a Consider light of free-space wavelength 1300 nm traveling in pure silica medium. Calculate the phase velocity and group velocity of light in this medium. Is the group velocity ever greater than the phase velocity?

b What is the Brewster angle (the polarization angle θp) and the critical angle (θc) for total internal reflection when the light wave traveling in this silica medium is incident on a silica/air interface. What happens at the polarization angle?

c What is the reflection coefficient and reflectance at normal incidence when the light beam traveling in the silica medium is incident on a silica/air interface?

d What is the reflection coefficient and reflectance at normal incidence when a light beam traveling in air is incident on an air/silica interface? How do these compare with part (c) and what is your conclusion?

Solution

a From n and Ng vs. λ curves, at λ = 1300 nm

n = 1.447 and Ng = 1.462

Phase velocity : v = c/n = (3×108 m s-1) / 1.447 = 2.073×108 m s-1

Group velocity: vg = c/Ng = (3×108 m s-1) / 1.462 = 2.052×108 m s-1

For glasses, dn/dλ is negative so that Ng > n and hence vg < v. Note that vg > v in a medium that will have a positive dn/dλ. For example, PbS, PbTe, PbSe in the region λ = 1 − 3.5 μm.

b The polarization (the Brewster) angle is

θp = arctan(n2/n1) = arctan(1/ 1.447) = 34. 65°

At this angle of incidence, r// = 0, the reflected wave has an E-field only perpendicular to the plane of incidence.

The critical angle for TIR is,

θc = arcsin(n2/n1) = arcsin(1/ 1.447) = 43.72°

c For light traveling in glass incident on the glass-air interface at normal incidence,

r = r // = r⊥ =n1 − n2n1 + n2

=1.447 −11.447 +1

= 0.183

Thus, R = r2 = (0.183)2 = 0.0335

d For light traveling in air incident on the air-glass interface at normal incidence,

r = r // = r⊥ =n1 − n2n1 + n2

=1−1.4471+1.447

= −0.183

i.e. R = r2 = (−0.183)2 = 0.0335

There is a 180° phase change as r is negative. Notice that in both cases the amount of reflection (3.35%) is the same.

_____________________________________________________________________ 1.3 The Sellmeier dispersion coefficient for pure silica (SiO2) and 86.5%SiO2-13.5 mol.% GeO2 re given in Table 1.2 Write a program on your computer or calculator, or use a math software package or even a spread sheet program (e.g. Excel) to obtain the refractive index n as a function of λ from 0.5 μm to 1.8 μm for both pure silica and 86.5%SiO2-13.5%GeO2. Obtain the group index, Ng, vs. wavelength for both materials and plot it on the same graph. Find the wavelength at which the material dispersion becomes zero in each material.

Solution

Excel program to plot n and differentiate and find Ng

Figure 1Q8-1 Refractive index n and the group index Ng of pure SiO2 (silica) glass as a function of wavelength (Excel). The minimum in Ng is around 1.3 μm. Note that the smooth line option used in

Excel to pass a continuous smooth line through the data points. Data points are exactly on the line and are not shown for clarity.

Refractive index n and the group index Ng of 86.5%SiO213.5%GeO as a function of wavelength

(Excel). The minimum in Ng is around 1.4 μm. Note that the smooth line option used in Excel to pass a continuous smooth line through the data points. Data points are exactly on the line and are not shown

for clarity. Material dispersion is proportional to derivative of group velocity over wavelength. The corresponding values are close to 1.3 and 1.4 μm. 1.15 (a) Suppose that frequency spectrum of a radiation emitted from a source has a central frequency υo and a spectral width Δυ. The spectrum of this radiation in terms of wavelength will have a central wavelength λo and a spectral width Δλ. Clearly, λo = c/υo. Since Δλ << λo and Δυ << υo, using λ = c/υ, show that the line width Δλ and hence the coherence length lc are

co

o

o2λ

υυλ

υλ Δ=Δ=Δ and λ

λΔ

=Δ=2o

c tcl

(b) Calculate Δλ for a lasing emission from a He-Ne laser that has λo = 632.8 nm and Δυ ≈ 1.5 GHz. Find its coherence time and length.

Solution (a) See Example 1.9.3 (b) Consider the width in wavelength, Δλ = Δυ( λ2/c) = (1.5×109 s-1 (632/8×10-9 m)2/(3×108 m s-1) = 3.16×10-6 m. The coherence time is Δt ≈ 1/ Δυ = 1/(1.5×109 Hz) = 0.666×10-9 s The coherence length is lc = cΔt = (3×108 m s-1)(0.666×10-9 s) = 0.20 m = 20 cm 2.2 Consider two parallel rays 1 and 2 interfering in the guide as in Figure 2.52. Given the phase difference

)()( mmm maymy φππ +−=Φ=Φ

between the waves at C, distance y above the guide center, find the electric field pattern E (y) in the guide. Recall that the field at C can be written as

)](cos[)cos()( ytAtAyE mΦ++= ωω . Plot the field pattern for the first three modes taking a planar dielectric guide with a core thickness 20 μm, n1 = 1.455 n2 = 1.440, light wavelength of 1.3 μm.

Rays 1 and 2 are initially in phase as they belong to the same wavefront. Ray 1 experiences total internal

reflection at A. 1 and 2 interfere at C. There is a phase difference between the two Solution The two waves interfering at C are out phase by Φ,

)](cos[)cos()( ytAtAyE mΦ++= ωω

where A is an arbitrary amplitude. Thus,

⎥⎦⎤

⎢⎣⎡ Φ⎥⎦

⎤⎢⎣⎡ Φ+= )(

21cos)(

21cos2 yytAE mmω

or ( )Φ′+⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡ Φ= tyAE m ωcos)(21cos2 = Eocos(ωt + Φ′)

in which Φ′ = Φm/2, and cos(ωt + Φ′) is the time dependent part that represents the wave phenomenon, and the curly brackets contain the effective amplitude. Thus, the amplitude Eo is

⎥⎦⎤

⎢⎣⎡ +−= )(

22cos2 mo m

aymAE φππ

To plot Eo as a function of y, we need to find φm for m = 0, 1 and 2 , the first three modes. From Example 2.1.1 in the textbook, the waveguide condition is

mm mka φπθ =−cos)2( 1

we can now substitute for φm which has different forms for TE and TM waves to find,

TE waves )(cos

sin

2costan

2/12

1

22

1 mTEm

m

m fnn

mak θθ

θπθ =

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎠⎞

⎜⎝⎛ −

TM waves )(

cos

sin

2costan 2

1

2

2/12

1

22

1 mTM

m

m

m f

nn

nn

mak θ

θ

θπθ =

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎠⎞

⎜⎝⎛ −

The above two equations can be solved graphically as in Example 2.1.1 to find θm for each choice of m. Alternatively one can use a computer program for finding the roots of a function. The above equations are functions of θm only for each m. Using a = 10 μm, λ = 1.3 μm, n1 = 1.455 n2 = 1.440, the results are: TE Modes m = 0 m = 1 m = 2

θm (degrees) 88.84 87.67 86.51

φm (degrees) 163.75 147.02 129.69

TM Modes m = 0 m = 1 m = 2

θm (degrees) 88.84 87.67 86.51

φm (degrees) 164.08 147.66 130.60 There is no significant difference between the TE and TM modes (the reason is that n1 and n2 are very close).

Field distribution across the core of a planar dielectric waveguide

We can set A = 1 and plot Eo vs. y using

⎥⎦⎤

⎢⎣⎡ +−= )(

22cos2 mo m

aymE φππ

with the φm and m values in the table above. This is shown in Figure 2Q4-1.

2.7 Consider a planar dielectric waveguide with a core thickness 10 μm, n1 = 1.4446, n2 = 1.4440. Calculate the V-number, the mode angle θm for m = 0 (use a graphical solution, if necessary), penetration depth, and mode field width, MFW = 2a + 2δ, for light wavelengths of 1.0 μm and 1.5 μm. What is your conclusion? Compare your MFW calculation with 2wo = 2a(V+1)/V. The mode angle θ0, is given as θ0 = 88.85° for λ = 1 μm and θ0 = 88.72° for λ = 1.5 μm for the fundamental mode m = 0.

Solution

λ = 1 μm, n1 = 1.4446, n2 = 1.4440, a = 5 μm. Apply

( ) 2/122

21

2 nnaV −=λπ

to obtain V = 1.3079

Solve the waveguide condition

)(cos

sin

2costan

2/12

1

22

1 mm

m

m fnn

mak θθ

θπθ =

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎠⎞

⎜⎝⎛ −

graphically as in Example 2.1.1 to find: θc = 88.35° and the mode angle (for m = 0) is θo = 88.85°.

Then use

λ

θπ

αδ

2/1

22

2

12 1sin2

1 ⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛

==m

mm

nnn

to calculate the penetration depth:

δ = 1/α = 5.33 μm.

∴ MFW = 2a + 2δ = 20.65 μm

We can also calculate MFW from

MFW = 2a(V+1)/V = 2(5 μm)(1.3079+1)/(1.3079) = 17.6 μm (Difference = 15%)

λ = 1.5 μm, V = 0.872, single mode. Solve waveguide condition graphically that the mode angle is θo = 88.72°.

δ = 1/α = 9.08 μm.

∴ MFW = 2a + 2δ = 28.15 μm.

Compare with MFW = 2a(V+1)/V = 2(5 μm)(0.872+1)/(0.872) = 21.5 μm (Difference = 24%)

Notice that the MFW from 2a(V+1)/V gets worse as V decreases. The reason for using MFW = 2a(V+1)/V, is that this equation provides a single step calculation of MFW. The calculation of the penetration depth δ requires the calculation of the incidence angle θ and φ.

Author's Note: Consider a more extreme case

λ = 5 μm, V = 0.262, single mode. Solve waveguide condition graphically to find that the mode angle is θo = 88.40°.

δ = 1/α = 77.22 μm.

∴ MFW = 2a + 2δ = 164.4 μm.

Compare with MFW = 2a(V+1)/V = 2(5 μm)(0.262 + 1)/(0.262) = 48.2 μm (Very large difference.)

2.8 Consider a multimode fiber with a core diameter of 100 μm, core refractive index of 1.4750, and a cladding refractive index of 1.4550 both at 850 nm. Consider operating this fiber at λ = 850 nm. (a) Calculate the V-number for the fiber and estimate the number of modes. (b) Calculate the wavelength beyond which the fiber becomes single mode. (c) Calculate the numerical aperture. (d) Calculate the maximum acceptance angle. (e) Calculate the modal dispersion Δτ and hence the bit rate × distance product.

Solution Given n1 = 1.475, n2 = 1.455, 2a = 100×10-6 m or a = 50 μm and λ = 0.850 μm. The V-number is,

( )μm) (0.850

)1.455μm)(1.475 (5022 1/2222/12

221

−=−=

πnnaVλπ = 89.47

Number of modes M,

247.89

2

22

==VM ≈ 4002

The fiber becomes monomode when,

( ) 405.22 2/122

21 <−= nnaV

λπ

or ( )

405.2)455.1475.1)(μm 50(2

405.22 2/1222/12

221 −

=−

>ππλ nna = 31.6 μm

For wavelengths longer than 31.6 μm, the fiber is a single mode waveguide.

The numerical aperture NA is

2/1222/122

21 )455.1475.1()( −=−= nnNA = 0.242

If αmax is the maximum acceptance angle, then,

)1/242.0arcsin(arcsinmax =⎟⎟⎠

⎞⎜⎜⎝

⎛=

onNAα = 14°

Modal dispersion is given by

1-821intermode

sm 103455.1475.1

×−

=−

=Δ

cnn

Lτ

= 66.7 ps m-1 or 67.6 ns per km

Given that σ ≈ 0.29Δτ, maximum bit-rate is

)km ns 7.66)(29.0(

25.025.025.01-

intermodetotal

=≈=σσ

LLBL

i.e. BL = 13 Mb s-1 km (only an estimate)

We neglected material dispersion at this wavelength which would further decrease BL. Material dispersion and modal dispersion must be combined by

2material

2intermode

2otal σσσ +=t

For example, assuming an LED with a spectral rms deviation σλ of about 20 nm, and a Dm ≈ −200 ps km-1 nm-1 (at about 850 nm)we would find the material dispersion as

σmaterial = −(200 ps km-1 nm-1)(20 nm)(1 km) ≈ 4000 ps km-1 or 4 ns km-1,

which is substantially smaller than the intermode dispersion and can be neglected.

