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1051-455-20073 Homework #6 Due 8 May 2008 (Th)
- “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
- 212 × 1014 Hz
= 3. 113 × 10
− 15 s
coherence length c ≡ c · ∆t =
c
∆ν
= 9.^334 ×^10
− 7 m ∼= 93. 34 μm
- 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 ¯ ¯
- 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
- 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
- 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
- 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
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
dα
μZ (^) +∞
−∞
RECT
β
exp
− 2 πiβ ·
y
λ 0 z 1
dβ
z 1
exp
+2πi
μ z 1
λ 0
− ν 0 t
¶¸ μZ
− 1
1 · exp
− 2 πiα ·
x
λ 0 z 1
dα
μZ
− 1
1 · exp
− 2 πiβ ·
y
λ 0 z 1
dβ
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
