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Solutions Manual for Optoelectronics and Photonics: Principles and Practices 5 March 2002

S.O. Kasap

1.1

Chapter 1 1 . 1 Gaussian beams Consider two identical spherical mirrors A and B that have been aligned to be focal directly face each other as in Figure 1Q1. The two mirrors and the space in between them (the optical cavity) form an optical resonator because only certain light waves with certain frequencies can exit in the optical cavity. The light beam in the cavity is a Gaussian beam. When it starts at A its wavefront is the same as the curvature of A. Sketch the wavefronts on this beam as it travels towards B, at B, as it is then reflected from B back to A. If R =25 cm, and the mirrors are of diameter 2.5 cm, estimate the divergence of the beam and its spot size (minimum waist) for light of wavelength 500 nm. L Spherical mirror

Spherical mirror Optical cavity F

A

B

R

R

Wave front

Two confocal spherical mirrors reflect waves to and from each other. F is the focal point and R is the radius. The optical cavity contains a Gaussian beam Figure 1Q1 Solution Let D = diameter of the mirror, from Figure 1Q1, tan = (D/2)/R gives = arctan(D/2R) = arctan(0.05) = 0.05 rad. Divergence is 2 or 0.1 rad. Divergence 2 and spot size 2wo are related by 2

4 (2 wo )

and depends on the wavelength of interest. Taking 4 (2 )

and,

2wo

1.2

Refractive index

4(500 10 9 m) ( 0. 1)

500 nm,

6 6. 4 10 m or about 6 micron.

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?

Solutions Manual for Optoelectronics and Photonics: Principles and Practices 5 March 2002

S.O. Kasap

1.2

b What is the Brewster angle (the polarization angle p) and the critical angle ( c) 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.

curves, at

= 1300 nm

n = 1.447 and Ng = 1.462 Phase velocity :

v = c/n = (3 10 8 m s-1) / 1.447 = 2.073 10 8 m s-1

Group velocity:

vg = c/N g = (3 10 8 m s-1) / 1.462 = 2.052 10 8 m s-1

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

The polarization (the Brewster) angle is p

= 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, c

c

For light traveling in glass incident on the glass-air interface at normal incidence, n1 n2 1 .447 1 r r r = 0.183 // n1 n2 1 .447 1 Thus,

d

i.e.

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

R = r2 = (0.183)2 = 0.0335

For light traveling in air incident on the air-glass interface at normal incidence, n1 n2 1 1 .447 = r r r 0.183 // n1 n2 1 1 .447 R = r 2 = ( 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 Refractive index and the Sellmeier dispersion equation The Sellmeier dispersion equation is an empirical expression for the refractive index of glass in terms of the wavelength , n2 1

G1 2

2

G2 2 1

2

2 2 2

G3 2

2 2 3

Sellmeier equation

(1)

Solutions Manual for Optoelectronics and Photonics: Principles and Practices 5 March 2002

S.O. Kasap

1.3

where G1, G2, G3 and 1, 2 and 3 are constants (called Sellmeier coefficients) that are determined by fitting this expression to the experimental data. The actual Sellmeier formula has more terms in the right 2 2 hand summation of the same type e.g. Gi 2/( i ) where i = 4, 5,... but these can generally be neglected in representing n vs. behavior over typical wavelengths of interest and ensuring that three terms included in Eq. (1) correspond to the most important or relevant terms in the summation. Table 1Q3 gives the Sellmeier coefficients for pure Silica (SiO2) and SiO2-13.5 mol.% GeO2. Write a program on your computer or calculator, or use a math software package (e.g. Mathcad, Matlab, Theorist, Mathview) or even a spread sheet program (e.g. Excel) to obtain the refractive index n as a function of from 0.5 m to 1.8 m for both pure silica and SiO2-13.5%GeO2. Obtain the group index, Ng, vs. wavelength for both materials and plot it on the same graph. Find the wavelengths where material dispersion becomes zero in the two materials. Table 1Q3 The Sellmeier coefficients for SiO2 and SiO2-13.5%GeO2. The

1

,

2

,

3

are in m.