________________________________________________________________________

2.9) Consider a fiber with a 86.5%SiO2-13.5%GeO2 core of diameter of 8 μm and refractive index of 1.468 and a cladding refractive index of 1.464 both refractive indices at 1300 nm where the fiber is to be operated using a laser source with a half maximum width of 2 nm. (a) Calculate the V-number for the fiber. Is this a single mode fiber? (b) Calculate the wavelength below which the fiber becomes multimode. (c) Calculate the numerical aperture. (d) Calculate the maximum acceptance angle. (e) Obtain the material dispersion and waveguide dispersion and hence estimate the bit rate × distance product (B×L) of the fiber.

Solution (a) Given n1 = 1.475, n2 = 1.455, 2a = 8×10-6 m or a = 4 μm and λ =1.3 μm. The V-number is,

( ))μm 3.1(

)464.1468.1)(μm 4(22 2/1222/12

221

−=−=

πλπ nnaV = 2.094

(b) Since V < 2.405, this is a single mode fiber. The fiber becomes multimode when

405.2)(2 2/122

21 >−= nnaV

λπ

or ( ) ( )

405.2464.1468.1μm) 4(2

405.22 2/1222/12

221 −

=−

<ππλ nna =1.13 μm

For wavelengths shorter than 1.13 μm, the fiber is a multi-mode waveguide.

(c) The numerical aperture NA is

( ) 2/1222/122

21 )464.1468.1( −=−= nnNA = 0.108

(d) If αmax is the maximum acceptance angle, then,

)1/108.0arcsin(arcsinmax =⎟⎟⎠

⎞⎜⎜⎝

⎛=

onNAα = 6.2°

so that the total acceptance angle is 12.4°.

(e) At λ =1.3 μm, from D vs. λ, Figure 2.22, Dm ≈ −7.5 ps km-1 nm-1, Dw ≈ −5 ps km-

1 nm-1.

2/12/1 λτ

Δ+=Δ

wm DDL

= |−7.5−5 ps km-1 nm-1|(2 nm) = 15 ps km-1 + 10 ps km-1

= 0.025 ns km-1

Obviously material dispersion is 15 ps km-1 and waveguide dispersion is 10 ps km-1

The maximum bit-rate distance product is then

1-2/1 km ns 025.0

59.059.0=

Δ≈

τLBL = 23.6 Gb s-1 km.

2.10) According to Question 1.3 (Ch.1), the Sellmeier dispersion equation provides n vs. λ for pure SiO2 and SiO2-13.5 mol.%GeO2. The refractive index increases linearly with addition of GeO2 to SiO2 from 0 to 13.5 mol.%. A single mode step index fiber for use at 1300 nm is required to have the following properties: NA = 0.1, core diameter of 9 μm and a core of SiO2-13.5% GeO2. What should be the cladding composition?

Solution Given

2a = 9×10-6 m or a = 4.5 μm. From Ch 1, Question 1.3, the Sellmeier equation is,

nG G G2 1

2

212

22

222

32

2321− =

−+

−+

−

λλ λ

λλ λ

λλ λ

where G1, G2, G3 and λ1, λ2 and λ3 are constants given below where λ1, λ2, λ3 are in μm.

Sellmeier constants G1 G2 G3 λ1 λ2 λ3

SiO2-13.5%GeO2 0.711040 0.451885 0.704048 0.0642700 0.129408 9.425478

The fiber is to operate at λ =1.3 μm, thus, using the Sellmeier equation above with the constants in the table we find

n1 = 1.4682

The V-number is,

V =2πaλ

NA =2π(4.5 μm)(1.3 μm)

(0.1) = 2.175

Apply NA = n12 − n2

2( )1 / 2≈ 2n1

2Δ( )1/ 2

or 0.1 ≈ 2(1.4682)2 Δ[ ]1 / 2

to obtain Δ = 0.002320

Apply Δ =n1 − n2

n1

, i.e. 0.0232 =1.4681 − n2

1.4681

Thus, the required cladding refractive index is

n2 = 1.4648

Pure silica has n = 1.4473, SiO2-13.5 mol.%GeO2 has n1 = 1.4682, by linear interpolation the composition corresponding to n2 = 1.4648 is 11.3 mol.% GeO2. Note, the refractive index n(x) of SiO2-x mol.%GeO2, assuming a linear relationship, can be written as

n(x) = n(0) 1−x

13.5⎛ ⎝

⎞ ⎠ + n(13.5)

x13.5

where n(0) = 1.4473; n(13.5) = n1 = 1.4682. Substituting n(x) = 1.4648 gives x = 11.3 .

________________________________________________________________________

2.12) Waveguide dispersion arises as a result of the dependence of the propagation constant on the V-number, which depends on the wavelength. It is present even when the refractive index is constant; no material dispersion. Let us suppose that n1 and n2 are wavelength (or k) independent. Suppose that β is the propagation constant of mode lm and k = 2π/λ in which λ is the free space wavelength. Then the normalized propagation constant b and propagation constant are related by

β = n2k[1 + bΔ] (1)

The group velocity is defined and given by

ββ

ωddkc

dd

==gv

Show that the propagation time, or the group delay time, τ of the mode is

dkkbd

cLn

cLnL )(22 Δ

+==gv

τ (2)

Given the definition of V,

2/12

2/122

21 )2(][ Δ≈−= kannnkaV (3)

and

[ ] )()2()2()( 2/12

2/12 bk

dVdanbkan

dVd

dVVbd

Δ=Δ= (4)

Show that

2

22 )(

dVVbdV

cLn

dd

λλτ Δ

−= (5)

and that the waveguide dispersion coefficient is

2

22 )(

dVVbdV

cn

LddDw λλτ Δ

−== (6)

Figure 2.53 shows the dependence of V[d2(Vb)/dV2] on the V-number. In the range 1.5 < V < 2.4,

22

2 984.1)(VdV

VbdV ≈

Show that,

221

22 984.1)(984.1

Vcnn

VcnDw λλ

−−=

Δ−≈ (7)

which simplifies to

λ2

2 2)2(984.1

nacDw π

−≈ (8)

i.e. 2

211

)]μm([)μm(76.83)kmnmps(

naDw

λ−≈−− Waveguide dispersion

coefficient (9)

Consider a fiber with a core of diameter of 8 μm and refractive index of 1.468 and a cladding refractive index of 1.464, both refractive indices at 1300 nm. Suppose that a 1.3 μm laser diode with a spectral linewidth of 2 nm is used to provide the input light pulses. Estimate the waveguide dispersion per kilometer of fiber using Eqs. (6) and (8).

0

0.5

1

1.5

0 1 2 3V-number

V[d2(Vb)/dV2]

Figure 2.53 d2(Vb)/dV2 vs V-number for a step index fiber. (Data extracted from W. A. Gambling et al.

The Radio and Electronics Engineer, 51, 313, 1981.)

Solution Waveguide dispersion arises as a result of the dependence of the propagation constant on the V-number which depends on the wavelength. It is present even when the refractive index is constant; no material dispersion. Let us suppose that n1 and n2 are wavelength (or k) independent. Suppose that β is the propagation constant of mode lm and k = 2π/λ where λ is the free space wavelength. Then the normalized propagation constant b is defined as,

22

21

22

2)/(nn

nkb−

−=

β (1)

Show that for small normalized index difference Δ = (n1 − n2)/n1, Eq. (1) approximates to

21

2)/(nn

nkb−

−=

β (2)

which gives β as,

β = n2k[1 + Δb] (3)

The group velocity is defined and given by

ββ

ωddkc

ddv ==g

Thus, the propagation time τ of the mode is

dkkbd

cLn

cLn

dkd

cL

vL )(22 Δ

+=⎟⎠⎞

⎜⎝⎛==

βτg

(4)

where we assumed Δ ≈ constant (does not depend on the wavelength). Given the definition of V,

2/12

2/112

2/1

1

21121

2/12121

2/122

21

)2(]2[

)(

)])([(][

Δ≈Δ≈

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −+≈

−+=−=

kannnka

nnnnnnka

nnnnkannkaV

(5)

From Eq. (5),

[ ] )()2()2()( 2/12

2/12 bk

dVdanbkan

dVd

dVVbd

Δ=Δ=

This means that τ depends on V as,

dVVbd

cLn

cLn )(22 Δ

+=τ (6)

Dispersion, that is, spread δτ in τ due to a spread δλ can be found by differentiating Eq. (6) to obtain,

2

22

2

222

)(

)()(

dVVbdV

cLn

dVVbdV

cLn

dVVbd

dVd

ddV

cLn

dd

λ

λλλτ

Δ−=

⎟⎠⎞

⎜⎝⎛−

Δ=

Δ=

(7)

The waveguide dispersion coefficient is defined as

2

22 )(

dVVbdV

cn

LddDw λλτ Δ

−== (8)

Figure 2.53 shows the dependence of V[d2(Vb)/dV2] on the V-number.

In the range 2 < V < 2.4,

22

2 984.1)(VdV

VbdV ≈

so that Eq. (8) becomes,

221

22 984.1)(984.1

Vcnn

VcnDw λλ

−−=

Δ−≈ (9)

We can simplify this further by using

2/1

2/12

22

2

)2(2984.1984.1

⎥⎦

⎤⎢⎣

⎡Δπ

Δ−≈

Δ−≈

ancn

VcnDw

λλλ

∴ λ2

2 2)2(984.1

nacDw π

−≈ (10)

Equation (6) should really have Ng2 instead of n2 in which case Eq. (10) would be

λ22

22

2)2(984.1

nacN

Dw π−≈ g (11)

Consider a fiber with a core of diameter of 8 μm and refractive index of 1.468 and a cladding refractive index of 1.464 both refractive indices at 1300 nm. Suppose that a1.3 μm laser diode with a spectral linewidth of 2 nm is used to provide the input light pulses. Estimate the waveguide dispersion per kilometer of fiber using Eqs. (8) and (11).

( ))μm 3.1(

)464.1468.1)(μm 4(22 2/1222/12

221

−=−=

πλπ nnaV = 2.094

and Δ = (n1 − n2)/n1 = (1.468−1.464)/1.468 = 0.00273.

From the graph, Vd2(Vb)/dV2 = 0.45,

)45.0()m 101300)(s m 103(

)1073.2)(464.1()(91-8

3

2

22

−

−

×××

−=Δ

−=dV

VbdVcnDw λ

∴ Dw ≈ −4.6×10-6 s m-2 or −4.6 ps km-1 nm-1

Using Eq. (10)

)]464.1(2]m 1042)[s m 103(

)m 101300(984.12)2(

984.1261-8

9

22 −

−

××π××

−=π

−≈ λnac

Dw

∴ Dw ≈ −4.6×10-6 s m-2 or −4.6 ps km-1 nm-1

For Δλ1/2 = 2 nm we have,

Δτ1/2 = |Dw|L Δλ1/2 = (4.6 ps km-1 nm-1)(2 nm) = 9.2 ps/km

2.14) Consider an optimal graded index fiber with a core diameter of 30 μm and a refractive index of 1.4740 at the center of the core and a cladding refractive index of 1.4530. Find the number of modes at 1300 nm operation. What is its NA at the fiber axis, and its effective NA? Suppose that the fiber is coupled to a laser diode emitter at 1300 nm and a spectral linewidth (FWHM) of 3 nm. The material dispersion coefficient at this wavelength is about −5 ps km-1 nm-1. Calculate the total dispersion and estimate the bit rate × distance product of the fiber. How does this compare with the performance of a multimode fiber with same core radius, and n1 and n2? What would the total dispersion and maximum bit rate be if an LED source of spectral width (FWHM) Δλ1/2 ≈ 80 nm is used?

