2014 2.71/2.710 Optics, Solution for HW4
1. Mach-Zehnder Interferometer
i1
a) In a Mach-Zehnder interferometer, the light source is split into two beams ( E1  E e
 E eik0OPL1 eit ,
E2  E ei2  E eik0OPL2 eit ). Since the two beams come from the same source, they have equal wavelength 
(therefore equal wavenumber k0  2 /  ) and equal time dependent phase eit . The magnitude of the fields do
not change the interference pattern except for its brightness, and thus can be set as a constant value (E).
The two beams go through two different light paths with a slightly different optical path length, and then they are
joined in the end to produce an interference pattern on the screen. The interference pattern on the screen can be
controlled by the rotation of the signal mirror and by a gradual change of the refractive index inside the
pressurized cell.
In this problem, the brightness change on the screen is produced by the gradual pressurization of the gas
cell, and all measurements are taken at x=0 on the screen. A difference in the optical phase between the two
beams arriving at x=0 can be calculated from:
  2  1
 k0 (OPL 2  OPL1 )
 k2

2

 n
2
dl2   n1 dl1

L(n2  n1 ).
When the cell is evacuated, n2  n1  1 and there is no phase difference between the two signals, resulting
in a constructive interference between two beams (brightness). As the cell is being pressurized, 10 brightness
cycles were recorded (m=10). When the cell is fully pressurized, the refractive index reaches n, and the phase
difference is equal to   2 L(n  1)  2  m . Therefore, the refractive index can be calculated from:
 
2

L(n  1)  20
 n  1  mL  1.0000633
1
2014 2.71/2.710 Optics, Solution for HW4
b) The interference pattern for 1 and 2 can be considered separately, since two beams with distinct
wavelengths are independent and therefore do not interfere with each other. (Recall that two sinusoidal function
are orthogonal when the period is not exactly equal.) For each wavelength, the interference pattern can be
computed from:
I  | E1  E2 |2  ( E1  E2 )  ( E1 *  E2 *)
 E 2 (ei1  ei (1   ) )  (ei1  e i (1   ) )
 E 2 (1  ei )  (1  e i )
 E 2 (2  2 cos( )).
Using the relationship   2 L(n  1) , we obtain:
I1  E 2 (2  2 cos( 2 L(n  1)))
1
2
2
I 2  E (2  2 cos(  L(n  1))).
2
Recall that the variable in the above equation is n, since the gas cell is being pressurized in time. The total
intensity is calculated by summing the two intensities I 1  I 2 :




Itot  I 1  I 2  2 E 2  2  cos 2 L(n  1) 1  cos 2 L(n  1) 1 
1
2 

 I 0 1  cos 2 L(n  1)( 1  1 ) cos 2 L(n  1)( 1  1 ) 
1
2
1
2


 I0

1  cos 2 L(n  1)


1  2
12
 
 cos  2 L(n 1)

2 1
12
 .

 

The result is a beating interference: a high frequency oscillation, cos 2 L(n  1) 1  2 , is modulated by a
1 2

 

slowly varying envelope, cos 2 L(n  1) 2  1 . The plot of I tot as a function of n, which is a function of time,
1 2
looks like the following figure.
Image removed due to copyright restrictions. Image found at http:
//hyperphysics.phy-astr.gsu.edu/hbase/sound/imgsou/beat4.gif.
How a beat is formed (http://hyperphysics.phy-astr.gsu.edu/hbase/sound/beat.html)
2
2014 2.71/2.710 Optics, Solution for HW4
Thus, the intensity at x=0 on the screen varies periodically in time as the cell is gradually pressured. Since
the change in the refractive index is very small, it is likely that only the high frequency oscillation can be counted,
rather than the envelope. The number of cycles for the high frequency oscillation can be counted to measure the
total change in the optical phase:
  2 L(n  1)
 n  1  L1
12

2 1
12
 2 m
2
m  1  L m.
The accuracy of the measurement can be determined by how much index difference is measurable from a single
brightness cycle (m=1), which gives
n 

