Assignment#1 Solution

1-10

(a) From Figure 1.8, at = 1300 nm, n = 1.447, Ng = 1.462, so that

The phase velocity is given by

v = c/n = (3108 m s

-1)/(1.447) = 2.07310

8 m s

-1.

The group velocity is given by

vg = c/Ng = (3108 ms

-1)/(1.462) = 2.05210

8 m s

-1.

The group velocity is about ~1% smaller than the phase velocity.

(b)

The Brewster angle p is given by

2

1

1tan 0.691

1.447p

n

n

1tan 0.691 34.64p

At the Brewster angle of incidence i = p, the reflected light contains only field oscillations

normal to the plane of incidence (paper).

The critical angle is

2

1

1sin 0.691

1.447c

n

n

1sin (0.691) 43.7c

(c) Given

1

2

1.447

1

n

n

1 2/ /

1 2

1.447 10.1827

1.447 1

n n

n n

r r

and 2 2

// / / 0.0333 R R r r

(d) Given

1

2

1

1.447

n

n

1 2/ /

1 2

1 1.4470.1827

1.447 1

n n

n n

r r

and 2 2

// / / 0.0333 R R r r

Reflection coefficients are negative, which means that in external reflection at normal

incidence there is a phase shift of 180°.

1-11

The problem is sketched in Figure 1Q12-1

Figure 1Q12-1 Light beam deflection through a glass plate of thickness L = 2 mm. The angle of

incidence is 40 and the glass has a refractive index of 1.570

io

ii

nnL

d

22 sin)/(

cos1sin

2 2

cos 40sin 40 1

(1.570 /1) sin 40

d

L

0.7660

0.6428 1 0.29862.46 0.4132

2986.0mm2

d

d = 0.60 mm

This is a significant displacement that can be easily measured by using a photodiode array.

1-13

(a) Snell's law at interfaces at A:

sin

sin 1

i

t

n

Snell's law at interfaces at B:

sin 1

sin

i

tn

(b) Consider the deflection angle

= 1 + 2 where 1 = i t and 2 = t i, i.e. = i t + t i. Now,

n

it

sinarcsin

and from Figure 1.55

n

itti

sinarcsin180

so that

nnn i

it

sinarcsinsinarcsinsinarcsin

and the deflection is,

nnn

n

iiii

sinarcsin

sinarcsinsinarcsin

sinarcsin

so that finally,

nn i

i

sinarcsinsinarcsin

Substituting the values, and keeping n as a variable, the deflection (n) as a function of n is

nnn

2

1arcsin60sinarcsin)6045()(

where sin(45) = 1/2.

The separation L for two wavelengths1 and 2 corresponding to n1 and n2 at the screen at a

distance L away is therefore

L = L[(n1) (n2)]

where the deflections must be in radians.

Consider the deflection of blue light

)4634.1(2

1arcsin60sin)4634.1(arcsin)6045(blue

blue = 34.115

Similarly, yellow = 33.709

The separation of blue and yellow beams at the screes is

Lblue-yellow = L(blue blue) = (1m)(/180)( 34.115 33.709) = 7.08 mm

Table 1Q13-1 summarizes the results of the calculations for blue, yellow and red light.

Table 1Q13-1 Deflection of blue, yellow and red light through a prism with apex angle 60.

The angle of incidence is 45.

Blue

486.1 nm

Yellow

589.2 nm

Red

656.3 nm

n 1.4634 1.4587 1.4567

Deflection angle)0.5954 rad

34.115

0.5883 rad

33.709

5853 rad

33.537

L between colors 7.08 mm 3.00 mm

L between blue and

rede 10.1 mm

1-20

(a) Apply

tanp n2

n1

with 1

2

1

1.33

n

n

1 1

2 1tan ( / ) tan (1.33 /1) 53.06p n n

(b) Given

and 1

2

1.33

1

n

n

The critical angle is

1 12

1

1sin sin 48.75

1.33c

n

n

1-22

1 12

1

3.40sin sin 70.81

3.60c

n

n

Clearly, 80i c

So, this results in total internal reflection.

