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Optical Coherence Length and Linewidth Calculation for Different Light Sources, Assignments of Typography

The calculations for the coherence length and linewidth of various light sources, including a sodium arc and a he:ne 'greenie' laser. The document also includes the calculation for laser light with a coherence length of 10 km.

Typology: Assignments

2009/2010

Uploaded on 03/28/2010

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1051-455-20073 Homework #6 Due 8 May 2008 (Th)
1. “White” light includes equal “amounts” of each wavelength in the interval 400 nm λ700 nm.
(a) Determine the frequency bandwidth for this wavelength range
ν|ν1ν2|=¯¯¯¯
c
λ1
c
λ2¯¯¯¯
=c¯¯¯¯
1
λ1
1
λ2¯¯¯¯
=2.99792458 ×108ms
1·¯¯¯¯
1
400 nm 1
700 nm ¯¯¯¯
ν
=3.212 ×1014 Hz
(b) Compute the associated coherence time and coherence length of white light.
coherence time t1
ν
=1
3.212 ×1014 Hz
=3.113 ×1015 s
cohe rence l eng th c·t=c
ν
=9.334 ×107m
=93.34 μm
2. The range of angular temporal frequencies by a light source is ω:
(a) Find an expression for ν
ω=2πν =ω=2πν=ν=ω
2π
(b) Derive an expression for the linewidth λ
λν =c=c=0=λ·ν+ν·λ
=
ν
ν=
λ
λ=λ=λ
νν=c
ν2ν=λ
(c) Use the result of (a) to find an expression for the coherence length of the source.
=c
ν=2πc
ω
(d) If the light source is a sodium arc that emits two narrow spectral lines with ω1=3.195×1015 radians
sec
and ω2=3.198 ×1015 radians
sec ,find the coherence length.
=2πc
ω=2π2.99792458 ×108ms
1
¯¯3.195 ×1015 radians
sec 3.198 ×1015 radians
sec ¯¯
=+62.8mm
(e) If the light source is a He:Ne “greenie” laser with ω1=3.171 ×1015 radians
sec and ω2=3.469 ×
1015 radians
sec ,find the coherence length.
=2πc
ω=2π2.99792458 ×108ms
1
|3.171 ×1015 s13.469 ×1015 s1|
=+6.32 μmmuc h shorter because λis much larger
1
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1051-455-20073 Homework #6 Due 8 May 2008 (Th)

  1. “White” light includes equal “amounts” of each wavelength in the interval 400 nm ≤ λ ≤ 700 nm.

(a) Determine the frequency bandwidth for this wavelength range

∆ν ≡ |ν 1 − ν 2 | =

c

λ 1

c

λ 2

= c

λ 1

λ 2

= 2.^99792458 ×^10

8 m s

− 1 ·

400 nm

700 nm

∆ν

= 3.^212 ×^10

14 Hz

(b) Compute the associated coherence time and coherence length of white light.

coherence time ∆t ≡

∆ν

  1. 212 × 1014 Hz

= 3. 113 × 10

− 15 s

coherence length c ≡ c · ∆t =

c

∆ν

= 9.^334 ×^10

− 7 m ∼= 93. 34 μm

  1. The range of angular temporal frequencies by a light source is ∆ω:

(a) Find an expression for ∆ν

ω = 2πν =⇒ ∆ω = 2π∆ν =⇒ ∆ν =

∆ω

2 π

(b) Derive an expression for the linewidth ∆λ

λν = c =⇒ ∆c = 0 = λ · ∆ν + ν · ∆λ

∆ν

ν

∆λ

λ

=⇒ ∆λ = −

λ

ν

∆ν = −

c

ν 2

∆ν = ∆λ

(c) Use the result of (a) to find an expression for the coherence length of the source.

c =

c

∆ν

2 πc

∆ω

(d) If the light source is a sodium arc that emits two narrow spectral lines with ω 1 = 3. 195 × 10

15 radians sec

and ω 2 = 3. 198 × 10

15 radians sec , find the coherence length.

c =

2 πc

∆ω

= 2π

2. 99792458 × 10

8 m s

− 1 ¯ ¯

  1. 195 × 10 15 radians sec

− 3. 198 × 10

15 radians sec

c

= +62.8 mm

(e) If the light source is a He:Ne “greenie” laser with ω 1 = 3. 171 × 10

15 radians sec and ω 2 = 3. 469 ×

15 radians sec , find the coherence length.

c =

2 πc

∆ω

= 2π

2. 99792458 × 10

8 m s − 1

| 3. 171 × 10

15 s − 1 − 3. 469 × 10 15 s − 1 |

c

= +6. 32 μm much shorter because ∆λ is much larger

  1. Determine the linewidth in nanometers and in Hertz for laser light whose coherence length is 10 km if

the mean wavelength is 632 .8 nm (He:Ne)

c =

c

∆ν

=⇒ ∆ν =

c

c

c = 10 km =⇒ ∆ν

2. 99792458 × 10

8 m s − 1

10 km

= 29 .979 kHz

= 3 × 10

4 Hz

= ∆ν

∆ν =

∆λ =

c

ν 2 0

∆ν

c ³ c λ 0

´ 2 ·^ ∆ν^ =^

λ

2 0

c

· ∆ν

(632.8 nm)

2

2. 99792458 × 10

8 m s − 1

· 29 .979 kHz

∆λ ∼= 4. 00 × 10 − 17 m = 4. 00 × 10 − 8 nm

  1. Michelson found that the cadmium red line (λ 0 = 643.8 nm) was the best available light source for

his interference experiment. With it, he could see fringes for optical path differences up to 300 mm.