Si O 2

G 0.696749 1

G 0.408218 2

G 0.890815 3

0.0690660 1

0.115662 2

9.900559 3

SiO 2 -13.5%GeO 2

0.711040

0.451885

0.704048

0.0642700

0.129408

9.425478

Solution

Silica; n vs ( m), thin curve; N g vs ( m), thick bold curve. N g vs ( m ) Minimum is around 1.3 m .

SiO 2 -13.5 mol.% GeO 2 : n vs ( m), thin curve; N g vs ( m), thick bold curve. N g vs ( m) Minimum is around 1.4 m .

1 . 4 Antireflection coating a Consider three dielectric media with flat and parallel boundaries with refractive indices n1, n2 and n3. Show that for normal incidence the reflection coefficient between layers 1 and 2 is the same as that between layers 2 and 3 if n2 = [n1n3]. What is the significance of this? b Consider a Si photodiode that is designed for operation at 900 nm. Given a choice of two possible antireflection coatings, SiO2 with a refractive index of 1.5 and TiO2 with a refractive index of 2.3 which would you use and what would be the thickness of the antireflection coating? The refractive index of Si is 3.5. Explain your decision.

Solutions Manual for Optoelectronics and Photonics: Principles and Practices 5 March 2002

S.O. Kasap

1.4

Solution For light traveling in medium 1 incident on the 1-2 interface at normal incidence,

r12

n1 n1

n2 n2

n1 n1

n1n 3 n1n 3

1 1

n3 n1 n3 n1

For light traveling in medium 2 incident on the 2-3 interface at normal incidence,

r23

thus,

n2 n2

n3 n3

n1n3 n1n3

n3 n3

n1 1 n3 n1 1 n3

1 1

n3 n1 n3 n1

r23 = r12

Significance? For an efficient antireflection effect, waves A (reflected at 1-2) and B (reflected at 23) in Figure 1Q4 below should interfere with near “total destruction”. That means they should have the same magnitude and that requires that the reflection coefficient between 1 and 2 should be the same as that between 2 and 3; r12 = r23. Thus, the layer 2 can act as an antireflection coating if its index n2 = (n1n3)1/2. This can be achieved by r12 = r23.

n1

d n2

n3

A B Surface

Antireflection Semiconductor of photovoltaic device coating

Illustration of how an antireflection coating reduces the reflected light intensity Figure 1Q4 The best antireflection coating has to have a refractive index n2 such that n2 = (n1n3)1/2 = [(1)(3.5)]1/2 = 1.87. Given a choice of two possible antireflection coatings, SiO2 with a refractive index of 1.5 and TiO2 with a refractive index of 2.3, both are close . The phase change for wave B going through the coating of thickness d is 2k2d where k2 = n2ko and ko = wavevector in free space = 2 / . This should be 180 or . Thus we need 2n2(2 / )d = or For SiO2

d

4 n2

900 10 9 m = 0.15 4 (1 .5)

m

Solutions Manual for Optoelectronics and Photonics: Principles and Practices 5 March 2002

For TiO2

d

4 n2

900 10 9 m = 0.10 4 (2 .3)

S.O. Kasap

1.5

m

1.5 Reflection at glass-glass and air-glass interfaces A ray of light which is traveling in a glass medium of refractive index n1 = 1.460 becomes incident on a less dense glass medium of refractive index n2 = 1.430. Suppose that the free space wavelength of the light ray is 850 nm a

What should be the minimum incidence angle for TIR?

b 90 ?

What is the phase change in the reflected wave when the angle of incidence i = 85 and when

c 90 ?

What is the penetration depth of the evanescent wave into medium 2 when i = 85 and when

=

i

i

=

d What is the reflection coefficient and reflectance at normal incidence ( i = 0 ) when the light beam traveling in the silica medium (n = 1.455) is incident on a silica/air interface? e What is the reflection coefficient and reflectance at normal incidence when a light beam traveling in air is incident on an air/silica (n = 1.455) interface? How do these compare with part (c) and what is your conclusion? Solution a What should be the minimum incidence angle for TIR? The critical angle for TIR is, c

= arcsin(n2/n1) = arcsin(1.430/ 1.460) = 78.4 .

b What is the phase change in the reflected wave when the angle of incidence = 90 ?

i

= 85 and when

i

n = n2/n1 = 1.430/1.460 = 0.97945,

tan

sin2

1 2

i

cos

n2

1/ 2

i

1.430 sin ( 85 ) 1.460 o cos(85 ) 2

2 1/ 2

o

= 2.0868 = tan[1/2(128.79 )] = 128.8 .