Solution The normalized refractive index difference Δ = (n1 − n2)/n1 = (1.4740 − 1.453)/1.474 = 0.01425

Modal dispersion for 1 km of graded index fiber is

28

21intermode )01425.0(

)103(320)474.1)(1000(

320 ×=Δ≈

cLnσ = 2.9×10-11 s or 0.029 ns

The material dispersion (FWHM) is

)nm 3)(km ns ps 5)(m 1000( 112/1)2/1(

−−−=Δ=Δ λτ mm LD = 0.015 ns

Assuming a Gaussian output light pulse shape, rms material dispersion is,

σm = 0.425Δτ1/2 = (0.425)(0.015 ns) = 0.00638 ns

Total dispersion is

2222intermodetotal 00638.0029.0 +=+= mσσσ = 0.0295 ns.

so that B = 0.25/σtotal = 8.5 Gb

If this were a multimode step-index fiber with the same n1 and n2, then the rms dispersion would roughly be

1-821

sm 103453.1474.1

×−

=−

≈Δ

cnn

Lτ = 70 ps m-1 or 70 ns per km

Maximum bit-rate is

)km ns 70)(28.0(

25.0)28.0(

25.025.01-

intermode

=Δ

≈≈τσ

LLBL

i.e. BL = 12.8 Mb s-1 km (only an estimate!)

The corresponding B for 1 km would be around 13 Mb s-1

With LED excitation, again assuming a Gaussian output light pulse shape, rms material dispersion is

)nm 80)(km ns ps 5)(m 1000)(425.0(

)425.0()425.0(11

2/1)2/1(

−−−=

Δ=Δ= λτσ mmm LD

= 0.17 ns Total dispersion is

2222intermodetotal 17.0029.0 +=+= mσσσ = 0.172 ns

so that B = 0.25/σtotal = 1.45 Gb

The effect of material dispersion now dominates intermode dispersion.

_______________________________________________________________________

3.3) Consider a GaAs pn junction which has the following properties. Na = 1016 cm-3 (p-side), Nd = 1016 cm-3 (n-side), B = 7.21×10-16 m3 s-1, cross sectional area A = 0.1 mm2. What is the diode current due to diffusion in the neutral regions at 300 K when the forward voltage across the diode is 1 V? See Question 3.2 and Table 3Q2 for GaAs properties.

Solution

De = kTμe/e ≈ (0.0259 V)(7000×10-4 m2 V-1 s-1) = 0.01813 m2 s-1

and Dh = kTμh/e ≈ (0.0259 V)( 310×10-4 m2 V-1 s-1) = 0.000803 m2 s-1

Recombination time of electrons diffusing in the p-region is

τe =1

BNa

=1

(7.21×10−16 m3s −1 )(1× 1016 ×106 m−3) = 138.7 ns

Recombination time of holes diffusing in the n-region is

τh =1

BNd

=1

(7.21 ×10−16 m3s−1)(1×1016 ×106 m −3 ) = 138.7 ns

The diffusion lengths are then:

Le = √[Deτe] = √[0.01813 m2 s-1)(138.7 × 10-9 s)] = 5.02 × 10-5 m, or 50.2 μm

and Lh = √[Dhτh] = √[0.000803 m2 s-1)(138.7 × 10-9 s)] =1.056 × 10-5 m, or 10.56 μm

The diffusion component of the current is

I = Idiff = Ιso[exp(eV/(kT)) − 1] ≈ Ιsoexp(eV/(kT)) for V >> kT/e (≈ 0.02586 V)

where Ιso = AJso = Aeni2[(Dh/(LhNd)) + (De/(LeNa))]

Given ni = 1.8×1012 m-3, A = 0.1×10-6 m2,

Thus

Iso =0.1×10−6 m2( )1.602 ×10−19 C( )1.8 ×1012 m−3( )2

0.000803 m2 s−1( )1.056 ×10−5 m( )1×1022 m−3( )

+0.1×10−6 m2( )1.602 ×10−19 C( )1.8 ×1012 m−3( )2

0.01813 m2 s−1( )5.02 ×10−5 m( )1×1022 m−3( )

∴ Iso = 2.27 × 10-21 A.

The forward current due to diffusion is

Idiff = Isoexp[(eV/(kT)] = (2.27 × 10-21 A)exp[(1 V)/(0.0259 V)]

∴ Idiff = 0.00013 A or 0.13 mA

________________________________________________________________________

3.5) An AlGaAs LED emitter for use in a local optical fiber network has the output spectrum shown in Figure 3.32 (b). It is designed for peak emission at about 820 nm at 25°C. (a) Why does the peak emission wavelength increase with temperature? (c) What is the bandgap of AlGaAs in this LED? (d) The bandgap, Eg, of the ternary AlxGa1-xAs alloys follows the empirical expression, Eg(eV) = 1.424 + 1.266x + 0.266x2. What is the composition of the AlxGa1-xAs in this LED?

Solution

Figure 3.32 (a) A typical output spectrum (relative intensity vs. wavelength) from an IR (infrared) AlGaAs LED. (b) The output spectrum of the LED in (a) at 3 temperatures: 25 °C, −40 °C and 85 °C.

Values normalized to peak emission at 25 °C. The spectral widths are FWHM.

(a) The bandgap decreases with temperature

(b) Use the peak emission wavelength to find Eg as follows:

At −40 °C (233 K), λpeak = 804 nm.

At 25 °C (298 K), λpeak = 820 nm.

At 85 °C (358 K), λpeak = 837 nm.

We first note that we need the required bandgap Eg at the wavelength of interest. The photon energy at peak emission is hc/λpeak = Eg + (1/2)kBT. Then,

TkhE Bog 21−≈ υ

and at λpeak = 820×109 m, taking T = 25 + 273K,

TkhcE Bo

g 21−≈

λ= 1.499 eV

(c) The bandgap Eg of the ternary alloys AlxGa1-xAs follows the empirical expression,

Eg(eV) = 1.424+1.266x+0.266x2.

∴ Eg(eV) = 1.499 = 1.424 + 1.266x + 0.266x2.

Solving for x we find, x = 0.05 or Al0.05Ga0.95As 3.6) Figure 3.52 represents the bandgap Eg and the lattice parameter a in a quarternary III-V alloy system. A line joining two points represents the changes in Eg and a with composition in a ternary alloy composed of the compounds at the ends of that line. For example, starting at GaAs point, Eg = 1.42 eV and a = 0.565 nm, Eg decreases and a increases as GaAs is alloyed with InAs, as we move along the line joining GaAs to InAs. Eventually at InAs, Eg = 0.35 eV and a = 0.606 nm. Point X in Figure 3.52 is composed

of InAs and GaAs and it is the ternary alloy In1-xGaxAs. At X, In0.53Ga0.47As (often called "in-gas" in telecom) has Eg = 0.73 eV and a = 0.587 nm, which is the same a as that for InP. In1-xGaxAs at X is therefore lattice matched to InP and can hence be grown on an InP substrate without creating defects at the interface. Further, In1-xGaxAs at X can be alloyed with InP to obtain a quaternary alloy1 In1-xGaxAs1-yPy whose properties lie on the line joining X and InP and therefore all have the same lattice parameter as InP but different bandgap. Layers of In1-xGaxAs1-yPy with composition between X and InP can be grown epitaxially on an InP substrate by various techniques such as liquid phase epitaxy (LPE) or molecular beam epitaxy (MBE). The grey shaded area between the solid lines represents the possible values of Eg and a for the quarternary III-V alloy system in which the bandgap is direct and hence suitable for direct recombination. The compositions of the quaternary alloy lattice matched to InP follow the line from X to InP.

(a) Given that the In1-xGaxAs at X is In0.53Ga0.47As show that quaternary alloys In1-

xGaxAs1-yPy are lattice matched to InP when y = 1−2.13x.

(b) The bandgap energy Eg, in eV for In1-xGaxAs1-yPy lattice matched to InP is given by the empirical relation, Eg (eV) = 0.75 + 0.46y + 0.14 y2. Find the composition of the quaternary alloy suitable for an LED emitter operating at 1.30 μm.

Figure 3.52 Bandgap energy Eg and lattice constant a for various III-V alloys of GaP, GaAs, InP and InAs. A line represents a ternary alloy formed with compounds from the end points of the line. Solid lines are for direct bandgap alloys whereas dashed lines for indirect bandgap alloys. Regions between lines represent quaternary alloys. The line from X to InP represents quaternary alloys In1-xGaxAs1-yPy made from In0.53Ga0.47As and InP which are lattice matched to InP.

Solution (a) The III−V quaternary alloy is made of (In0.53Ga0.47As) alloyed with InP. Suppose that 1−z fraction of In0.53Ga0.47As is alloyed with z fraction of InP, which means that z = 0

1 Some books other formats for the chemical composition e.g. GaxIn1-xAsyP1-y. The present notation In1-

xGaxAs1-yPy was chosen to reflect the common vernacular for InGaAs ("in-gas").

for In0.53Ga0.47As, and z = 1 for InP. The formula for the quaternary alloy is In1−xGaxAs1-

yPy. Thus,

(1−z)(In0.53Ga0.47As) + zInP = In1−xGaxAs1-yPy (1)

This equation of course must hold at point X where we have (In0.53Ga0.47As). We can now balance the contents for each element in Equation (1):

For P z = y

For Ga (1 − z)(0.47) = x ∴ y = 1 − 2.13x

For In (1 − z)(0.53) + z = 1 − x ∴ z = 1 − 2.13x or y = 1 − 2.13x

For As 1 − z = 1 − y ∴ z = y

(b) Consider a lattice matched InGaAsP to InP. Given λ = 1.30 μm.

Eg = 1.24/λ = 0.954 eV

Thus, Eg (eV) = 0.75 + 0.46y + 0.14 y2 = 0.954 eV

Solving the quadratic equation,

y = 0.39, and hence x = 0.47(1−y) = 0.28

∴ In0.72Ga0.28As0.61P0.39

3.8) Experiments carried out on various direct bandgap semiconductor LEDs give the output spectral linewidth (between half intensity points as in listed in Table 3.5. From Figure 3.31 we know that a spread in the wavelength is related to a spread in the photon energy, ph

2ph )/( EEhc Δ≈Δλ where Eph = hυ is the photon energy. Suppose that we write

Eph = hc/λ and ΔEph = Δ(hυ) ≈ mkBT where m is a numerical constant. Therefore,

2)/( λλ hcTkBm≈Δ . (1)

By appropriately plotting the data in Table 3.5, and assuming T = 300 K, find m.

Table 3.6 gives the linewidth Δλ1/2 for various visible LEDs based on GaAsP. Radiative recombination is obtained by appropriately doping the material. Using m = 3.0 in Eq. (1), T = 300 K, calculate the expected spectral width for each and compare with the experimental value. What is your conclusion?

Table 3.5 Linewidth Δλ1/2 between half points in the output spectrum (intensity vs. wavelength) of four LEDs using direct bandgap semiconductors.

Peak wavelength of emission (λ) nm

650 810 820 890 950 1150 1270 1500

Δλ1/2 nm 22 36 40 50 55 90 110 150

Material (Direct Eg) AlGaAs AlGaAs AlGaAs GaAs GaAs InGaAsP InGaAsP InGaAsP

Table 3.6 Linewidth Δλ1/2 between half points in the output spectrum (intensity vs. wavelength) of various visible LEDs using GaAsP

Peak wavelength of emission (λ) nm

565 583 600 635

Δλ1/2 nm 28 36 40 40

Color Green Yellow Orange Red

Material GaP(N) GaAsP(N) GaAs (N) GaAsP

Solution

0

20

40

60

80

100

120

140

160

0 1000000 2000000 3000000

AlGaAs

AlGaAsAlGaAsGaAs

GaAs

InGaAsP

InGaAsP

InGaAsP

Δλ1/2 (nm)

λ2 (nm)2

Figure 3Q32-1 Plot of Δλ1/2 vs. λ2. The data can be fitted with a straight line.

The best line forced through zero is

Δλ1/2 = [6.57×10-5 (nm)-1]λ2

and Slope = 6.57×10-5 (nm)-1.

This slope is mkT/hc; thus m ≈ 3.15.

Table 3Q23-1 below shows the calculated spectral widths Δλ1/2 using m = 3. The actual observed widths are substantially larger than the expected Δλ1/2 using m = 3, as we found above for the direct bandgap materials. The observed spread is much more than m = 3, and is related to the energy of the localized defect center. It is possible to give a semiquantitative plausible explanation as follows. A captured electron will have a wavefunction that is localized and hence a smaller uncertainty in its position than in the band; i.e. Δx will be small. That means the uncertainty Δp in its momentum will be higher and hence the uncertainly in its energy will also be higher. We would expect that the spread of photon energies will be more than from that in band to band recombinations

Table 3Q23-1 Linewidth Δλ1/2 between half points in the output spectrum (intensity vs. wavelength) of various visible LEDs using GaAsP

Peak wavelength of emission (λ) nm

565 583 600 635

Δλ1/2 nm 28 36 40 40

Expected Δλ1/2 nm using m = 3 20.0 21.3 22.5 25.2

Color Green Yellow Orange Red

Material GaP(N) GaAsP(N) GaAs (N) GaAsP

3.10) a) It is found that approximately 200 μW is coupled into a multimode step index fiber from a surface emitting LED when the current is 75 mA and the voltage across the LED is about 1.5 V. What is the overall efficiency of operation?

