +z2 ) ei λ {cos 2θ(z2 +zc )+sin 2θx}
Δφ = �
z�0 + z2 + cos 2θ(z2 + zc ) + sin 2θx − �
z�0 − 2z1 − zc
= z2 (1 + cos 2θ) − 2z1 + zc (cos 2θ − 1) + sin 2θx
= 2z2 cos2 θ − 2z1 − 2zc sin2 θ + sin 2θx
= 2(cos2 θz2 − sin2 θzc − z1 ) + sin 2θx
3
Note that if θ = 0, Δφ = 2(z2 − z1 ), which only depends on z1 and z2 .
Neglecting attenuation due to reflection, we obtain the field as E(x) = E1 (x) + E2 (x).
The intensity of the interference is
I(x) = |E(x)|2 = |E1 (x)|2 + |E2 (x)|2 + 2Re{E1∗ E2 }
�
�
2π
= 1 + 1 + 2 cos
Δφ
λ
�
�
��
�
2π �
2
2
= 2 1 + cos
sin 2θx + 2(cos θz2 − sin θzc − z1 )
λ
The normalized intensity is:
∴ The period of the fringe is dependent on θ. If θ increases, the period decreases (finer
fringes). Due to the phase shift by λ2 (cos2 θz2 − sin2 θzc − z1 ), the whole fringe shifts
as z1 and z2 change.
4. Consider the 4-f system shown below,
4
(a) The pupil mask can be implemented by placing two pinholes (small apertures),
one centered with respect to the optical axis and the second one at 1 cm off-axis.
The 2nd pinhole is phase delayed by a piece of glass of thickness t, where
φ=π=
2π
t(1.5 − 1) ⇒ t = λ
λ
(b) The input transparency is
∞
qx
1 �
��=
�
gin = gt · �
gillumination
α
sinc(αq)e
i2π Λ
��
q=−∞
At the Fourier plane,
∞
�
�
�
q
�
��
Gin = α
sinc(αq)δ u −
Λ
�
u= x��
q=−∞
λf
�
�
��
∞
�
�
1
�
q
λf
��
��
Gin (x ) =
sinc
δ
x − q
,
2 q=−∞
2
Λ
λf
0.5 × 10−4 · 10
=
= 1
Λ
5 × 10−4
After the pupil mask, only the 0th and +1 orders pass. The +1 order gets phase
delayed by eiπ = −1.
�
�
1
1
1
��
��
Gout (x ) = δ(x ) − sinc
δ(x�� − 1)
2
2
2
1
1
=
δ(x�� ) − δ(x�� − 1)
2
π
�
1
1
��
gout (u� ) = − e−i2πu ��
2 π
x�
u� = λf
�
�2
�
�
� 1 1
−i 2πx� �
1
1
1
2π
x�
�
�
λf
Iout =
�
−
e
�
=
4
+
π 2 − π cos
λ f
2 π
The contrast is v = 0.906 =
1
π
1
4
+
1
π2
=
4π
.
4 + π2
5. (a) To compute the OTF, we first need the ATF:
5
u1 =
x��
1cm
=
= 0.2µm−1 = 200mm−1
λf
0.5µm × 10cm
est.
� �� �
δx��
0.2cm
≈
δu =
= 0.025µm−1 = 25mm−2
λf
0.5µm × 10cm
H(u) = H(u) ⊗ H(u)
(autocorrelation)
� ��
� �
�� �
� �
�
� �
��
� �
u
u − u1
u −u
u − u1 − u
=
rect
− rect
rect
− rect
du�
δu
δu
δu
δu
�
�
=
rect
�
�
−
rect
�
�
−
rect
�
�
+
rect
u�
δu
�
u�
δu
�
�
rect
u� − u
δu
�
− rect
u� − u1
δu
�
u� − u1
δu
�
du� →
u� − u1 − u
δu
�
�
u� − u
δu
�
u� − u1 − u
δu
rect
rect
�
�
du� →
du� →
6
�
du� →
So the resultant OTF and MTF are (after normalization):
(b) Only the DC and the ± 1st harmonics at u = ±200mm−1 (period = 5µm), i.e.
�
�
1 1 1
2πx�
�
I(x ) = − × × 2 cos
2 2 π
5µm
�
�
2πx�
−2
= DC term − H(200mm ) × 1st harmonic × 2 cos
5µm
(c) Solution:
F
−→
sinc2 (δux), so the iPSF is:
= sinc2
7
�
x
40µm
�
�
�
��
x
× 1 − cos 2π 5µm
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Spring 2009
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