Exercise 1.2-6The results of this problem are somewhat general and apply to any rectangular parallelepiped with source located at any position inside. One can see this as follows. The direction of an arbitrary ray emerging from the source can be described by

r = xx + yy + zzwhere

x2 + y2 + z2 = 1corresponding to points on the unit sphere. Without loss of generality, we assume that the coordinate axes are normal to the faces of the parallelepiped. Obeying the law of reflection, if the ray is reflected from a face, the corresponding component of the ray undergoes a sign change, so that after an arbitrary set of reflections the ray direction still satisfies

r = ±xx ± yy ± zzThe angle this ray makes with the coordinate axes is

cosθx = r ⋅ x = ±x cosθy = r ⋅ y = ±y cosθz = r ⋅ z = ±zso no combination of reflections will change the angles the ray makes with the face normals, other than to change which of two parallel faces it is heading toward (or away from). If one of these three angles is less than the critical angle, then eventually the ray will strike the corresponding face and escape. If none of these angles is small enough, it will never escape. (Though not described by ray optics, trapped light will eventually dissipate through absorptive processes and ultimately be converted to heat.) If the parallelepiped is not rectangular the analysis is much trickier. I dare say one would use numerical methods. (a) The half angle of the cones (inside the material) is just given by the critical angle

sinθc = 1 / nFor GaAs

n = 3.6θc = 0.281 = 16.1

The cone is quite narrow.

(b)With an isotropic radiator, the power is proportional to the solid angle, or equivalently the surface area on a unit sphere. The set of directions with angle of incidence less than the critical angle (and thus escape) comprise a cone that intersects the unit sphere in circularly shape spherical cap. There are six such caps. The area of one cap is given by

A1 = 2π sinθ dθ0

θc

∫ = −2π cosθ[ ]0θc = 2π 1− cosθc( )

where

cosθc = 1−1 / n2

The fraction of power extracted isPextractedPtotal

=6A14π

=12π 1− cosθc( )

4π= 3 1− 1−1 / n2( )

Where it is assumed that the spherical caps are small enough that they do not intersect. If they do, the area (and power) will be double counted (middle figure). Intersection will not occur when

θc < π / 4 ⇒ sinθc <12

⇒ 1n< 1

2⇒ n > 2

For GaAs not much power is extracted!n = 3.6PextractedPtotal

= 0.118

For what values of n does all the power escape (left figure)?

cosθc <1

12 +12 +12= 1 / 3

1−1 / n2 < 1 / 3

n < 3 / 2 ≈ 1.22

The problem of extraction is a major issue in the LED industry!

Exercise 1.3-1To find the focal length one sends in parallel rays and measures the distance to the focus. (Imperfect lenses, with spherical aberration, will have a range of focal lengths for different regions of the lens aperture. A single quoted focal length will always refer to the paraxial focal length, so the paraxial approximation can always be used to calculate the focal length.) Setting θ0 to zero in (1.3-11) and (1.3-12) we obtain

y z( ) = y0 cosαzθ z( ) = y0α sinαz

Where I have chosen to define the downward angle as positive. The ray follows these trajectories over the thickness, d, and then refracts out. Using paraxial Snell’s law

n0θ d( ) ≈θ2The tangent of the angle that the ray makes with the optic axis is given by

y0f= tanθ2 ≈θ2

Solving for the focal length we find

f ≈y0θ2

=y0

n0θ d( ) =y0

n0y0α sinαdobtaining the correct result

f ≈ 1n0α sinαd

The result (1.3-13) must be wrong since we know the focal length will be periodic in d. (Though if αd<<1 one can take variables in and out of the sine function at will, but then why have the sine function at all?) Next, find the distance to the principle point, H

AH = f − AFUsing a triangle similar to the on used above we find

y d( )AF

= tanθ2 ≈θ2

Solve for “back focal length”

AF ≈y d( )θ2

Calculate the required distance

AH = f − AF ≈ 1n0α sinαd

−y d( )θ2

= 1n0α sinαd

−y0 cosαd

n0y0α sinαd= 1n0α

1− cosαdsinαd

Using a trigonometric identity

AH ≈ 1n0α

tanαd2

For the special cases d = π/α, π/2α the rays appear as follows. The focal length is infinity and zero respectively.