2
L
.

3
2014 2.71/2.710 Optics, Solution for HW4
b) Alternate interpretation for the calculation of accuracy
Let us consider a comparison between two separate interference measurements with 1 and 2 . While this
situation is physically different from the first interpretation and is also hard to implement in reality, the accuracy
of the refractive index measurement can be derived in a different way. For the same change in the refractive
index, two different number of cycles can be measured for 1 and 2 .
m1  L(n  1) 1
1
1
m2  L(n  1) 
2
The smallest change in the refractive index can be measured when m1 and m2 differ by one:
m  m2  m1  L(n  1)
m  1  n 
 

1
2

 1  L(n  1)
1

1 2
21

2
L
4
2014 2.71/2.710 Optics, Solution for HW4
2. Fabry-Perot Interferometer
The relationship between the frequency and the wavelength is f  c /  ,, where c is the speed of light. Using
1  435.8 nm, 2  546.1 nm, c  3 108 m / s , we get:
3 108
 6.9 1014 Hz,
9
435.8 10
3 108
 5.5 1014 Hz.
f2 
9
546.110
f1 
The thickness d of the etalon can now be calculated from:
2d  m
m1 
m2 
d
2d
1
2d
2
 m1  150
150 
1
1




2  435.8nm 546.1nm 
1
 161.8 m.
5
Michael Pasqual
2.710 – Optics
Homework #4
April 2, 2014
Problem 3. Fourier Transforms
I assume here that each function is defined as follows (Section 2.1.6 in Goodman):
1
x <1
2

rect(x) =  1
x =1
2
2
 0 otherwise

sinc(x) =
sin(πx )
πx
1

circ( x 2 + y 2 ) =  1
2
0

comb(x) =
x2 + y2 <1
x2 + y2 =1
otherwise
∝
∑ δ (x − n)
n =−∝
Part (a)
Separability Theorem:
Scaling Theorem:
F
g(x) ⋅ h( y) ←→
G( f X ) ⋅ H ( f Y ) Eq. 2-21 in Goodman
1  fX 
F
Eq. 2-12 in Goodman
g(ax) ←→
G

a  a 
F
rect(x) ←→
sinc( f X )
FT Pair:
Table 2.1 in Goodman
Apply
 y 
 x 
g(x,y) = rect 

 ⋅ rect 
 2W 
 2W 
g(x,y) = rect(ax) ⋅ rect(ay)
g(x,y) = h(ax) ⋅ h(ay)
1
2W
where h(x) = rect(x)
where a =
1  f X   f
Y 
H
Separability & Scaling Theorems
 ⋅ K 
a2  a   a 
1
f 
f 
G( f X ,f Y ) = 2 sinc X  ⋅ sinc Y 
FT Pair
a
 a 
 a 
G( f X ,f Y ) = 4W 2 sinc(2Wf X ) ⋅ sinc(2Wf Y )
G( f X ,f Y ) =
 y 
 x 
F
2
rect 
 ←→ 4W ⋅ sinc(2Wf X ) ⋅ sinc(2Wf Y )
 ⋅ rect 
 2W 
 2W 
Page 6 of 12
Michael Pasqual
2.710 – Optics
Homework #4
April 2, 2014
Part (b)
Separability Theorem:
Scaling Theorem:
F
g(x) ⋅ h( y) ←→
G( f X ) ⋅ H ( f Y ) Eq. 2-21 in Goodman
1  fX 
F
Eq. 2-12 in Goodman
g(ax) ←→
G

a  a 
F
exp(−πx 2 ) ←→
exp(−πf X2 )
FT Pair:
Table 2.1 in Goodman
Apply
 x 2 + y 2


g(x,y) = exp
2 
 2W 
 x2 
 y2 
 ⋅ exp 
g(x,y) = exp 2 
2 
 2W 
 2W 
  y  2 
  x 2 



g(x,y) = exp -
 
 ⋅ exp -
  2W  
  2W  




  y 2 
  x 2 
g(x,y) = exp -π 
 
  ⋅ exp -π 
  2πW  
  2πW  




g(x,y) = exp(− π (ax) 2 ) ⋅ exp(− π (ay) 2 )
g(x,y) = h(ax) ⋅ h(ay)
G( f X ,f Y ) =
1  f X  1  fY 
H ⋅ H 
a  a  a  a 
G( f X ,f Y ) =
1  f X   fY 
H ⋅ H 
a2  a   a 
1
2πW
where h(x ) = exp(- πx 2 )
where a =
Separability & Scaling Theorems
  f X 2 
  fY  2 
1


