ECEN5645 Introduc0ontoOptoelectronics $ …ecee.colorado.edu/~ecen5645/pdfLectures/Class9.pdf · Optical Resonator! Fabry-Perot! ... Optoelectronics, Homeworkproblem$1.32$presenta0on$

ECEN  5645  Introduc0on  to  Optoelectronics  

Class  Mee0ng  9  

Fabry-­‐Perot  Interferometers  

Today’s  Topics  

•  Fabry  and  Perot  •  Problem  1.32  –  Kyle  Mize  •  The  Fabry  –  Perot  Interferometer  •  Problem  1.33  –  Tobias  Bothwell  •  Mul0ple  Bounces  in  a  Cavity  •  Problem  1.34  –  Alec  Herr  •  Examples  of  Fabry-­‐Perot  resonators    •  A  Fabry-­‐Perot  Interferometer  •  Problem  1.36  –  Isaac  Khader  •  Thin  Films  Op0cs  (TFs  as  F-­‐P  etalons  

Optical Resonator���

Fabry-Perot���Optical Cavity

This is a tunable large aperture (80 mm) etalon with two end plates that act as reflectors. The end plates have been machined to be flat to λ/110. There are three piezoelectric transducers that can tilt the end plates and hence obtain perfect alignment. (Courtesy of Light Machinery)

Problem  Set  4  Problem  1.32    

•  This  is  not  the  same  problem  as  in  the  “interna0onal”  edi0on.  What  gives?    

•  Next  is  problem  from  the  interna0onal  edi0on  

Optoelectronics  

Homework  problem  1.32  presenta0on    

By  Kyle  Mize  

Problem  

•  Coherence  Length  – A  narrow  band  pass  filter  transmits  wavelength  in  the  range  of  500  ±  0.05nm.  If  this  filter  is  placed  in  front  of  a  source  of  white  light,  what  is  the  coherence  length  of  the  transmi^ed  light?  

Ques2on  Parameters  

•  Given:  –  Frequency  Width    –  Center  Frequency    

•  Solve  for:  

•  Equa0ons  Used  –           

Solu2on  (spectral  width)  

•  Solve  for  spectral  width  

Solu2on  (coherence  2me)  

•  Solve  for  coherence  0me  – Δ𝑡≈1/∆𝑣  

– ∆𝑡= 1/(0.12∗10↑12 ) [𝑠]

– ∆𝑡=8.33  ∗10↑−12 [𝑠]  

– ∆𝑡=8.33  [𝑝𝑠]  

Solu2on  (Coherence  Length)  •  Solve  for  Coherence  Length  

–  [s]  

•  The  coherence  length  of  the  transmi^ed  light  is  2.5  mm    

Optical Resonator

Fabry-Perot Optical Cavity

Schematic illustration of the Fabry-Perot optical cavity and its properties. (a) Reflected waves interfere. (b) Only standing EM waves, modes, of certain

wavelengths are allowed in the cavity. (c) Intensity vs. frequency for various modes. R is mirror reflectance and lower R means higher loss from the cavity.

Note: The two curves are sketched so that the maximum intensity is unity

Each  allowed  EM  oscilla0on  is  a  cavity  mode  

Optical Resonator

Fabry-Perot Optical Cavity

Problem  Set  4  Problem  1.33  

•  Solu0on  by  Tobias  Bothwell  

Problem 1.33

Part a

�0 =c⌫0

��0 = �c�⌫0⌫20

Substituting �0 =c⌫0

and converting to �

�� =

�⌫�0⌫0

=

�⌫�20c .

We can rearrange to get that

1�⌫ =

�20��c = �t. Then

lc = c�t = �20��.