(b) Experiments are carried out on coupling light from a 1310 nm ELED (edge emitting LED) into multimode and single mode fibers. (i) At room temperature, when the ELED current is 120 mA, the voltage is 1.3 V and light power coupled into a 50 μm multimode fiber with NA (numerical aperture) = 0.2 is 48 μW. What is the overall efficiency? (ii) At room temperature, when the ELED current is 120 mA, the voltage is 1.3 V and light power coupled into a 9 μm single mode fiber is 7 μW. What is the overall efficiency?

Solution

(a) ηoverall =Po

IV=

200 ×10−6 W(75 ×10−3 A)(1.5 V)

= 1.8×10-3 = 0.18 %

(b) (i) ηoverall =Po

IV=

48 ×10−6 W(120 ×10−3 A)(1.3 V)

= 0.0307%

(ii) ηoverall =Po

IV=

7 ×10−6 W(120 ×10−3 A)(1.3 V)

= 0.0045%

3.11) For a particular AlGaAs LED emitting at 850 nm, the nonradiative recombination lifetime is τnr = 50 ns. The recombination occurs in the active region, which has been doped with acceptors of concentration 1017 cm-3, and the direct recombination coefficient B is 2×10-16 m3 s-1. What is the internal optical power generated at a current of 100 mA?

Solution The radiative lifetime τr = 1/BNa = 1/(2×10-16 m3 s-1)(1×1023 m-3) = 5.0 ×10-8 s or 50 ns. IQE is,

50.0)ns50()ns50(

)ns50(11

1

11

1

IQE =+

=+

= −−

−

−−

−

nrr

r

τττ

η = 50%

Equation (3) gives eIhP

eIo

//

/secondper lost carriers Totalsecondper emitted Photons (int)ph

IQE

υη =

Φ== where Po(int) is

the optical power generated internally (not yet extracted). Therefore,

υη heIPo IQE(int) = = 72.9 mW

ρ(hυ) is the energy of the electromagnetic radiation per unit volume per unit frequency due to photons with energy hυ = E2 − E1. Suppose that there are nph photons per unit volume. Each has an energy hυ. The frequency range of emission is Δυ. Then,

ρ(hυ) =nphhυ

Δυ

Consider the Ar ion laser system. Given that the emission wavelength is at 488 nm and the linewidth in the output spectrum is 5×109 Hz (between half intensity points, that is Δυ = 2×5×109 Hz), estimate the photon concentration necessary to achieve more stimulated emission than spontaneous emission.

Solution

Suppose that there are nph photons per unit volume. Each has an energy hυ. The frequency range of emission is Δυ as is the linewidth between half intensity points. Then,

ρ(hυ) =nphhυ

Δυ

Now, R21(stim)R21(spon)

=c3

8πhυ 3 ρ(hυ) >1

∴ ρ(hυ) >8πhυ 3

c3 =8πhλ3 =

8π(6.62 ×10−34 J s)(488 ×10−9 m)3 = 1.43×10-13 J s m-3

∴ nph =Δυρ(hυ)

hυ=

(2 × 5 ×109 s-1 )(1.43 ×10−13 J s m-3)

(6.62 ×10−34 J s)(3 ×108 m s-1

433 ×10−9 m)

∴ nph =3.5×1015 Photons m-3.

Note that this represents the critical photon concentration for stimulated emission to just exceed spontaneous emission in the absence of any photon losses. It does not represent the photon concentration for laser operation. In practice, the photon concentration is much greater during laser operation.

________________________________________________________________________

Consider the energy diagram of a forward biased GaAs laser diode as in Figure 4Q9 For simplicity we assume a symmetrical device (n = p) and we assume that population inversion has been just reached by A and B overlapping as illustrated in Error! Reference source not found. which results in EFn − EFp = Eg. Estimate the minimum carrier concentration n = p for population inversion in GaAs at 300K. The intrinsic carrier concentration in GaAs is of the order of 107 cm-3. Assume for simplicity that

n = niexp[(EFn − EFi)/kBT)] and p = niexp[(EFi − EFn)/kBT)]

(Note: The analysis will only be an order of magnitude as the above equations do not hold in degenerate semiconductors. A better approach is to use the Joyce-Dixon equations as can be found in advanced textbooks)

p+

Eg

V

n+

EFp

EFne−

h+

A

BEv

Ec

Ev

Ec

The energy band diagram of a degenerately doped p-n junction with asufficiently large forward bias to just cause population inversion where A andB overlap.

Figure 4Q9

Solution

From Figure 4Q9, the potential barrier from Ec(n-side) to Ec(p-side) is ΔEc. Similarly, Ev(p-side) − Ev(n-side) = ΔEc.

It can be seen that to get any population inversion, the Fermi level EFp must be at least 1/2ΔEc below Ev(p-side) or 1/2ΔEc above EFv(n-side) and EFn must be at least 1/2ΔEc above Ec(n-side).

Thus population inversion occurs when,

EFn − EFp = [Ec(n-side)+ 1/2ΔEc] − [Ev(n-side) + 1/2ΔEc] = Eg. (1)

Now, if EFi is the Fermi-level in the intrinsic material, then for n = p, we must have

EFn − EFi = EFi − EFp. (2)

Substituting for EFp = EFn − Eg from Eq. (1) into (2) we find,

EFn − EFi = EFi − (EFn − Eg)

2EFn − 2EFi = Eg,

giving EFn − EFi = Eg/2 = (1.43 eV)/2 = 0.715 eV.

From, n/ni = exp[(EFn − EFi)/kT)] (3)

we have n = (107 cm-3)exp(0.715/0.0259) =1019 cm-3.

This confirms degenerate doping.

Author's note: EFn and EFp are quasi-Fermi levels or imrefs, and are not the true Fermi levels - hence the estimation.

4.10

a Consider the rate equations and their results in Section 4.10 It takes Δt = nL/c second for photons to cross the laser cavity length L, where n is the refractive index. If Nph is the coherent radiation photon concentration, then only half of the photons, 1/2Nph, in the cavity would be moving towards the output face of the crystal at any instant. Given that the active layer has a length L, width W and thickness d, show that the coherent optical output power and intensity are

Po =

hc2 NphdW2nλ

⎡

⎣ ⎢ ⎤

⎦ ⎥ 1 − R( ) and I =

hc2 Nph

2nλ⎡

⎣ ⎢ ⎤

⎦ ⎥ 1 − R( )

where R is the reflectance of the semiconductor crystal face.

b If α is the attenuation coefficient for the coherent radiation within the semiconductor active layer due to various loss processes such as scattering and R is the reflectance of the crystal ends then the total attenuation coefficient αt is,

αt = α +

12L

ln1

R2⎛ ⎝

⎞ ⎠

Consider a double heterostructure InGaAsP semiconductor laser operating at 1310 nm. The cavity length L ≈ 60 μm, width W ≈ 10 μm, and d ≈ 0.25 μm. The refractive index n ≈ 3.5. The loss coefficient α ≈ 10 cm-1. Find αt, τph.

c For the above device, threshold current density Jth ≈ 500 A cm-2 and τsp ≈ 10 ps. What is the threshold electron concentration? Calculate the lasing optical power and intensity when the current is 5 mA.

Solution a If Nph is the coherent radiation photon concentration, then only half of the photons, 1/2Nph, in the cavity would be moving towards the output face of the crystal at any instant. It takes Δt = nL/c seconds for photons to cross the laser cavity length L.

Po = Energy flow per unit time in cavity towards face × Transmittance

=

hcλ

⎛ ⎝

⎞ ⎠

12 Nph( )dWL( )

Δt

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥ Transmittance

=

hcλ

⎛ ⎝

⎞ ⎠

12 Nph( )dWL( )

Lnc

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

1− R( )

=hc2 NphdW

2nλ⎡

⎣ ⎢ ⎤

⎦ ⎥ 1 − R( )

and I =

Optical PowerArea

=Po

dW=

hc2 Nph

2nλ⎡

⎣ ⎢ ⎤

⎦ ⎥ 1− R( )

where R is the reflectance of the crystal face.

b Consider one round trip through the cavity. The length L is traversed twice and there is one reflection at each end. The overall attenuation of the coherent radiation after one-round trip is

RRexp[−α(2L)]

where R is the reflectance of the crystal end.

Equivalently we can represent this reduction by an effective or a total loss coefficient αt such that after one round trip, the reduction factor is

exp[−αt(2L)]

Equating the two,

RRexp[−α(2L)] = exp[−αt(2L)]

and rearranging,

αt = α +

12L

ln1

R2⎛ ⎝

⎞ ⎠

c The reflectance is

R =

n −1n +1

⎛ ⎝

⎞ ⎠

2

=3.5 −13.5 +1

⎛ ⎝

⎞ ⎠

2

= 0.309

The total loss coefficient is

αt = α +

12L

ln1

R2⎛ ⎝

⎞ ⎠ =1000 m−1 +

160 ×10−6 m−1 ln

10.3092

⎛ ⎝

⎞ ⎠ = 2.06×104

m-1.

∴ τ ph =

ncαt

=3.5

(3 ×108 m s−1)(2.06 × 108 m−1)= 5.7×10-13 s (0.57 ps)

Coherent radiation is lost from the cavity after, on average, 0.57 ps.

For the above device, threshold current density Jth ≈ 500 A cm-2 and τsp ≈ 10 ps, d ≈ 0.25 μm,

From Jth =nthedτ sp

we have, nth ≈Jthτ sp

ed≈

(500 ×104 A m-2 )(10 ×10−9 s)(1.6 ×10−19C)(0.25 ×10−6 m)

≈ 1.25×1024 m-3 or 1.2×1018

cm-3

Now, the current density corresponding to I = 30 mA is

J = I/(WL) = (0.05 A)/[10×60×10-610-6 m2)] = 833×104 A m-2.

And, N ph ≈τ ph

edJ − Jth( ) ≈

(5.7 ×10−13s)(1.6 ×10−19C)(0.25 ×10−6 m)

(833 −500) ×104 A m-2

≈ 4.7×1019 photons m-3 The optical power is

Po =hc 2N phdW

2nλ

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ 1−R( )

= (6.62 ×10−34 J s)(3×108 m s-1)2(4.7 ×1019 m-3)(0.25 ×10−6 m)(10 ×10−6 m)2(3.5)(1310 ×10−9 m)

× 1− 0.309( )

≈ 0.00053 W or 0.53 mW.

Intensity = Optical Power / Area = Po / (dW)

= (0.00053)/[0.25×10×10-310-3 mm2)] = 223 W mm-2.

This intensity is right at the crystal face over the optical cavity cross section. As the beam diverges, the intensity decreases away from the laser diode.

4.11)Consider a InGaAsP-InP laser diode which has an optical cavity of length 250 microns. The peak radiation is at 1550 nm and the refractive index of InGaAsP is 4. The optical gain bandwidth (as measured between half intensity points) will normally depend on the pumping current (diode current) but for this problem assume that it is 2 nm.

a What is the mode integer m of the peak radiation?

b What is the separation between the modes of the cavity?

c How many modes are there in the cavity?

d What is the reflection coefficient and reflectance at the ends of the optical cavity (faces of the InGaAsP crystal)?

e What determines the angular divergence of the laser beam emerging from the optical cavity?

Solution

a The wavelength λ of a cavity mode and length L are related by

m

λ2n

= L

so that m =

2nLλ

=2(4)(250 ×10−6 )

(1550 ×10−9 )= 1290.3.

When m = 1290, λ = 2nL/m = 1550.39 nm so that the peak radiation has m = 1290.

b Mode separation is given by,

Δλm =

λ2

2nL=

(1550 ×10−9 )2

2(4)(250 ×10−6 )= 1.20 nm

c The linewidth is 2 nm

Let the optical linewidth Δλ be between λ1 and λ2. Then λ1 = λ − (1/2)Δλ = 1549 nm and λ2 = λ + (1/2)Δλ = 1551 nm and the mode numbers corresponding to these are

m1 =

2nLλ1

=2(4)(250 ×10−6 )

(1549 ×10−9 )=1291.15

m2 =

2nLλ2

=2(4)(250 ×10−6 )

(1551×10−9 )= 1289.49

Now m must be an integer and the corresponding wavelength must fit into the optical gain curve.