G( f X ,f Y ) = 2 exp -π   ⋅ exp -π   




a
  a  
  a  
FT Pair
 f X2 + f Y2 
1

 -π
exp
a2
a 2 

G( f X ,f Y ) = 2πW 2 exp(-2π 2W 2 ( f X2 + f Y2 ) )
G( f X ,f Y ) =
 x2 + y2
exp 2
 2W

F
 ←→
2πW 2 exp(-2π 2W 2 ( f X2 + f Y2 ) )

Page 7 of 12
Michael Pasqual
2.710 – Optics
Homework #4
April 2, 2014
Part (c)
Scaling Theorem for Circular Symmetry:
F
g(ar) ←→
1  ρ
G 
a2  a 
Eq. 2-34 in Goodman
where r = x 2 + y 2 and ρ =
F
circ(r) ←→
FT Pair:
f X2 + f Y2
J 1 ( 2πρ)
ρ
Eq. 2-35 in Goodman
where r = x 2 + y 2 , ρ =
kind, order 1
f X2 + f Y2 , and J1 is a Bessel function of the 1st
Apply
 x2 + y2 

g(x,y ) = circ


W


 r 
g(r) = circ 
 W 
g (r ) = circ(ar )
G( ρ) =
1  ρ
G 
a2  a 
1
G ( ρ) = 2
a
ρ
J 1 ( 2π )
a
ρ
where r = x 2 + y
2
where a =
1
W
Scaling Theorem
FT Pair
a
ρ
J 1 (2π )
1
a
G( ρ) =
a
ρ
J ( 2πWρ)
G( ρ) =
W 1
ρ
 x2 + y2
circ

W

(

J 1 2πW f X2 + f Y2
J 1 ( 2πWρ)
F
 ←→
W
=W

ρ
f X2 + f Y2

Page 8 of 12
)
Michael Pasqual
2.710 – Optics
Homework #4
April 2, 2014
Part (d)
Separability Theorem:
Scaling Theorem:
Convolution Theorem:
FT Pair:
F
g(x) ⋅ h( y) ←→
G( f X ) ⋅ H ( f Y ) Eq. 2-21 in Goodman
1  fX 
F
Eq. 2-12 in Goodman
g(ax) ←→
G