Estimate the linewidth ∆λ and coherence time ∆t of this light source.

OP D = 300 mm = c = c · ∆t =

c

∆ν

=⇒ ∆ν =

c

c

2. 99792458 × 10

8 m s

− 1

300 mm

9 Hz = 1000 MHz

∆t =

∆ν

− 9 s = 1 ns = ∆t

∆λ =

λ

2 0

c

· ∆ν =

(643.8 nm)

2

2. 99792458 × 10

8 m s − 1

9 Hz

= 1. 382 × 10

− 12 m

∆λ = 1. 382 × 10

− 3 nm

  1. The light diffracted by an object of the form f [x, y] and observed at a distance z 1 in the Fraunhofer

diffraction region has the “shape” of the squared magnitude of the Fourier transform of the object after

appropriate rescaling of the coordinates back to the space domain

g [x, y] ∝ |F [ξ, η]|

2

ξ→ x λ 0 z 1 ,η→^

y λ 0 z 1

where the 2-D Fourier transform is defined:

F [ξ, η] ≡ F 2 {f [x, y]} ≡

Z Z +∞

−∞

f [x, y] exp [− 2 πi (ξx + ηy)] dx dy

The object f [x, y] satisfies the following conditions:

f [x, y] = 1 if |x| ≤ 1 AND |y| ≤ 1

f [x, y] = 0 if otherwise

(a) Sketch f [x, y];

This is a rectangle function of width 2 units in each direction

(b) Calculate the diffraction pattern in the Fraunhofer diffraction region if f [x, y] is illuminated by

light with wavelength λ 0 ;

The observation plane is in the Fraunhofer diffraction region, so the output is approximately:

g [x, y]

z 1

exp

+2πi

μ z 1

λ 0

− ν 0 t

¶¸+Z Z∞

−∞

RECT

α

β

exp

− 2 πi

μ

α ·

x

λ 0 z 1

  • β ·

y

λ 0 z 1

dα dβ

z 1

exp

+2πi

μ z 1

λ 0

− ν 0 t

¶¸ μZ +∞

−∞

RECT

h α

i

exp

− 2 πiα ·

x

λ 0 z 1

μZ (^) +∞

−∞

RECT

β

exp

− 2 πiβ ·

y

λ 0 z 1

z 1

exp

+2πi

μ z 1

λ 0

− ν 0 t

¶¸ μZ

− 1

1 · exp

− 2 πiα ·

x

λ 0 z 1

μZ

− 1

1 · exp

− 2 πiβ ·

y

λ 0 z 1

z 1

exp

+2πi

μ z 1

λ 0

− ν 0 t

Ã

− 2 πi

x λ 0 z 1

exp

− 2 πiα ·

x

λ 0 z 1

α=+

α=− 1

Ã

− 2 πi

y λ 0 z 1

exp

− 2 πiβ ·

y

λ 0 z 1

β=+

β=− 1

z 1

exp

+2πi

μ z 1

λ 0

− ν 0 t

Ã

− 2 πi x λ 0 z 1

μ

exp

− 2 πi ·

x

λ 0 z 1

− exp

+2πi ·

x

λ 0 z 1

Ã

− 2 πi

y λ 0 z 1

μ

exp

− 2 πi ·

y

λ 0 z 1

− exp

+2πi ·

y

λ 0 z 1

z 1

exp

+2πi

μ z 1

λ 0

− ν 0 t

Ã

πx λ 0 z 1

μ

sin

+2π

x

λ 0 z 1

! Ã

πy λ 0 z 1

μ

sin

+2π

y

λ 0 z 1

z 1

exp

+2πi

μ z 1

λ 0

− ν 0 t

Ã

2 πx λ 0 z 1

μ

sin

+2π

x

λ 0 z 1

! Ã

2 πy λ 0 z 1

μ

sin

+2π

y

λ 0 z 1

z 1

exp

+2πi

μ z 1

λ 0

− ν 0 t

Ã

SIN C

x ¡ λ 0 z 1 2

#! Ã

SINC

y ¡ λ 0 z 1 2

The observed irradiance is:

|g [x, y]|

|z 1 |

2

SIN C

2

x ¡ λ 0 z 1 2

y ¡ λ 0 z 1 2