Taking m = 1290, gives λ = 2nL/m = 1550.39 nm; within optical gain 1549 − 1551 nm

Taking m = 1291, gives λ = 2nL/m = 1549.18 nm; within optical gain 1549 − 1551 nm

Taking m = 1289, gives λ = 2nL/m = 1551.59 nm; just outside optical gain 1549 − 1551 nm

There are 2 modes.

d r = (n − 1)/(n + 1) = (4 − 1)/(4 + 1) = 0.6

R = r2 = 0.36 or 36%.

e Diffraction at the active region cavity end.

4.12 a There are several laser diode efficiency definitions as follows:

The external quantum efficiency ηEQE, of a laser diode is defined as

ηEQE =Number of output photons from the diode per unit second( )Number of injected electrons into diode per unit second( )

The external differential quantum efficiency, ηEDQE, of a laser diode is defined as

ηEDQE =Increase in number of output photons from diode per unit second( )

Increase in number of injected electrons into diode per unit second( )

The external power efficiency, ηEPE, of the laser diode is defined by

ηEPE =Optical output powerElectical input power

If Po is the emitted optical power, show that

ηEQE =ePo

Eg I

ηEDQE =e

Eg

⎛

⎝ ⎜ ⎞

⎠ ⎟ dPo

dI

ηEPE = ηEQE

Eg

eV⎛ ⎝

⎞ ⎠

b A commercial laser diode with an emission wavelength of 670 nm (red) has the following characteristics. The threshold current at 25°C is 76 mA. At I = 80 mA, the output optical power is 2 mW and the voltage across the diode is 2.3 V. If the diode current is increased to 82 mA, the optical output power increases to 3 mW. Calculate the external QE, external differential QE and the external power efficiency of the laser diode.

c Consider an InGaAsP laser diode operating at λ = 1310 nm for optical communications. At I = 40 mA, the output optical power is 3 mW and the voltage across the diode is 1.4 V. If the diode current is increased to 45 mA, the optical output power increases to 4 mW. Calculate external quantum efficiency (QE), external differential QE, external power efficiency of the laser diode.

Solution

a The external quantum efficiency ηEQE, of a laser diode is defined as

ηEQE =Number of output photons from the diode per unit second( )Number of injected electrons into diode per unit second( )

∴ ηEQE =Optical Power / hυDiode Current / e

=Po /Eg

I / e=

ePo

IEg

The external differential quantum efficiency, ηEDQE, of a laser diode is defined as

ηEDQE =Increase in number of output photons from diode per unit second( )

Number of injected electrons into diode per unit second( )

∴ ηEDQE =(Change in Optical Power) / hυ

(Change Diode Current) / e=

ΔPo /Eg

ΔI / e=

eEg

dPo

dI⎛ ⎝

⎞ ⎠

The external power efficiency, ηEPE, of the laser diode is defined by

ηEPE =Optical ouput powerElectical input power

=Po

IV=

Po

IVeEg

eEg

=ePo

IEg

⎛

⎝ ⎜ ⎞

⎠ ⎟

Eg

eV⎛ ⎝

⎞ ⎠

∴ ηEPE =ePo

IEg

⎛

⎝ ⎜ ⎞

⎠ ⎟

Eg

eV⎛ ⎝

⎞ ⎠ = ηEQE

Eg

eV⎛ ⎝

⎞ ⎠

b 670 nm laser diode

Eg ≈ hc/λ = (6.626×10-34)(3×108)/[(670×10-9)(1.6×10-19)] = 1.85 eV,

so that ηEQE =ePo

Eg I

∴ ηEQE =(1.6 ×10−19 C)(2 ×10−3 Js−1)

(80 ×10−3 A)(1.85 eV ×1.6 ×10−19 eV/J)= 0.0135 or 1.35%

ηEDQE =e

Eg

dPo

dI⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

(1.6 ×10−19 C)(1.85 eV ×1.6 ×10−19 C)

3 ×10−3 − 2 ×10−3 Js−1

82 ×10−3 − 80 ×10−3 A

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

= 0.270 or 27%.

ηEPE =Po

IV=

2 ×10−3 W(80 ×10−3 A )(2.3 V)

= 0.009 or 1.09%

c 1310 nm laser diode

Eg ≈ hc/λ = (6.626×10-34)(3×108)/[(1310×10-9)(1.6×10-19)] = 0.9464 eV,

so that ηEQE =(1.6 × 10−19 C)(3 ×10−3 Js−1 )

(40 ×10−3 A)(0.9464 eV ×1.6 ×10−19 J/eV)= 0.079 or 7.9%

∴

ηEDQE =e

Eg

dPo

dI⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

(1.6 ×10−19 C)(0.9464 eV ×1.6 ×10−19 C)

4 ×10−3 − 3 ×10−3 Js−1

45 ×10−3 − 40 ×10−3 A

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

= 0.211 or 21%.

and ηEPE =Po

IV=

3 ×10−3 W(40 ×10−3 A )(1.4 V)

= 0.0535 or 5.3%

4.15) Consider a SQW (single quantum well) laser which has an ultrathin active InGaAs of bandgap 0.70 eV and thickness 10 nm between two layers of InAlAs which has a bandgap of 1.45 eV. Effective mass of conduction electrons in InGaAs is about 0.04me and that of the holes in the valence band is 0.44me where me is the mass of the electron in vacuum. Calculate the first and second electron energy levels above Ec and the first hole energy level below Ev in the QW. What is the lasing emission wavelength for this SQW laser? What is this wavelength if the transition were to occur in bulk InGaAs with the same bandgap?

Solution The lowest energy levels with respect to the CB edge Ec in InGaAs are determined by the energy of an electron in a one-dimensional potential energy well

εn =h2n2

8me*d2

where n is a quantum number 1, 2, …, εn is the electron energy with respect to Ec in InGaAs, or εn = En − Ec.

Using d = 10×10-9 m, me* = 0.04me and n = 1 and 2, we find the following electron energy levels

n =1 εn =h2n2

8me*d2 =

(6.626 ×10−34 )2 (1)2

8(0.04 × 9.11 ×10−31)(10 ×10−9 )2 = 1.51×10-20 J = 0.094 eV

ε1 = 0.094 eV

n = 2 ε2 = 0.376 eV

Using d = 10×10-9 m, mh* = 0.44me and n = 1, the hole energy levels below Ev is

n =1 ′ ε n =h2n2

8mh*d2 =

(6.626 ×10−34 )2 (1)2

8(0.44 × 9.11 ×10−31)(10 ×10−9 )2 = 1.37×10-21 J = 0.00855

eV

The wavelength light emission from the QW laser with Eg = 1.45 eV is

λQW =hc

Eg + ε1 + ′ ε 1=

(6.626 ×10−34)(3×108 )(0.70 + 0.094 + 0.00855)(1.602 ×10−19)

= 1545×10 -9 m (1545 nm)

The wavelength of emission from bulk InGaAs with Eg = 1.45 eV is

λg =hcEg

=(6.626 ×10−34 )(3 ×108)

(0.70)(1.602 ×10−19) = 1771×10 -9 m (1771 nm)

The difference is λg − λQW = 1771 - 1545 = 226 nm.

4.16 Effective mass of conduction electrons in GaAs is 0.07me where me is the electron mass in vacuum. Calculate the first three electron energy levels for a quantum well of thickness 8 nm. What is the hole energy below Ev if the effective mass of the hole is 0.47me? What is the change in the emission wavelength with respect to bulk GaAs which has an energy bandgap of 1.42 eV.

Solution

The lowest energy levels with respect to the CB edge Ec in GaAs are determined by the energy of an electron in a one-dimensional potential energy well

εn =h2n2

8me*d2

where n is a quantum number 1, 2, …, εn is the electron energy with respect to Ec in GaAs, or εn = En − Ec. Thus,

n =1 εn =h2n2

8me*d2 =

(6.626 ×10−34 )2 (1)2

8(0.07 ×9.11×10−31)(8 ×10−9 )2 = 0.0839 eV

ε1 = 0.0839 eV

n = 2 ε2 = 0336 eV

n = 3 ε3 = 0.755 eV

Note: Whether ε3 is allowed depends on the depth of the QW and hence on the bandgap of the sandwiching semiconductor.

The hole energy levels below Ev is

′ ε n =h2n2

8mh*d2 =

(6.626 ×10−34)2(1)2

8(0.47 × 9.11×10−31)(8 ×10−9)2 = 0.0125 eV

The wavelength of emission from bulk GaAs with Eg = 1.42 eV is

λg =hcEg

=(6.626 ×10−34 )(3 ×108 )

(1.42)(1.602 ×10−19) = 874×10 -9 m or 874 nm.

Whereas from the GaAs QW, the wavelength is,

λQW =hc

Eg + ε1 + ′ ε 1=

(6.626 ×10−34)(3×108 )(1.42 + 0.0839 + 0.0125)(1.602 ×10−19)

= 818×10 -9 m or 818 nm.

The difference is λg − λQW = 874 − 818 =56 nm.

5.1 Bandgap and photodetection

(a) Determine the maximum value of the energy gap which a semiconductor, used as a photoconductor, can have if it is to be sensitive to yellow light (600nm).

(b) A photodetector whose area is 5×10-2 cm2 is irradiated with yellow light whose intensity is 2 mW cm–2. Assuming that each photon generates one electron-hole pair, calculate the number of pairs generated per second.

(c) From the known energy gap of the semiconductor GaAs (Eg = 1.42 eV), calculate the primary wavelength of photons emitted from this crystal as a result of electron-hole recombination. Is this wavelength in the visible?

(d) Will a silicon photodetector be sensitive to the radiation from a GaAs laser? Why?

Solution (a) Given, λ = 600 nm, we need Eph = hυ = Eg so that,

Eg = hc/λ = (6.626×10-34 J s)(3×108 m s-1)/(600×10-9 m)

= 2.07 eV

(b) A = 5×10-2 cm2 and Ilight = 20×10-3 W/cm2. The received optical power is

Po = AIlight = (5×10-2 cm2)(20×10-3 W/cm2) = 10-3 W

and Nph = number of photons arriving per second = Po/Eph

= (10-3 W)/(2.07×1.60218 ×10-19 J/eV)

= 2.98×1015 Photons s-1.

= 2.98×1015 EHP s-1.

(c) For GaAs, Eg = 1.42 eV and the corresponding wavelength is

λg = hc/ Eg = (6.626×10-34 J s)(3×108 m s-1)/(1.42 eV × 1.6 ×10-19 J/eV)

= 873 nm

The wavelength of emitted radiation due to EHP recombination is therefore 873 nm.

It is not in the visible region (it is in the IR).

(d) For Si, Eg = 1.1 eV and the corresponding cut-off wavelength is,

λg = hc/ Eg = (6.626×10-34 J s)(3×108 m s-1)/(1.1 eV × 1.6 ×10-19 J/eV)

= 1120 nm

Since the 873 nm wavelength is shorter than the cut-off wavelength of 1120 nm, the Si photodetector can detect the 873 nm radiation (Put differently, the photon energy corresponding to 873 nm, 1.42 eV, is larger than the Eg, 1.1 eV, of Si which mean that the Si photodetector can indeed detect the 873 nm radiation)

5.2 Absorption coefficient

(a) If d is the thickness of a photodetector material, Io is the intensity of the incoming radiation, show that the number of photons absorbed per unit volume of sample is

υ

αdh

dn o )]exp(1[ph

−−=

I

(b) What is the thickness of a Ge and In0.53Ga0.47As crystal layer that is needed for absorbing 90% of the incident radiation at 1.5 μm?

(c) Suppose that each absorbed photon liberates one electron (or electron hole pair) in a unity quantum efficiency photodetector and that the photogenerated electrons are immediately collected. Thus, the rate of charge collection is limited by rate of photon generation. What is the external photocurrent density for the photodetectors in (b) if the incident radiation is 100 μW mm-2?