a  a 
F
g(x, y) ⊗ h(x, y) ←→
G( f X , f Y ) ⋅ H ( f X , f Y ) Eq. 2-15
F
sinc(x) ←→
rect( f X )
F
comb(x) ←→ comb( f X )
Table 2.1 in Goodman
Table 2.1 in Goodman
Apply
x
 y
g(x,y ) = sinc(x) ⋅ comb( y) ⊗ comb  ⋅ sinc 
2
 4 
g(x,y ) = h(x)k( y) ⊗ k(ax)h(by)
G( f X ,f Y ) = H ( f X ) ⋅ K( f Y ) ⋅
1  fX
K
a  a
where a =
1
1
, b = , h(x) = sinc(x), k(x) = comb(x)
2
4
 1 f 
⋅ H Y 
 b  b 
1
f 
f 
comb X  ⋅ rect  Y 
ab
 a 
 b 
G( f X ,f Y ) = 8 ⋅ rect( f X ) ⋅ comb( f Y ) ⋅ comb(2 f X ) ⋅ rect (4 f Y )
G( f X ,f Y ) = rect( f X ) ⋅ comb( f Y ) ⋅
Convolution & Scaling Theorems
FT Pair
 x
 y
F
sinc(x) ⋅ comb( y) ⊗ comb  ⋅ sinc  ←→
8 ⋅ rect( f X ) ⋅ comb( fY ) ⋅ comb(2 f X ) ⋅ rect (4 fY )
2
4
Page 9 of 12
Michael Pasqual
2.710 – Optics
Homework #4
April 2, 2014
Part (e)
Linearity Theorem:
F
g(x, y) + h(x, y) ←→
G( f X , f Y ) + H ( f X , f Y ) page 8 in Goodman
Shift Theorem:
F
g(x − a, y − b) ←→
G( f X , f Y ) ⋅ exp(− j2π (af X + bf Y ) )
FT Pair:
F
circ(r) ←→
J 1 ( 2πρ)
ρ
Eq. 2-13
Eq. 2-35 in Goodman
where r = x 2 + y 2 and ρ =
1st kind, order 1
f X2 + f Y2 and J1 is a Bessel function of the
Apply
g ( x, y ) = circ( ( x − a ) 2 + y 2 ) + circ( ( x + a ) 2 + y 2 )
where h( x, y ) = circ( x 2 + y 2 )
g ( x, y ) = h ( x − a , y ) + h ( x + a , y )
G ( f X , f Y ) = H ( f X , f Y ) ⋅ exp( − j 2πaf X ) + H ( f X , f Y ) ⋅ exp( + j 2πaf X )
G ( f X , f Y ) = H ( f X , f Y ) ⋅ (exp( − j 2πaf X ) + exp( + j 2πaf X ) )
G ( f X , f Y ) = H ( f X , f Y ) ⋅ (cos( −2πaf X ) + j sin( −2πaf X ) + cos(2πaf X ) + j sin(2πaf X ) )
G ( f X , f Y ) = H ( f X , f Y ) ⋅ (cos(2πaf X ) − j sin(2πaf X ) + cos(2πaf X ) + j sin(2πaf X ) )
G ( f X , f Y ) = H ( f X , f Y ) ⋅ 2 cos(2πaf X )
G( f X , f Y ) = 2
J 1 ( 2πρ)
⋅ cos(2πaf X )
ρ
circ( ( x − a ) 2 + y 2 ) + circ( ( x + a ) 2 + y 2 )
F
←→
2
(
)
J 1 2π f X2 + f Y2
J 1 ( 2πρ)
cos(2πaf X ) = 2
cos(2πaf X )
ρ
f X2 + f Y2
Page 10 of 12
Michael Pasqual
2.710 – Optics
Homework #4
April 2, 2014
Problem 4. Fourier Transform with MIT Seal
Part (a)
Matlab Commands
g
G
magG
argG
=
=
=
=
imread('MIT Seal','png');
fftshift(fft2(g));
abs(G);
atan2(imag(G),real(G));
g(x,y)
%g(x,y)
%G(u,v)
%|G(u,v)|
%arg(G(u,v))
log10(|G(u,v)|)
arg(G(u,v))
Part (b)
Matlab Commands
G_MagOnly
g_MagOnly
G_ArgOnly
g_ArgOnly
=
=
=
=
magG;
ifft2(ifftshift(G_MagOnly));
exp(sqrt(-1)*argG);
ifft2(ifftshift(G_ArgOnly));
%|G(u,v)|
%F-1{|G(u,v)|}
%exp(j·arg(G(u,v)))
%F-1{exp(j·arg(G(u,v)))}
log10(F-1{|G(u,v)|})
F-1{exp(j·arg(G(u,v))}
The image recovered from the phase information is closer to the original image, because the
phase information determines the spatial location of frequencies in the image.
Page 11 of 12
Michael Pasqual
2.710 – Optics
Homework #4
April 2, 2014
Part (c)
Here, we applied a low-pass filter (LPF) to the image. The result, g1(x,y), is a blurry image
containing low frequencies only.
Matlab Commands
N
mid
S
mid_inds
T
T(mid_inds,mid_inds)
g1
=
=
=
=
=
=
=
size(g,1);
ceil(N/2);
5;
(0:S-1)+mid-floor(S/2);
zeros(size(g));
1;
ifft2(ifftshift(G.*T));
%size of image
%middle index of image
%filter size
%indices of filter
%initialize T
%T(u,v)  Low-Pass Filter
%g1(x,y) = F-1{G(u,v)T(u,v)}
T(u,v)
g1(x,y)
Low Pass Filter (LPF)
Filtered Image
Part (d)
Here, we applied a high-pass filter (HPF) to the image. The result, g2(x,y), is a more detailed
image with high frequency edges and features.
Matlab Commands
H = 1 – T;
g2 = ifft2(ifftshift(G.*H));
%H(u,v)  High-Pass Filter
%g2(x,y) = F-1{G(u,v)H(u,v)}
H(u,v)
g2(x,y)
High Pass Filter (HPF)
Filtered Image
Page 12 of 12
MIT OpenCourseWare
http://ocw.mit.edu
2.71 / 2.710 Optics
Spring 2014
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.