Conducting waveguides and slab optical waveguides. PH 317 January 12, 2007 rev Jan 18

The z-direction is parallel to the axis of the waveguide

Energy in the form of electromagnetic waves is confined within the waveguide

Assume waves with z-dependence of exp(ik z z –i
 t)

The x and y dependence is that of superposed plane waves

These combine to form standing waves between the boundaries in x and y

Solutions will be things like sin (k x x) cos(k y y) exp(ik z z –i
 t)
 k x
 k z
, k y
will be determined by boundary conditions
will be determined from k z
=

(
 2
/c
2
– k x
2

Separate two distinct kinds of waves,
– k y
2
) (or v
2
instead of c

First write down all 6 components of both curl equations in free space

Then solve for x and y components of E and B in terms of E z
2
in glass)
and B z
or their derivatives o TE in which E z
is zero so E components are transverse to z o TM in which B z
is zero, so B components are transverse to z

For TE waves, find the form of B z
which will satisfy the boundary conditions

For TM waves, find the form of E z
which will satisfy the boundary conditions

This will fix the values of k x
and k y

In a slab waveguide, waves travel in the x-z plane and boundaries are at x= -a/2, +a/2

For waves to be confined in the slab, n outside
<n inside
and

>
 critical

E and B vectors outside the slab must be exponentially decreasing for the wave to be confined

For both types of waveguides, solve separately for even and odd symmetries within TE and TM modes
Slab optical waveguide.
A slab of width a and index n
1
is the waveguide, surrounded on both sides by material of index n
2
<n
1
. For modes to propagate in the slab waveguide we need to launch waves at an angle greater than the critical angle. This causes the fields in n
2
to exponentially decay. We will separately solve for EVEN waveguide modes where the function of x is cos (k x x) and ODD waveguide modes where the function of x is sin(k x x)
The sketch shows x a TE ray (E z
=0) in index n
1 reflecting from a boundary, and showing a decaying field in n
2
. We will want the n
2
< n
1
 t
H t n
1
H i

H r
z
x=+a/2
All E vectors are in the -y direction
(into the paper) parallel component of E and the parallel component of H to be continuous across the boundary.
The waves travel in the x-z plane. x= -a/2
1

The net E field in n
1
is
E net, n1,y
= -E i
exp(ik x x + ik z z - i
 t) - E r
exp(-ik x x +ik z z -i
 t), or
E net, n1,y
= - [A cos(k x x) +B sin (k x x)] exp( ik z z -i
 t).
We will tackle these one at a time: The EVEN TE mode in the slab waveguide and the ODD TE mode in the slab waveguide. For the TM modes (H z
= 0) we must also work out both even and odd solutions.
EVEN TE mode conditions.
For the parallel component of E at the boundary we have
E y
= -A cos (k x
a/2) = -E t
exp(i k tx
a/2) .
Snell's law says n
1
sin

= n imaginary. k tx
= k t
cos
 t
2
sin
 t
. With n
1
> n
2
, when

is so large that sin
 t
>1, cos

, so when

>
 critical
, k tx
becomes pure imaginary, so we say t
is pure
(1) k tx
= i

(

is real,

>
 critical
) (2)
The propagation vectors are k
1
=

/v
1
=

/c c/v
1
= n
1

/c, and k
2
=

/v
2
=

/c c/v
2
= n
2

/c. For the boundary conditions to be satisfied at all z along the boundary, we must have k
1z
= k tz
, or k
1z
= k
1
sin

= k tz
= k
2
sin
 t
= k t
sin
 t
, (this is the same thing as snell's law)
Then in n
1
we have k
1
2
= k
1x
2
+ k z
2
=(n
1

/c)
2
, and in n
2
k
2
2 = k t
2 = k tx
2 + k z
2 = -
 2 + k z
2 =(n
2

/c) 2 .
(2a)
(2b)
For matching the H field parallel components, we want H z
. To obtain H z
we appeal to the curl of E.
(curl E) z
= -
 
H z
/
 t =

E y
/
 x -

E x
/
 y = i

H z
.
We apply this both inside and outside
In n
1
: (then evaluated at x=a/2) +k x
A sin(k x a/2) = i

H z
In n
2
: (then evaluated at x=a/2) -E t
i k tx
exp(i kt x a/2) = i

H tz
.
Setting parallel components of H (namely H z
) equal at the boundary gives
+k x
A sin(k x a/2) = -E t
i k tx
exp(ik tx
a/2) .
Dividing (3) by (1) and using (2), then multiplying both sides by a/2 we find k x
a/2 tan (k x
a/2) = | k tx
| a/2 =

a/2.
(3)
(4)
2

3
This is of the form u tan u = q, where u = k x
a/2 and q = | k tx
| a/2 =

a/2 .
Squaring both sides and adding u
2
to both sides gives u 2 tan(u) 2 + u 2 = (u/cos(u)) 2 = q 2 + u 2 .
Next we write out k
2
on both sides of the boundary, as in Eqs (2a) and (2b) k t
2
= k tz
2
+ k tx
2
= k tz
2
+ -
 2
= n
2
2  2
/c
2
and k 2 = k z
2 + k x
2 = n
1
2  2 /c 2 .
Because k z
is the same on both sides (for constant phase along the boundary) and (

)
2
= -k tx
2
, n
1
2  2
/c
2
- n
2
2  2
/c
2
= k x
2
+ (

)
2
Multiplying both sides by (a/2)
2
, we get
(5) u
2
+ q
2
= (k x
2
+
 2
)(a/2)
2
= R
2
, where R
2
= (a/2)
2
(n
1
2  2
/c
2
- n
2
2  2
/c
2
)
Now we want to solve
(u/cos(u))
2
= q
2
+ u
2
= R
2
, where R
2
depends only on the slab width and indices.
(6)
(7)
In particular we want solutions of cos u =
 u/R .
But because u tan u = q, we only want solutions where tan u is positive, namely the first and third quadrants (indeed, all odd quadrants). { u = k
1
a/2 cos

. q = k
2
a/2 |cos
 t
| }
(8)
One can plot simultaneously cos u and +u/R and -u/R, then accept solutions in the odd quadrants.
1. Show that when u/R = 1 we are at the critical angle from n
1
to n
2
.
2. Work out the condition for odd TE modes (analogous to cos u =
 u/R , odd quadrants).
3. For a slab whose thickness is 10 microns, at an omega of 1 x 10
15
rad/s, and n
1
= 1.5 and n
2
= 1.46, determine the valid even TE solutions. Determine the launch angle for each of these (theta in n
1
). {This is readily done in Maple. Could also be done on a spreadsheet. }