Solution (a) If Io is the intensity of incoming radiation (energy flowing per unit area per second), Ioexp(−αd) is the transmitted intensity through the specimen with thickness d (Figure 5.26) and thus Io[1− exp(−αd)] is the “absorbed” intensity. If Γph is the number of photons arriving per unit area per unit second (the photon flux), then Γph = Io/hυ where hυ is the energy per photon. Thus the number of photons absorbed per unit volume per unit second of sample is

[ ][ ]

υαυ

α

dhd

dh

d

AdA

n )exp(1)exp(1

phph

−−=

−−

=Γ

= o

oI

I

(b) For Ge, α ≈ 5 × 105 m-1 at 1.5 μm incident radiation (from Figure 5.5).

∴ 1− exp(−αd) = 0.9

∴ 65 1060.4

9.011ln

1051

9.011ln1 −×=⎟

⎠⎞

⎜⎝⎛

−×=⎟

⎠⎞

⎜⎝⎛

−=

αd m = 4.60 μm.

For In0.53Ga0.47As, α ≈ 7.5 × 105 m-1 at 1.5 μm incident radiation (Figure 5.5).

65 1007.3

9.011ln

105.71 −×=⎟

⎠⎞

⎜⎝⎛

−×=d m = 3.07 μm.

(c) The quantum efficiency is unity. Therefore the collected electrons per unit area per unit second is equal to the absorbed photons per unit area per unit second. So, the current density (current per unit area),

[ ] [ ]hc

deh

deJ )exp(1)exp(1ph

αλυ

α −−=

−−= oo II

Given, oI = 100 μW mm-2 = 100 × 10-6 × 106 W m-2 = 100 W m-2,

8.108100.310626.6

9.0105.11001060218.1834

619

ph =×××

×××××= −

−−

J A m-2 =10.9 mA cm-2.

Note: We neglected any light reflection from the surface of the semiconductor crystal (100% efficient AR coating assumed).

5.5 Ge Photodiode Consider a commercial Ge pn junction photodiode which has the responsivity shown in Figure 5.47. Its photosensitive area is 0.008 mm2. It is used under a reverse bias of 10 V when the dark current is 0.3 μA and the junction capacitance is 4 pF. The rise time of the photodiode is 0.5 ns.

(a) Calculate its quantum efficiency at 850, 1300 and 1550 nm.

(b) What is the light intensity of light at 1.55 μm that gives a photocurrent equal to the dark current.

(c) What would be the effect of lowering the temperature on the responsivity curve?

(d) Given that the dark current is in the range of microamperes, what would be the advantage in reducing the temperature?

(e) Suppose that the photodiode is used with a 100 Ω resistance to sample the photocurrent. What limits the speed of response?

Figure 5.47 The responsivity of a commercial G pn junction photodiode

Solution

0

0.1

0.20.3

0.40.5

0.6

0.70.8

0.5 1 1.5 2Wavelength (µm)

Responsivity(A/W)

Figure 5Q05-1 Responsivity vs. wavelength for a commercial Ge photodiode

(a) At λ = 850×10-9 m, from the responsivity vs. wavelength curve we have R = 0.275 A/W as shown in Figure 5Q05-1. From the definitions of quantum efficiency (QE) η and responsivity we have,

%1.04)m10850()C1060218.1(

)A/W275.0()ms103(Js)10626.6(919

1834

=×××××××

== −−

−−

λη

ehcR

Similarly, we can calculate quantum efficiency at other wavelengths. The results are summarized in Table 5Q05.

Table 5Q05

(b) Given, photocurrent Iph = Id = 0.3 μA = 0.3×10-6 A and area, A = 8×10-3 mm2 = 8×10-9 m2, the incident optical power,

Po = Iph/R = (0.3×10-6 A)/(0.73 A W-1) = 4.12 × 10-7 W

Light intensity, Io = Po/A = (4.12 ×10-7 W)/(8×10-9 m2) = 51.5 W m-2 or 5.15 mW cm-2.

(c) From Chapter 3, we know that dEg/dT is negative, that is, Eg increases with decreasing temperature.

Stated differently, α vs. λ curve shifts towards shorter λ with decreasing T.

Wavelength, (nm) 850 1300 1550

Responsivity R, (A/W) 0.275 0.57 0.73

Quantum efficiency η, (%) 40.1 54.3 58.4

The change in α with T means that the amount of optical power absorbed in the depletion region and hence the quantum efficiency will change with temperature. The peak responsivity will shift to lower wavelengths with decreasing temperature. If maximum photogeneration requires a certain absorption depth 1/αmax, corresponding to some αmax, will occur at a shorter wavelength at lower temperatures. In Figure 5Q05-2, the maximum responsivity corresponds to αmax which occurs at λmax at high T and at λ′max at lower T. Clearly the peak responsivity shifts to lower wavelengths as the temperature decreases. (See Figure 5.50)

λ

Absorption coefficient

Low T High T

λmaxλ′max

αmax

Figure 5Q05-2 The absorption coefficient depends on the temperature

(d) Dark current (∝ ni2 ∝ exp(−Eg/kT)) will be drastically reduced if we decrease the

temperature. Reduction of dark current improves SNR.

(e) The RC time constant = 100 Ω × (4×10-12 F) = 400 ps, neglecting stray capacitances. The rise time, 0.5 ns or 500 ps. The manufacturer does not state whether the rise time is and RC limitation or intrinsic device limitation (e.g. carrier diffusion and drift). It is likely to be an intrinsic limitation. The speed of response will depend on both the rise time and RC time constant. (It is not simply 400 ps + 500 ps)

5.7 InGaAs pin Photodiodes Consider a commercial InGaAs pin photodiode whose responsivity is shown in Figure 5.49. Its dark current is 5 nA.

(a) What optical power at a wavelength of 1.55 μm would give a photocurrent that is twice the dark current? What is the QE of the photodetector at 1.55 μm?

(b) What would be the photocurrent if the incident power in (a) was at 1.3 μm? What is the QE at 1.3 μm operation?

Figure 5.49 The responsivity of an InGaAs pin photodiode

Solution (a) At λ = 1.55×10-6 m, from the responsivity vs. wavelength curve we have R ≈ 0.87 A/W as in Figure 5Q07-1. From the definition of responsivity,

o

ph

PI

==(W) Power OpticalIncident

(A)nt PhotocurreR

we have Po = Iph/R = 2Idark/R = (2×5×10-9 A)/(0.87 A W-1) = 1.15×10-8 W or 11.5 nW.

From the definitions of quantum efficiency (QE) ηe and responsivity we have

hce

he

eeλη

υη ==R

∴ )m 1055.1)(C 106.1(

)W A 87.0)(s m 103)(sJ 1062.6(619

-1-1834

−−

−

××××

==λ

ηe

hce

R≈ 0.70 (70%)

Note the following dimensional identities: A = C s-1 and W = J s-1 so that A W-1 = C J-1. Thus, responsivity in terms of photocurrent per unit incident optical power is also charge collected per unit incident energy.

Figure 5Q17-1 Responsivity of InGaAs photodiode at two wavelengths

(b) At λ = 1.3×10-6 m, from the responsivity vs. wavelength curve we have R = 0.82 A/W as in Figure 5Q07-1. Since Po is the same and 11.5 nW as in (a),

Iph = R Po = (0.82 A W-1)(1.15×10-8 W) = 9.43×10-9 A or 9.43 nA.

The QE at λ = 1.3 μm is

)m 103.1)(C 106.1(

)W A 82.0)(s m 103)(sJ 1062.6(619

-1-1834

−−

−

××××

==λ

ηe

hce

R≈ 0.78 (78%)

5.13 Transient photocurrents in a pin photodiode Consider a reverse biased Si pin photodiode As shown in Figure 5.51. It is appropriately reverse biased so that the field in the depletion region (i-Si layer) is E = Vr/W is the saturation field. Thus, photogenerated electrons and holes in this layer drift at saturation velocities vde and vdh . Assume that the field E is uniform and that the thickness of the p+ is negligible. A very short light pulse (infinitesimally short) photogenerates EHPs in the depletion layer as shown in Figure 5.51 which results in an exponentially decaying EHP concentrations across W. Figure 5.51 shows the photogenerated electron concentration at time t = 0 and also at a later time t when the electrons have drifted a distance Δx = vdeΔt. Those that reach the back electrode B become collected. The electron distribution shifts at a constant velocity until the initial electrons at A reach B which represents the longest transit time τe = W/vde. Similar argument apply to holes but they drift in the opposite direction and their transit time τh = W/vdh where vdh is their saturation velocity. The photocurrent density at any instant is

dhhdeeheph eNeNtjtjj vv +=+= )()(

where Ne and Nh are the overall electron and hole concentration in the sample at time t. Assume for convenience that the cross sectional area A = 1 below (derivations are not affected as we are interested in the photocurrent current densities).

(a) Sketch the hole distribution at time τh > t > 0 where τh = hole drift time = W/vdh. (b) The electron concentration distribution n(x) at time t corresponds to that at t = 0 shifted by vdet. Thus the total electrons in W is proportional to integrating this distribution n(x) from A at x = vdet to B at x = W.

Given n(x) = noexp(−αx) at t = 0, where no is the electron concentration at x = 0 at t = 0 we have

∫ −−=W

tdeo

de

dxtxntv

v )](exp[ at time electronsnumber Total α

and Volume

at time electronsnumber Total)( ttNe =

Then

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−−=

−−= ∫

e

o

W

tdeoe

tWWn

dxtxnW

tNde

τα

α

α

1exp1

)](exp[1)(v

v

Ne(0) is the initial overall electron concentration at time t = 0, that is,

)]exp(1[)exp(1)0(0

WWndxxn

WN o

W

oe αα

α −−=−= ∫

We note that no depends on the intensity I of the light pulse so that no ∝ I. Show that for holes,

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

−= 11exp)exp()(

h

oh

tWW

WntNτ

αα

α

(c) Given W = 40 μm, α = 5×104 m-1, vde = 105 m s-1, vdh = 0.8×105 m s-1, no = 1013 cm-3, calculate the electron and hole transit time, sketch the photocurrent densities je(t) and jh(t) and hence jph(t) = as a function of time, and calculate the initial photocurrent. What is your conclusion?

Solution (a) The hole distribution is shown in Figure 5Q13-1.

Figure 5Q13-1 An infinitesimally short light pulse is absorbed throughout the

depletion layer and creates an EHP concentration that decays exponentially

(b) For holes we have

[ ]

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

−−−=

+−= ∫−

11exp)exp(

)exp()exp(

)](exp[1)(0

h

o

dho

tW

dhoh

tWW

Wn

WtWn

dxtxnW

tN dh

τα

αα

ααα

α

v

vv

Having described Ne(t) and Nh(t) we can find the individual transient photocurrent densities as follows,

je(t) = eNe(t)vde

∴ ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−−=

e

deoe

tWW

entjτ

αα

1exp1)( v t < τe

and jh(t) = eNh(t)vdh

∴ ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

−= 11exp)exp()(

h

deoh

tWW

Wentjτ

αα

αv t < τe

(c) Given W = 40 μm, α = 5×104 m-1, vde = 105 m s-1, vdh = 0.8×105 m s-1, no = 1013 cm-3, we have

τe = W/vde = 4×10-10 s = 400 ps

and τh = W/vdh = 4×10-10 s = 500 ps

Using these values we can plot the photocurrents as in the figure below.

At time t = 0, Ne(0) = Nh(0)

{ })]cm 500)(cm 1040(exp[1

)cm 500)(cm 1040()cm 10(

)]exp(1[)0(

1-41-4

3-13−

− ×−−×

=

−−= WWnN o

e αα

= 4.3×1012 cm-3.

The initial photocurrent densities are

je(0) = eNe(0)vde = (1.6×10-19 C)(4.3×1012 cm-3)(1×107 cm s-3)

= 6.9 A cm-2 or 69 mA/mm2

and,

jh(0) = eNh(0)vdh = (1.6×10-19 C)(4.3×1012 cm-3)(0.8×107 cm s-3)

= 5.5 A cm-2 or 55 mA/mm2

so that the total initial photocurrent is 55 + 69 = 124 mA/mm2.

The photocurrent densities je(t) and jh(t) are shown in Figure 5Q13-2

Figure 5Q13-2 Transient photocurrent from a pin photodiode pulsed with a very

short duration light absorbed throughout the device.

Clearly the response is determined by the slowest transit time. There is a kink in the photocurrent waveform when all the electrons have been swept out at τe = 400 ns.