Part B

�0 = 632.8nm �⌫ = 1.5GHz

�� = 2.002 ⇤ 10�12m�t = 2

3 ⇤ 10�9s

lc = .2m

1

Optical Resonator Fabry-Perot���Optical Cavity

A + B = A + Ar 2exp(-j2kL)

Ecavity = A + B + … = A + Ar2exp(-j2kL) + Ar4exp(-j4kL) + Ar6exp(-j6kL) + …

Maxima at kmL = mπ

)2exp(1 2cavity kLjAE−−

=r

)(sin4)1( 22cavity kLII o

RR +−= 2max )1( R−

= oII

m = 1,2,3,…integer

θ

sinθ

π2π

3ππ/2

Optical Resonator Fabry-Perot Optical Cavity

Maxima at kmL = mπ

)(sin4)1( 22cavity kLII o

RR +−= 2max )1( R−

= oII

m = 1,2,3,…integer

m(λm/2) = L

(2π/λm)L = mπ

υm = m(c/2L) = mυf = Mode frequency

m = integer, 1,2,…

υf =free spectral range = c/2L = Separation of modes

€

δυm =υ f

F

€

F =πR 1/ 2

1−RF = Finesse���R = Reflectance (R > 0.6)

Fused silica etalon (Courtesy of Light Machinery)  

A 10 GHz air spaced etalon with 3 zerodur spacers. (Courtesy of Light Machinery)  

Fabry-Perot etalons can be made to operate from UV to IR wavelengths with optical cavity spacings from a few microns to many centimeters (Courtesy of IC Optical Systems Ltd.)  

Quality factor Q is similar to the Finesse F

mFQm

m ===δυυ

widthSpectralfrequencyResonant

Optical Resonator is also an optical filter

Only certain wavelengths (cavity modes) are transmitted

)(sin4)1()1(

22

2

incidentdtransmitte kLII

RRR

+−−

=

Piezoelectric transducer controlled Fabry-Perot etalons. Left has a 70 mm and the right has 50 mm clear aperture. The piezoelectric controller maintains the reflecting plates parallel while the cavity separation is scanned. (The left etalon has a reflection of interference fringes that are on the adjacent computer display) (Courtesy of IC Optical Systems Ltd.)  

A scanning Fabry-Perot interferometer (Model SA200), used as a spectrum analyzer, that has a free spectral range of 1.5 GHz, a typical finesse of 250, spectral width (resolution) of 7.5 MHz. The cavity length is 5 cm. It uses two concave mirrors instead of two planar mirrors to form the optical cavity. A piezoelectric transducer is used to change the cavity length and hence the resonant frequencies. A voltage ramp is applied through the coaxial cable to the piezoelectric transducer to scan frequencies. (Courtesy of Thorlabs)  

Problem  Set  4  Problem  1.34  

•  Solu0on  by  Alec  Herr  

Fabry  Perot  Examples  

•  Op0cal  Resonator  in  air  •  Op0cal  Resonator  in  semiconductor  

Example: An Optical Resonator in Air Consider a Fabry-Perot optical cavity in air of length 100 microns with mirrors that have a reflectance of 0.90. Calculate the cavity mode nearest to the wavelength 900 nm, and corresponding wavelength. Calculate the separation of the modes, the finesse, the spectral width of each mode and the Q-factor

Thus, m = 222 (must be an integer)

λm = 900.90 nm ≈ 900 nm (very close)

Solution

2.222)10900()10100(22

9

6

=××

== −

−

λLm nm9.900

)222()10100(22 6

=×

==−

mL

mλ

The frequency corresponding to λm is

υm = c/λm = (3×108)/(900.9×10-9) = 3.33×1014 Hz

Find the mode number m corresponding to 900 nm and then take the integer

Example: An Optical Resonator in Air

υf = c/2L = separation of modes

= (3×108) / [2(100×10-6)] = 1.5×1012 Hz.

8.2990.0190.0

1

2/12/1

=−

=−

=ππ

RRF GHz 3.50

8.29105.1 12

=×

==Ff

m

υδυ

nm 136.0)1003.5()1033.3(

)103( 10214

8

2 =×××

=−=⎟⎟⎠

⎞⎜⎜⎝

⎛= m

mmm

ccδυ

υυδδλ

Solution: Continued

The Q-factor is

Q = mF = (222)(29.8) = 6.6×103

Example: Semiconductor Optical Cavity Consider a Fabry-Perot optical cavity of a semiconductor material of length 250 microns with mirrors, each with a reflectance of 0.90. Calculate the cavity mode nearest to 1310 nm. Calculate the separation of the modes, finesse, the spectral width of each mode, and the Q-factor. Take n = 3.6 for the semiconductor medium.