2

1

3.400.9444

3.60

nn

n

1/2 1/2

2 2 2 2

1

2

sin sin (80 ) (0.9444)tan 1.6078

cos cos(80 )

i

i

n

12tan (1.6078) 116.24

1 1 1 1

// 2 22 2 2 2

1 1tan tan tan 116.24 1.8027

0.9444n

1 1( //2

) tan (1.8027) 60.98

( / / ) 121.96

/ / 121.96 180 58.04

2-12

(a) Given n1 =1.475, n2=1.455, 2a= 8*10-6

m or a=4µm and λ=1.3 µm. The V- number is,

V= 2πa/ λ (n12

–n22)1/2

= ( )( )

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

V= 2πa/ λ (n12

–n22)1/2

> 2.405

Or

λ< 2πa(n12

–n22)1/2

/ 2.405= ( )( )

For the wavelength shorter than 1.13 , the fiber is a multimode waveguide.

(c) The numerical aperture NA is

(

) ( )

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

(

)

So that the total acceptance angle is 12.4 .

(e)

At λ=1.3µm, form D vs. λ, Figure 2.22, Dm = -7.5 psKm-1

nm-1

, Dw = -5psKm-1

nm-1

| |

| | ( )

025 ns

Obviously material dispersion is 15 ps and waveguide dispersion is 10 ps

The maximum bit-rate distance product is then

BL

2-13

(a)

Apply, ( )

(

) , and solving for n2 we find n2 = 1.4474.

The V-number is given by

V= 2πa/ λ (n12

–n22)1/2

= ( )( )

Single mode fiber.

(b) V= 2πa/ λ (n12

–n22)1/2

= ( )( )

λ< 1.205µm.

(c) The numerical aperture NA is

(

) ( )

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

(

)

So that the total acceptance angle is 11.8 .

(e) Given, Dw = -5 ps km-1

nm-1

and Dm = 15 ps km-1

nm-1

Laser diode spectral width (FWHM) = 3 nm

Material dispersion

= | | (15 ps km

-1 nm

-1)(3 nm)

=45 psKm-1

Waveguide dispersion

= | | (-5 ps km

-1 nm

-1)(3 nm)

=-15 psKm-1

Chromatic dispersion,

= | | (-5 ps km

-1 nm

-1 + 15 ps km

-1 nm

-1)(3 nm)

=30 psKm-1

(f) Maximum bit-rate would be

BL

(g) To find the maximum diameter for SM operation solve,

V= 2πa/ λ (n12

–n22)1/2

= ( )( )

2a = 11.5µm.

(h) The mode filed diameter 2w is

( )

2.15

From example 2.3.4, we have

b = 0.3860859, K= 14188790 m-1

[ ] ( )( )[ ( )

]

Group Velocity

( )

( )

over 1Km.

Comparing to Example 2.3.4

%diff=| |

2.18

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,

( )

Show that for small normalized index difference Eq. (1) approximates

to

( )

which gives β as,

[ ]

The group velocity is defined and given by

=

=c

Thus, the propagation time of the mode is

(

)

( )

where we assumed constant (does not depend on the wavelength). Given the

definition of V,

[

] [( )( )]

[( ) (

)]

(5)

[ ]

( )

From Eq. (5),

( )

[ ( )

]

( )

This means that depends on V as,

( )

(6)

Dispersion, that is, spread in due to a spread can be found by differentiating Eq.

(6) to obtain

( )

(

) ( )

( )

The waveguide dispersion coefficient is defined as

( )

(8)

Figure 2.53 shows the dependence of ( )

on the V- number.

In the range 2<V<2.4,

( )

so that Eq. (8) becomes,

( )

We can simplify this further by using

[

( ) ]

( ) (10)

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

( )

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).

V= 2πa/ λ (n12

–n22)1/2

= ( )( )

( )

From the graph, ( )

( )

( )( )

( )( )( )= -4.6

Using Eq. (10)

( )

( )

( )( ) ( ) =-4.6

For we have

| | ( )( )

2-19

(

)

( )( )

( )

From the graph in Figure 2.53, when V = 2.03, Vd2(Vb)/dV

2= 0.50,

Profile dispersion:

(

( )

)(

)

( )( )

( )

( )

( )( ) ( )

2-20

( )

( )

( )

( )

(

) ( )

( )( )

( )

( )( )