5.14 Fiber attenuation and InGaAs pin Photodiode Consider the commercial InGaAs pin photodiode whose responsivity is shown in Figure 5.49. This is used in a receiver circuit that needs a minimum of 5 nA photocurrent for a discernible output signal (acceptable signal to noise ratio for the customer). Suppose that the InGaAs pin PD is used at 1.3 μm operation with a single mode fiber whose attenuation is 0.35 dB km-1. If the laser diode emitter can launch at most 2 mW of power into the fiber, what is the maximum distance for the communication without a repeater?

Solution

Figure 5Q14-1 The responsivity of an InGaAs pin photodiode at 1.3 μm is 0.81 A/W

Given, photocurrent Iph = 5 nA = 5×10-9 A.

Responsivity, R = 0.81 A/W at 1.3 μm wavelength from Figure 5.49, reproduced in Figure 5Q14-1

Power absorbed by photodiode, Po= Iph/R = (5×10-9 A)/(0.81 A/W)= 6.173×10-9 W.

If the maximum distance for the communication without a repeater = L in kilometers, then, from the definition of the attenuation coefficient in dB/km

(1/L)10log10(Po/Pin) = αdB/km

or (1/L)10log10(Po/Pin) = (− 0.35 dB/km)

or, ( ) km157=⎥⎦

⎤⎢⎣

⎡×

×== −

−

W10173.6W102log

35.010log

35.010

9

3

1010 oin PPL

5.21 Photoconductive detector An n-type Si photoconductor has a length L = 100 μm and a cross sectional area A = 10-4 mm2. The applied bias voltage to the photoconductor is 10 V. The recombination time is roughly 1 μs. [Author's note: The last sentence was missed in the text and will be included in the reprint.]

(a) What are the transit times, te and th, of an electron and a hole across L? What is the photoconductive gain?

(b) It should be apparent that as electrons are much faster than holes, a photogenerated electron leaves the photoconductor very quickly. This leaves behind a drifting hole and therefore a positive charge in the semiconductor. Secondary (i.e. additional electrons) then flow into the photoconductor to maintain neutrality in the sample and the current continues to flow. These events will continue until the hole has disappeared by recombination, which takes on average a time τ. Thus, more charges flow through the contact per unit time than charges actually photogenerated per unit time. What will happen if the contacts are not ohmic, i.e. they are not injecting ?

(c) What can you say about the product Δσ and the speed of response which is proportional to 1/τ.

Solution (a) We are given, length L = 100 μm, and applied bias, V = 10 V. The electric field,

E = V/L= 10 V/(100×10-6 m)= 105 V/m.

From Table 3.1, electron and hole drift nobilities are: μe = 1450 cm2 V-1 s-1, and μh = 490 cm2 V-1 s-1 (see inside front cover of textbook).

Electron and hole transit times are

( ))V/m10)(sVm101450(

m1010051-124

6

−−

−

××

==E

Lte

e μ= 6.9 ns.

and ( ))V/m10)(sVm10490(

m1010051-124

6

−−

−

××

==E

Lth

h μ= 22.4 ns.

hole lifetime τ = 1 μs The photoconductive gain,

( ) ( )( )( )m10100

)V/m10(sVm10490sVm101450s1016

5-1124-11246

−

−−−−−

××+××

=+

=L

EG he μμτ =

194

(b) If the contacts are not ohmic, secondary electrons cannot flow into the photoconductor to maintain neutrality. So, only the photogenerated charges can flow through the external circuit; no excess charge can flow and we will not get photoconductive gain. If the contacts cannot inject carriers, then there will be no photocurrent gain, G = 1. However, there will still be a photocurrent as photogereated carriers will be drifting. The situation is similar to photogeneration inside the i-layer of a reverse biased pin detector.

(c) The expression for Δσ is given by,

( )hcd

e he μμλτησ +=Δ

I

The speed of response is proportional to 1/τ. For example, if we suddenly switch off the light, it will take, on average, τ seconds for the excess (photogenerated) carriers to disappear by recombination. Therefore, the product of Δσ and the speed of response is,

( )hcd

e he μμλητ

σ +=Δ

I)1)((

which is a constant for given light intensity (I) and given device structure (d).

7.1 Polarization Suppose that we write the Ey and Ex components of a light wave generally as:

Ex = Exocos(ωt − kz ) and Ey = Eyocos(ωt − kz + φ)

Show that at any instant Ex and Ey satisfy the ellipse equation on the Ey vs Ex coordinate system:

Ex

Exo

⎛

⎝ ⎜ ⎞

⎠ ⎟

2

+Ey

Eyo

⎛

⎝ ⎜ ⎞

⎠ ⎟

2

−2Ex

Exo

⎛

⎝ ⎜ ⎞

⎠ ⎟ Ey

Eyo

⎛

⎝ ⎜ ⎞

⎠ ⎟ cosφ = sin2 φ

Sketch schematically what this ellipse looks like assuming Exo = 2 Eyo. When would this ellipse form an (a) ellipse with its major axis on the x-axis, (b) a linearly polarized light at 45°, (c) right and left circularly polarized light ?

Solution

Consider the LHS,

Ex

Exo

⎛ ⎝ ⎜ ⎞

⎠ ⎟

2

+Ey

Eyo

⎛

⎝ ⎜

⎞

⎠ ⎟

2

−2 Ex

Exo

⎛ ⎝ ⎜ ⎞

⎠ ⎟ Ey

Eyo

⎛

⎝ ⎜

⎞

⎠ ⎟ cosφ

=Exo cos(ωt − kz)

Exo

⎛ ⎝ ⎜ ⎞

⎠ ⎟

2

+Eyo cos(ωt − kz + φ)

Eyo

⎛

⎝ ⎜

⎞

⎠ ⎟

2

−2Exo cos(ωt − kz)

Exo

⎛ ⎝ ⎜ ⎞

⎠ ⎟ Eyo cos(ωt − kz + φ)

Eyo

⎛

⎝ ⎜

⎞

⎠ ⎟ cosφ

= cos2 (ωt − kz) + cos2 (ωt − kz + φ) − 2cos(ωt − kz)cos(ωt − kz + φ)cosφ

= cos2 (ωt − kz) + [cos(ωt − kz)cosφ − sin(ωt − kz)sinφ]2

−2cos(ωt − kz)[cos(ωt − kz)cosφ − sin(ωt − kz)sinφ]cosφ= cos2 (ωt − kz) + cos2 (ωt − kz)cos2 φ + sin2(ωt − kz)sin2 φ

−2cos(ωt − kz)sin(ωt − kz)cosφ sinφ

−2cos2(ωt − kz)cos2 φ + 2cos(ωt − kz)sin(ωt − kz)sinφ cosφ

= cos2 (ωt − kz) − cos2(ωt − kz)cos2 φ + sin2 (ωt − kz)sin2 φ= cos2 (ωt − kz)[1− cos2 φ] + sin2 (ωt − kz)sin2 φ

= cos2 (ωt − kz)sin2 φ + sin2(ωt − kz)sin2 φ = [cos2(ωt − kz) + sin2(ωt − kz)]sin2 φ

= sin2 φ

Which is the desired result.

Consider the general expression

Ex

Exo

⎛

⎝ ⎜ ⎞

⎠ ⎟

2

+Ey

Eyo

⎛

⎝ ⎜ ⎞

⎠ ⎟

2

−2Ex

Exo

⎛

⎝ ⎜ ⎞

⎠ ⎟ Ey

Eyo

⎛

⎝ ⎜ ⎞

⎠ ⎟ cosφ = sin2 φ

This is a quadratic equation in Ey (or Ex),

aEy2 + bEy + c = 0

where

a = 1Eyo

2 ; b = −2 Ex

Exo

⎛ ⎝ ⎜ ⎞

⎠ ⎟ 1

Eyo

⎛

⎝ ⎜

⎞

⎠ ⎟ cosφ; c = Ex

Exo

⎛ ⎝ ⎜ ⎞

⎠ ⎟

2

− sin2 φ

Thus, for a given φ and given Exo and Eyo, we can plot Ey vs. Ex since Ex determines b and c and Ey is given by

Ey = −b ± b2 − 4ac

2a

Figure 7Q1 shows the plots of Ey vs Ex for various selections of Exo, Eyo and φ. Note that the angle α of the major axis a is not φ, though related to it. (a). Setting φ = π/2, Eyo = 1 and Exo = 2 we obtain an ellipse with its major axis on the x-axis. (b) Setting φ = 0, Eyo = 1 and Exo = 1 we obtain a line (linear polarization) at an angle π/4 (= 45°) to the x-axis.

(c) Setting φ =π/2, Eyo = 1 and Exo = 1 we obtain a circle. If φ = π/2 right circular polarization; φ = −π/2 left circular polarization.

φ = 90 °

Ex

Ey

Major axis

Minor axis

α

Eyo = 1

Exo = 2

Ey

Ex

Polarization ellipse for φ = 30°.

φ = 30 °

Ey

Ex

φ = 90 °

Major axis

Polarization ellipse with major axis on Ex.

φ = 0 °Eyo = Exo = 1

α = 45 °

Ex0 2-2

1

0

-1

1

0

-1

1

0

-1

0 2-2 0 2-2

1

0

-1

Ey

Linear polarization at 45° to the x-axis.

Eyo = Exo = 1

-1 0 1Circular polarization.

Figure 7Q1

7.2 Linear and circular polarization Show that a linearly polarized light wave can be represented by two circularly polarized light waves with opposite rotations. Consider the simplest case of a wave linearly polarized along the y-axis. What is your conclusion?

Solution As shown in Figure 7Q2, two opposite circular polarizations in (a) and (b), with the same field magnitude, and obviously with the same frequency, can be added to generate the linearly polarized light in (c).

E

y

x

Exo = 0Eyo = 1φ = 0

y

x

Exo = 1Eyo = 1φ = π/2

E

y

x

Exo = 1Eyo = 1φ = −π/2

(c)(a) (b)

Opposite circular polarizations in (a) and (b) add to give thelinearly polarized light in (c)

Figure 7Q2 Alternatively, consider the following: We can represent the right circular polarization as

ERx = cos(ωt − kz); ERy = cos(ωt − kz + 12 π )

and the left circular polarization as

ELx = cos(ωt − kz); ELy = cos(ωt − kz − 12 π )

Thus, the total x and y components due to combining these two opposite circular polarizations is

Ex = 2cos(ωt − kz)Ey = cos(ωt − kz + 1

2 π ) + cos(ωt − kz − 12 π )

∴ Ex = 2cos(ωt − kz)Ey = 2cos(ωt − kz)cos(π /2) = 0

which is linear polarization since only Ex exists.

7.5 Jones Matrices When we represent the state of polarization of a light wave using a matrix, called a Jones matrix2 (or vector) then various operations on the polarization state correspond to multiplying this matrix with another matrix that represents the optical operation. Consider a light wave traveling along z with field components Ex and Ey along x and y. These components are orthogonal and, in general, would be of different magnitude and have a phase difference φ between them . If we use the exponential notation then

Ex = Exoexp j(ωt − kz + φx) and Ey = Eyoexpj(ωt − kz + φy)

Jones matrix is a column matrix whose elements are Ex and Ey without the common expj(ωt − kz) factor

2 R. Clark Jones introudced these matrices circa 1941.

E =Ex

Ex

⎡ ⎣ ⎢

⎤ ⎦ ⎥ =

Exo exp( jφx )Eyo exp( jφy )

⎡

⎣ ⎢ ⎤

⎦ ⎥ (1)

Usually Eq. (1) is normalized by dividing the total amplitude Eo = (Exo2 + Eyo

2 )1/ 2 . We can further factor out exp(jφx) to further simplify to obtain the Jones matrix:

J =1Eo

Exo

Eyo exp( jφ)⎡

⎣ ⎢ ⎤

⎦ ⎥ (2)

where φ = φy − φx. We can further arbitrarily set Exo = 1 and adjust the size of Eyo so that the total Eo remains the same.

a Table 7T5 shows Jones vectors for various polarizations. Identify the state of polarization for each matrix.

b Passing a wave of given Jones vector Jin through an optical device is represented by multiplying Jin by the transmission matrix T of the device. If Jout is the Jones vector for the output light through the device, then Jout = T Jin. Given

T =1 00 j

⎡ ⎣ ⎢

⎤ ⎦ ⎥ (3)

determine the polarization state of the output wave and the optical operation represented by T.