Thus,

For the Er,// component, the phase change is tan

so that

which gives

1 2

tan(1/2

//

=

//

//

1 2

sin 2 i n2 n2 cos i

1/ 2

1 tan n2

1 2

+ 1/2 ) = (n1/n2)2tan( /2) = (1.460/1.430)2tan(1/2128.8 ) = tan(1/2130.6 )

49.4 .

Solutions Manual for Optoelectronics and Photonics: Principles and Practices 5 March 2002

When c 90 ?

= 18

= 90 we have

i

and

//

S.O. Kasap

=

What is the penetration depth of the evanescent wave into medium 2 when i = 85 and when When

i

2 n2 o

i.e.

n1 n2

=

1/ 2

2

sin

2 1.430 850 10 9 m

2

The penetration depth is = 1/ i

i

= 85 ,

2

When

1.6

2

2

i

1.460 1.430

1

1/ 2

2

sin (85 ) 1 2

= 1.96 10 6 m-1 .

o

= 5.09 10-7 m.

= 90 , 2 1.430 850 10 9 m

2

The penetration depth is = 1/

2

1.460 1.430

1 /2

2 o sin (90 ) 1 2

= 2.18 10 6 m-1 .

= 4.59 10-7 m.

d What is the reflection coefficient and reflectance at normal incidence ( i = 0 ) when the light beam traveling in the silica medium (n = 1.455) is incident on a silica/air interface? Thus n1 = 1.460 and n2 = 1. Then, r // and

r

n1 n 2 n1 n 2

1. 455 1 = 0.185 1. 455 1

R = r //2 = (0.185) 2 = 0.034 (3.4%)

e What is the reflection coefficient and reflectance at normal incidence when a light beam traveling in air is incident on an air/silica (n = 1.455) interface? How do these compare with part (d) and what is your conclusion? The light travels in air and becomes partially reflected at the surface of the glass which corresponds to external reflection. Thus n1 = 1 and n2 = 1.455. Then, n1 n 2 1 1. 455 = 0.185 r // r n1 n 2 1 1. 455 It is important to note that this is negative which means that there is a 180 phase shift. The reflectance (R) which gives the fractional reflected power is 2 2 R = r // = ( 0.185) = 0.034 (3.4%).

In both cases 3.4% of light intensity is reflected. 1.6

TIR and polarization at water-air interface

a Given that the refractive index of water is about 1.33, what is the polarization angle for light traveling in air and reflected from the surface of the water?

Solutions Manual for Optoelectronics and Photonics: Principles and Practices 5 March 2002

S.O. Kasap

1.7

b Consider a diver in sea pointing a flash light towards the surface of the water. What is the critical angle for the light beam to be reflected from the water surface? Solution a

For light traveling in air towards the sea surface, n1 = 1 and n2 = 1.33 p

= arctan(n2/n1) = arctan(1.33/ 1) = 53.1 .

At this angle of incidence, r// = 0, the reflected wave has an E-field only perpendicular to the plane of incidence. b For light traveling in the water, n1 = 1.33 and n2 = 1. The critical angle for TIR for light traveling under water and hitting the surface of water is c

= arcsin(n2/n1) = arcsin(1/ 1.33) = 48.75 .

1 . 7 Phase changes on TIR Consider a light wave of wavelength 870 nm traveling in a semiconductor medium (GaAs) of refractive index 3.6. It is incident on a different semiconductor medium (AlGaAs) of refractive index 3.4, and the angle of incidence is 80 . Will this result in total internal reflection? Calculate the phase change in the parallel and perpendicular components of the reflected electric field. Solution a First calculate the critical angle: c = arcsin(3.4/3.6) = 70.8 . The angle of incidence i is greater than c and hence there will be TIR. Since the incidence angle i > c there is a phase shift in the reflected wave. The phase change in Er, is given by . With n1 = 3.6, n2 = 3.4 and i = 80 ,

tan

sin 2

1 2

i

cos

n2

1/ 2

i

3.4 sin ( 80 ) 3.6 o cos( 80 ) 2

2 1/ 2

o

1

= 1.607= tan[ /2(116.21 )] so that the phase change is = 116.21 . For the Er,// component, the phase change is tan

1 2

//

tan(1/2

so that which gives

//

//

=

1 2

sin 2 i n 2 n2 cos i

1/ 2

58.1

i

1 2

+ 1/2 ) = (n1/n2)2tan( /2) = (3.6/3.4)2tan(1/2116.21 ) (or 121.9 )

We can repeat the calculation with i = 90 to find Note that as long as changes.