Exercise 1.4-11One round trip in the cavity is comprised of a propagation of distance d, a plane mirror reflection, a propagation of distance d and another plane mirror reflection. So the total matrix for one period of the periodic systems is

M = M 4M 3M 2M1

where

M1 = M 3 =1 d0 1

⎛⎝⎜

⎞⎠⎟

and M 2 = M 4 = 1 00 1

⎛⎝⎜

⎞⎠⎟

Working out the simple matrix product

M = 1 00 1

⎛⎝⎜

⎞⎠⎟

1 d0 1

⎛⎝⎜

⎞⎠⎟

1 00 1

⎛⎝⎜

⎞⎠⎟

1 d0 1

⎛⎝⎜

⎞⎠⎟= 1 2d

0 1⎛⎝⎜

⎞⎠⎟

Which yields for the parameters b and F

b = A + D2

= 1 and F2 = det M[ ] = AD − BC = 1

Then the quadratic equation in the parameter h can be easily factored.h2 − 2bh + F2 = 0h2 − 2h +1= 0

h −1( )2 = 0There is only one root, which leads to the solution that y is constant

ym = y0which is not very interesting, and is not a complete general solution to the second order difference equation, since it does not have two unknown coefficients. One can show that the linear equation

ym = α + mβis a solution to the difference equation by plugging it in.

ym+2 = 2bym+1 − F2ym

α + m + 2( )β = 2 α + m +1( )β( )− α + mβ( )α + m + 2( )β = α + m + 2( )β

The solution is, of course, unstable since it increases without bound as a function of m (except for the special case that y is constant). Only a ray bouncing perpendicularly back and forth between a pair of parallel mirrors will remain confined.Exercise 2.2-7The intensity is the same for a wave and its complex conjugate, since a complex number’s magnitude is unaffected by conjugation.

I r( ) = U r( ) 2 = U * r( ) 2

The equation for the wavefronts isarg U r( )⎡⎣ ⎤⎦ = 2πq q ∈0,±1,±2,...

The argument of complex number changes sign when conjugated, but this sign change does not affect the wavefronts since both signs of the enumerating integer are included in defining the set of wavefronts. One can just redefine the integer.

arg U * r( )⎡⎣ ⎤⎦ = 2π −q( ) = 2π p p ∈0,±1,±2,...Since a wavefront is a surface of constant phase, a wavefront normal is given by the gradient of that phase function. This is also called the wave vector

k = −∇arg U r( )⎡⎣ ⎤⎦(I have changed the above equation to show the correct sign for our assumed time dependence.) The wave vector changes sign when the wave is conjugated, thus the waves move in the opposite direction.

−∇arg U * r( )⎡⎣ ⎤⎦ = ∇ − arg U * r( )⎡⎣ ⎤⎦( ) = ∇arg U r( )⎡⎣ ⎤⎦ = −kConjugate waves are important technologically for recovering signals that traverse strongly scattering environments. Specific example, a plane wave and its conjugate

U r( ) = Aexp − j x + y( ) / 2⎡⎣ ⎤⎦ U * r( ) = A* exp + j x + y( ) / 2⎡⎣ ⎤⎦The intensity is given by

U r( ) 2 = A 2 U * r( ) = A*2= A 2

The wavefronts are given byarg U r( )⎡⎣ ⎤⎦ = arg A[ ]− x + y( ) / 2 = 2πq

The wavefronts of the conjugate wave are given byarg U * r( )⎡⎣ ⎤⎦ = arg A*⎡⎣ ⎤⎦ + x + y( ) / 2 = − arg A[ ]+ x + y( ) / 2 = 2π p = 2π −q( )

arg A[ ]− x + y( ) / 2 = 2πqThe wave vector is given by

−∇arg U r( )⎡⎣ ⎤⎦ = x + y( ) / 2and the wave vector of the conjugate wave is opposite in sign

−∇arg U * r( )⎡⎣ ⎤⎦ = − x + y( ) / 2Specific example, a spherical wave and its conjugate

U r( ) = Arexp − jkr[ ] U * r( ) = A*

rexp + jkr[ ]

The intensity is given by

U r( ) 2 = A 2

r2U * r( ) =

A*2

r2=A 2

r2

The wavefronts are given byarg U r( )⎡⎣ ⎤⎦ = arg A[ ]− kr = 2πq

The wavefronts of the conjugate wave are given byarg U * r( )⎡⎣ ⎤⎦ = arg A*⎡⎣ ⎤⎦ + kr = − arg A[ ]+ kr = 2π p = 2π −q( )

arg A[ ]− kr = 2πqThe wave vector is given below and the waves are propagating outwardly.

−∇arg U r( )⎡⎣ ⎤⎦ = k∇r = krwhere we use the (perhaps) well known expression for the gradient of the radial distance, r

∇r = ∂∂x, ∂∂y, ∂∂z

⎛⎝⎜

⎞⎠⎟x2 + y2 + z2( )1/2 = 12 x2 + y2 + z2( )−1/2 2x,2y,2z( ) = r

r= r

and the wave vector of the conjugate wave is opposite in sign and the waves are propagating inwardly.−∇arg U * r( )⎡⎣ ⎤⎦ = −k∇r = −kr