Table 7T5

Jones vectors

Jones vector 10

⎡ ⎣ ⎢

⎤ ⎦ ⎥

12

11

⎡ ⎣ ⎢

⎤ ⎦ ⎥

cosθsinθ

⎡ ⎣ ⎢

⎤ ⎦ ⎥

12

1j

⎡ ⎣ ⎢

⎤ ⎦ ⎥

12

1− j

⎡ ⎣ ⎢

⎤ ⎦ ⎥

Polarization

Transmission

matrix

1 00 0

⎡ ⎣ ⎢

⎤ ⎦ ⎥

e jφ 00 e jφ

⎡

⎣ ⎢ ⎤

⎦ ⎥ 1 00 j

⎡ ⎣ ⎢

⎤ ⎦ ⎥

1 00 −1

⎡ ⎣ ⎢

⎤ ⎦ ⎥

1 00 e− jΓ

⎡ ⎣ ⎢

⎤ ⎦ ⎥

Optical operation

Solution

Jones vectors

Jones vector 10

⎡ ⎣ ⎢

⎤ ⎦ ⎥

12

11

⎡ ⎣ ⎢

⎤ ⎦ ⎥

cosθsinθ

⎡ ⎣ ⎢

⎤ ⎦ ⎥

12

1j

⎡ ⎣ ⎢

⎤ ⎦ ⎥

12

1− j

⎡ ⎣ ⎢

⎤ ⎦ ⎥

Polarization Linear; horizontal E

Linear; E at

45° to x-axis

Linear; E at

θ to x-axis

Right circularly

Left circularly

polarized polarized

Transmission

matrix

1 00 0

⎡ ⎣ ⎢

⎤ ⎦ ⎥

e jφ 00 e jφ

⎡

⎣ ⎢ ⎤

⎦ ⎥ 1 00 j

⎡ ⎣ ⎢

⎤ ⎦ ⎥

1 00 −1

⎡ ⎣ ⎢

⎤ ⎦ ⎥

1 00 e− jΓ

⎡ ⎣ ⎢

⎤ ⎦ ⎥

Optical operation Linear polarizer; horizontal transmission axis

Isotropic phase changer

Quarter-wave plate

Half-wave plate

Wave retarder; fast axis along x

7.11 Glan-Foucault prism Figure 7Q11-1 shows the cross section of a Glan-Foucault prism which is made of two right angle calcite prisms with a prism angle of 38.5°. Both have their optic axes parallel to each other and to the block faces as in the figure. Explain the operation of the prisms and show that the o-wave does indeed experience total internal reflection.

Optic axis

38.5°

Absorber

Air-gap

Calcitee-ray

o-ray

The Glan-Foucault prism provides linearly polarized light Figure 7Q11-1

Solution As shown in Figure 7Q11-1 the light in the left prism travel as o- and e-waves (with field Eo and Ee, with refractive indices no = 1.658 and ne = 1.486. The critical angles for TIR for the o- and e-waves at the calcite/air interface are

θc(o-wave) = arcsin(1/no) = 37.1°.

and θc(e-wave) = arcsin(1/ne) = 42.3°.

If the angle of incidence is θ at the calcite/air interface then from Figure 7Q11-2,

90° + (90° − θ ) + 38.5° = 180°

or θ = 38.5° > θc(o-wave)

but θ = 38.5°< θc(e-wave)

Thus, the o-wave suffers TIR while the e-wave does not. Hence the beam that emerges is the e-wave, with a field Ee along the optic axis.

Optic axis

38.5°

Absorber

Air-gap

Calcite

e-ray

o-ray

The Glan-Foucault prism provides linearly polarized light

Eo

Ee

θ

38.5°

Figure 7Q11-2

7.12 Faraday Effect Application of a magnetic field along the direction of propagation of a linearly polarized light wave through a medium results in the rotation of the plane of polarization. The amount of rotation θ is given by

θ = ϑBL

where B is the magnetic field (flux density), L is the length of the medium, and ϑ is the so-called Verdet constant. It depends on the material and the wavelength. In contrast to optical activity, sense of rotation of the plane of polarization is independent of the direction of light propagation. Given that glass and ZnS have Verdet constants of about 3 and 22 minutes of arc Gauss-1 meter-1 at 589 nm respectively, calculate the necessary magnetic field for a rotation of 1° over a length 10 mm. What is the rotation per unit magnetic field for a 10 mm medium? (Note: 60 minutes of arc = 1° and 1 Gauss = 10-4 Tesla).

Solution Rotation per unit magnetic field

Glass: B =θ

ϑL=

6 ′ 0 ( ′ 3 Gauss−1m−1)(10 ×10−3 m)

= 2000 Gauss = 0.2 Tesla

ZnS: B =θ

ϑL=

6 ′ 0 (22 Gauss−1m−1)(10 ×10−3 m)

= 273 Gauss = 0.027 Tesla

NOTE: There is a typo in the Verdet constant units, which should have been in Gauss not Tesla. reprint will have this corrected.

7.15 Transverse Pockels cell with LiNbO3 Suppose that instead of the configuration in Figure 7.20, the field is applied along the z-axis of the crystal, the light propagates along the y-axis. The z-axis is the polarization of the ordinary wave and x-axis that of the extraordinary wave. Light propagates through as o- and e-waves. Given that Ea = V/d, where d the crystal length along z, the indices are

′ n o ≈ no + 12 no

3r13Ea and ′ n e ≈ ne + 12 ne

3r33Ea

Show that the phase difference between the o- and e-waves emerging from the crystal is,

Δφ = φe − φo =2πLλ

ne − no( )+2πLλ

12

ne3r33 − no

3r13( )Vd

where L is the crystal length along the y-axis.

Explain the first and second terms. How would you use two such Pockels cells to cancel the first terms in the total phase shift for the two cells.

If the light beam entering the crystal is linearly polarized in the z-direction, show that

Δφ =2πne L

λ+

2πLλ

(ne3r33)2

Vd

Consider a nearly monochromatic light beam of the free-space wavelength λ = 500 nm and polarization along z-axis. Calculate the voltage Vπ needed to change the output phase Δφ by π given a LiNbO3 crystal with d/L = 0.01 (see table 7.2).

Solution Consider the phase change between the two electric field components,

Δφ = φe − φo =2πLλ

ne − no( )+2πLλ

12

ne3r33 − no

3r13( )Vd

The first term is the natural birefringence of the crystal, just as in the calcite crystal, and occurs all the time, even without an applied field. The second term is the Pockels effect, applied field inducing a change in the refractive indices. Figure 7Q15 shows how two Pockels cells may be used to cancel the first terms in the combined system.

Two tranverse Pockels cell phase modulators together cancel the naturalbirefringence in each crystal.

y

x

dL

Ea

Δφ

y

Ea

z

x

d

L

zLight Light

Figure 7Q15

If the light beam is linearly polarized with its field along z, we only need to consider the extraordinary ray, thus we can set φo = 0.

Δφ = φe =2πLne

λ+

2πLλ

12

ne3r33( )V

d

The first term does not depend on the voltage. The voltage V that changes the output phase by π is

2πL

λ12

ne3r33( )V

d= π

or V =dL

λne

3r33

⎛

⎝ ⎜ ⎞

⎠ ⎟ = (0.01)

500 ×10−9 m(2.187)3 (30.8 ×10−12 m/V)

⎛

⎝ ⎜ ⎞

⎠ ⎟

∴ V = 15.5 V

7.21 Optical Kerr effect Consider a material in which the polarization does not have the second order term:

P = εoχ1E + εoχ3E3 or P/(εoE) = χ1 + χ3E2

The first term with the electric susceptibility χ1 corresponds to the relative permittivity εr and hence to the refractive index no of the medium in the absence of the third order term, i.e. under low fields. The E2 term represents the irradiance I of the beam. Thus, the refractive index depends on the intensity of the light beam, a phenomenon called the optical Kerr effect:

n = no + n2I and n2 =3ηχ3

4no2

and η = (μo/εo)1/2 = 120π = 377 Ω, is the impedance of free space.

a Typically, for many glasses, χ3 ≈ 10-21 m2/W; for many doped glasses, χ3 ≈ 10-18 m2/W; for many organic substances, χ3 ≈ 10-17 m2/W; for semiconductors, χ3 ≈ 10-14 m2/W. Calculate n2 and the intensity of light needed to change n by 10-3 for each case.

b The phase φ at a point z is given by

φ =ωot −

2πnλ

z = ωot −2π [no + n2I]

λz

It is clear that the phase depends on the light intensity I and the change in the phase along Δz due to light intensity alone is

Δφ =

2πn2Iλ

Δz

As the light intensity modulates the phase, this is called self-phase modulation. Obviously light is controlling light.

When the light intensity is small n2I << no, obviously the instantaneous frequency

ω = ∂φ/∂t = ωo.

Suppose we have an intense beam and the intensity I is time dependent I = I(t). Consider a pulse of light traveling along the z-direction and the light intensity vs. t shape is a “Gaussian” (this is approximately so when a light pulse propagates in an optical fiber,

for example). Find the instantaneous frequency ω. Is this still ωo? How does the frequency change with “time”, or across the light pulse? The change in the frequency over the pulse is called chirping. Self-phase modulation therefore changes the frequency spectrum of the light pulse during propagation. What is the significance of this result?

c Consider a Gaussian beam in which intensity across the beam cross section falls with radial distance in a Gaussian fashion. Suppose that the beam is made to pass through a plate of nonlinear medium. Explain how the beam can become self-focused? Can you envisage a situation where diffraction effects trying to impose divergence are just balanced by self-focusing effects?

Solution a The fractional change in the refractive index is

δn =

n − no

no

=n2Ino

=3ηχ3I

4no3

∴ I =

δ n 4no3

3ηχ3

We need δn = 10-3 or 0.1%,

Glasses: no ≈ 1.5, χ3 = 10-21, I = 1.2×1016 W m-2

Doped glasses: no ≈ 1.5, χ3 = 10-18, I = 1.2×1013 W m-2

Organics: no ≈ 1.5, χ3 = 10-17, I = 1.2×1012 W m-2

Semiconductors: no ≈ 2.5, χ3 = 10-14, I = 5.5×109 W m-2

Large intensities.

b The phase φ at a point z is given by

φ = ωot −

2πnλ

z = ωot −2π[no + n2I]

λz

so that the instantaneous frequency is

ω(t) =

∂φ∂t

=ωo −2πzn2

λ∂I∂t

When the light intensity is small n2I << no, obviously ω = ∂φ/∂t = ωo. However, suppose we have an intense beam and the intensity I is time dependent I = I(t). Consider a pulse of light traveling along the z-direction and the light intensity vs. t shape is a “Gaussian” (this is approximately so when a light pulse propagates in an optical fiber, for example). Then the instantaneous frequency is,

ω(t) =

∂φ∂t

=ωo −2πzn2

λ∂I∂t

so that the frequency changes with “time”, or across the light pulse. Since ∂I/∂t is rising at the leading edge and falling in the trailing edge, the two ends of the pulse contain

different frequencies! The change in the frequency over the pulse is called chirping. Self-phase modulation therefore changes the frequency spectrum of the light pulse during propagation. The second term above results in the broadening of the frequency spectrum and hence leads to more dispersion for a Gaussian pulse propagating in an optical fiber.

c If an intense Gaussian optical beam is transmitted through a plate of nonlinear medium, it will change the refractive index of the medium with the maximum change of the refractive index in the center of the beam. Therefore the plate will act as a graded-index medium and change the curvature of the wavefront. Under certain conditions the plate can act as a lens with a power-dependent focal length, producing a co-called self-focusing of the beam. Similarly, if an intense Gaussian beam propagates through a nonlinear medium, the medium can act as a graded-index waveguide. In this case, under certain conditions, the self-focusing effect can compensate the divergence of the beam due to diffraction, and the beam will be confined to its self-created waveguide. Such self-guided beams are called spatial solitons.

I

yNonlinear medium

z

Intensity across beamcross section

The intensity variation acrossthe beam cross section leads toa similar refractive indexvariation in the nonlinearmedium. Thus, the mediumresembles a graded index guideor a GRIN rod and can focusthe beam.

Figure 7Q21

"If your experiment needs statistics, you ought to have done a better experiment."

Ernest Rutherford (1871-1937)