1 tan n2

= 180 and

//

=0 .

> c, the magnitude of the reflection coefficients are unity. Only the phase

For information: The amplitude of the evanescent wave as it penetrates into medium 2 is

Solutions Manual for Optoelectronics and Photonics: Principles and Practices 5 March 2002

Et, (y,t) Eto, exp(

2

S.O. Kasap

1.8

y)

We ignore the z-dependence, expj( t kzz), as this only gives a propagating property along z. The field strength drops to e-1 when y = 1/ 2 = , which is called the penetration depth. The attenuation constant 2 is 2 n2 2

i.e.

n1 n2

1/ 2

2

sin

i

2 3.4 870 10 9 m

2

1

2

3.6 3.4

1/ 2

2 o sin (80 ) 1 2

= 7.26 106 m-1.

so that the penetration depth is, 1/ 1.8

2

= 1/(7.26 106 m) = 1.38 10-7 m, or 0.14 m.

Thin film coating and multiple reflections: A1

A0

A2

A3

n1 B1

B3 B2

B5 B4

n2 B6

n3 C1

C2

C3

Thin film coating of refractive index n2 on a semiconductor device

Figure 1Q8 Consider a thin film coating on an object as in Figure 1Q8. Suppose that the incident wave has an amplitude of A0, then there are various transmitted and reflected waves as shown in Figure 1Q8. We then have the following amplitudes based on the definitions of the reflection and transmission coefficients, A 1 = A0r12

B 1 = A0t 12

B 2 = A0t 12r23

A 2 = B 3 = A0t 12r23r21 B 4 = A0t 12r23r21r23 A 3 = A0t 12r23r21r23t 21 and so on.

B 5 = A0t 12r23r21r23r21

C1 = A0t12t23 C2 = A0t 12r23r21t 23

B 6 = A0t 12r23r21r23r21r23

C3 = A0t 12r23r21r23r21t 23

Solutions Manual for Optoelectronics and Photonics: Principles and Practices 5 March 2002

S.O. Kasap

1.9

Assume that n1 < n2 < n3 and that the thickness of the coating is d. For simplicity, we will assume normal incidence. The phase change in traversing the coating thickness d is = (2 )n2d where is the free space wavelength. The wave has to be multiplied by exp( j ) to account for this phase difference. The reflection and transmission coefficients are given by,

and a

r1 = r12

n1 n 2 n1 n 2

t 1 = t 12

2 n1 , n1 n 2

r21,

r2 = r23

n2 n3 n2 n3

t2 = t21

2n2 , n1 n2

Show that 1 t1 t2 = r1 2

b

Show that the reflection coefficient is r

Areflected A0

t1t2 r1

r1

rre

k

j2

1 2

k 1

which can be summed to

r c

r1 r 2 e j 2 1 r1r2 e j 2

Show that the transmission coefficient is t

t1t 2e j r1 r2

Ctransmitted A0

r1r2e

j2

k 1

k

t1t2 e j r1 r2

r1r2e j 2 1 r1r2 e j2

which can be summed to t

t1t 2e j 1 r1r 2e

j2

Solution Assume that n1 < n2 < n3 and that the thickness of the coating is d. For simplicity, we will assume normal incidence. The phase change in traversing the coating thickness d is = (2 )n2d where is the free space wavelength. The wave has to be multiplied by exp( j ) to account for this phase difference. The coefficients are given by,

and Consider 1

r1 = r12

n1 n 2 n1 n 2

t 1 = t 12

2 n1 , n1 n 2

t 1t 2,

r21,

r2 = r23

n2 n3 n2 n3

t2 = t21

2n2 , n1 n2

Solutions Manual for Optoelectronics and Photonics: Principles and Practices 5 March 2002

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