Given, L = 250×10-6 m, n = 3.6, R = 0.90

Δυm = υf = c/2nL = Separation of modes = 1.67×1011 Hz

€

F =πR 1/2

1−R =π0.91/2

1− 0.9= 29.8

GHz 59.58.291067.1 11

=×

==Ff

m

υδυ

Solution

05.1374)101310()10250)(6.3(22

9

6

=××

== −

−

λnLm

Mode number m corresponding to 1310 nm is

which  must  be  an  integer  (1374)  so  that  the  actual  mode  wavelength  is  

nm04.1310)1374(

)10250)(6.3(22 6

=×

==−

mnL

mλ

For  all  prac2cal  purposes  the  mode  wavelength  is  1310  nm  

Mode  frequency  is  

Hz103.2)101310(

)103( 149

8

×=××

== −m

mcλ

υ

Solution: Continued Example: Semiconductor Optical Cavity

nm 136.0)1003.5()1033.3(

)103( 10214

8

2 =×××

=−=⎟⎟⎠

⎞⎜⎜⎝

⎛= m

mmm

ccδυ

υυδδλ

Spectral  width  of  a  mode  in  wavelength  is  

The Q-factor is

Q = mF = (1374)(29.8) = 4.1×104

Solution: Continued Example: Semiconductor Optical Cavity

Fabry-Perot Interferometer

Bright  rings  

2nLcosθ  =  mλ;  m  =  1, 2, 3 ...  

Oblique Incidence on a Fabry-Perot Cavity Assume external reflection within the cavity. The point P on wave A propagates to the right reflector at Q where it is reflected. At Q, the wave experiences a phase change π. Then it propagates to R, and gets reflected again and experiences another π phase change.

Right after reflection, the phase at R must be the same as that at the start point P; R and P are on the same wavefront.

Δφ  =  Phase  difference  from  P  to  Q  to  R  =  m(2π)  

Some  Shop  Tes0ng  Examples  

•  Classic  text  for  applica0ons  and  pictures  of  results,  aberra0ons,  etc.  Now  online…  

Using  an  FP  for  surface  evalua0on  

•  Results  are  not  so  picturesque  

F-­‐P  flatness  test    

Problem  Set  4  Problem  1.36  

•  Solu0on  by  Isaac  Khader  

Kasap 1.36: Ruby Fabry-PerotR1 = 0.99

R2 = 0.95

L = 0.1 m

n = 1.78Fig. 1.32

Allowed standing waves in cavity: Near 694.3 nm:m = 512747

⌫m =c

�m=

mc

2nL

m

✓�

2

◆= nL

(m=1,2,3...)=) �m =

2Ln

m

�� = �m+1 � �m ⇡ 1.354 pm

�⌫ = ⌫m+1 � ⌫m ⇡ 8.427 GHz

�m = 694.299528 nm

Mode spacing:

Finesse and Q factor:

Q = mF = 5.25 ⇤ 107F ⇡ ⇡(R1R2)1/4

1�pR1R2

= 102.42

Thin Films Optics

Assume normal incidence φ  =  2×(2π/λ)n2d  

2121

21121 rrr −=

+−

==nnnn

32

32232 nn

nn+−

== rr

21

1121

2nnn+

== tt21

2211

2nnn+

==ʹ′ tt

32

2232

2nnn+

== tt

2

3

1

1-  t1tʹ′1  =  r12    

Thin Films Optics

φ  =  2×(2π/λ)n2d  Assume normal incidence

Areflected/A0  =  r1  +  t1tʹ′1r2e-jφ - t1tʹ′1r1r22e-j2φ

+ t1tʹ′1r12r23e-j3φ + …  

Areflected    =  A1  +  A2  +  A3  +  A4  +  …  

φ = 2×(2π/λ)n2d

Corrected

Thin Films Optics

t1 = t12 =2n1

n1 + n2t 2 = t 21 =

2n2n1 + n2

33 23

2 3

2nt tn n

= =+

φ

φ

j

j

ee−

−

++

=21

21

1 rrrrr

φ

φ

j

j

ee

−

−

+=

21

2/21

1 rrttt

Corrected

φ

φ

j

j

ee−

−

++

=21

21

1 rrrrr

Reflection Coefficient

r = 0

d = mλ4n2

exp(-jφ) = -1

r1 = r2

Choosen2 = (n1n3)1/2

∴ r1 = r2

n2 = (n1n3)1/2

φ  =  2×(2π/λ)n2d  =  mπ  m  is  an  odd  integer  

Corrected

φ

φ

j

j

ee

−

−

+=

21

2/21

1 rrttt

Transmission Coefficient

t = Maximum d = mλ

4n2

exp(-jφ) = -1

φ  =  2×(2π/λ)n2d  =  mπ  m  is  an  odd  integer  

Corrected

Minimum and Maximum Reflectance 2

3122

3122

min ⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

=nnnnnnR

2

13

13max ⎟⎟

⎠

⎞⎜⎜⎝

⎛

+−

=nnnnRn1 < n2 < n3

n1  <  n3  <  n2  then  Rmin  and  Rmax  equa0ons  are  interchanged  

While  Rmax  appears  to  be  independent  from  n2,  the  index  n2  is  nonetheless  s0ll  involved  in  determining  maximum  reflec0on  inasmuch  as  R  reaches  Rmax  when  φ  =  2(2π/λ)n2d    =  π(2m);  when    φ = π×(even  number)    

Reflectance and Transmittance of a Thin Film Coating

(a)  Reflectance  R  and  transmi^ance  T  vs.  φ  =  2×(2π/λ)n2d,  for  a  thin  film  on  a  substrate  where  n1  =  1  (air),  n2  =  2.5  and  n3  =  3.5,  and    n1  <  n2  <  n3.    (b)  R  and  T vs.  φ  for  a  thin  film  on  a  substrate  where  n1  =  1  (air),  n2  =  3.5  and  n3  =  2.5,  and  n2  >  n3  >    n1  

Corrected

EXAMPLE: Transmission spectra through a thin film (a-Se) on a glass substrate

Absorption region

Thin film interference fringes

Substrate

Example:  Thin  Film  Op0cs  Consider  a  semiconductor  device  with  n3  =  3.5  that  has  been  coated  with  a  transparent  op0cal  film  (a  dielectric  film)  with  n2  =  2.5,  n1  =  1  (air).  If  the  film  thickness  is  160  nm,  find  the  minimum  and  maximum  reflectances  and  transmi^ances  and  their  corresponding  wavelengths  in  the  visible  range.  (Assume  normal  incidence.)  

Solu0on:  We  have  n1  <  n2  <  n3.  Rmin  occurs  at  φ =  π  or  odd  mul0ple  of  π  ,  and  maximum  reflectance  Rmax  at  φ =  2π  or  an  integer  mul0ple  of  2π  .    

%0.8or080.0)5.3)(1(5.2)5.3)(1(5.22

2

22

3122

3122

min =⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

=⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

=nnnnnnR

%31or31.015.315.3 22

13

13max =⎟

⎠

⎞⎜⎝

⎛+−

=⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

=nnnnR

Tmax  =  1  –  Rmin  =  0.92  or  92%    

Tmin  =  1  –  Rmax  =  0.69  or  69%  

Multiple Reflections in Plates and Incoherent Waves

Tplate  =  (1-R)2      +  R2(1 - R)2        +  R4(1 - R)2          +  …    

 Tplate  =  (1 - R)2[1  +  R2  +  R4  +  …]  

2

2

plate 1)1(

RRT

−−

= 22

21

21plate

4nnnn+

=T22

21

221

plate)(nnnn+−

=R