EXERCISE 2.4-4

Given a paraboloidal wave centered at z = -z1

U1 r( ) = A1z + z1

exp − jk z + z1( )⎡⎣ ⎤⎦exp − jkx2 + y2

2 z + z1( )⎡

⎣⎢

⎤

⎦⎥

and a paraboloidal wave centered at z = z2

U2 r( ) = A2z − z2

exp − jk z − z2( )⎡⎣ ⎤⎦exp − jkx2 + y2

2 z − z2( )⎡

⎣⎢

⎤

⎦⎥

show that the transfer function for a lens

t x, y( ) = h0 exp jkx2 + y2

2 f⎡

⎣⎢

⎤

⎦⎥

applied at z = 0 transforms U1 into U2

U1 z = 0( )t x, y( ) =U2 z = 0( )Plugging in from aboveA1z1exp − jkz1[ ]exp − jk

x2 + y2

2z1

⎡

⎣⎢

⎤

⎦⎥h0 exp jk

x2 + y2

2 f⎡

⎣⎢

⎤

⎦⎥ = −

A2z2exp jkz2[ ]exp jk

x2 + y2

2z2

⎡

⎣⎢

⎤

⎦⎥

combining the exponential terms that appear on the left hand sideA1h0z1exp − jkz1[ ]exp jk

x2 + y2

21f−1z1

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ = −

A2z2exp jkz2[ ]exp jk

x2 + y2

2z2

⎡

⎣⎢

⎤

⎦⎥

and using1z1+1z2

=1f

we findA1h0z1exp − jkz1[ ]exp jk

x2 + y2

2z2

⎡

⎣⎢

⎤

⎦⎥ = −

A2z2exp jkz2[ ]exp jk

x2 + y2

2z2

⎡

⎣⎢

⎤

⎦⎥

canceling the common factor we obtainA1h0z1exp − jkz1[ ] = −

A2z2exp jkz2[ ]

so that if the amplitude of the second wave is given by

A2 = −A1h0z2z1exp − jk z1 + z2( )⎡⎣ ⎤⎦

the desired equation is satisfied identically.

EXERCISE 2.5-2

For a single paraboloidal wave we have

U0 r( ) = A0zexp − jkz( )exp − jk

x2 + y2

2z⎡

⎣⎢

⎤

⎦⎥

and

I0 = U0 (x, y,d)2 =

A02

d 2

For two waves, one at x = a and one at x = -a the total wave is

U r( ) = A0zexp − jkz( ) exp − jk

x + a( )2 + y22z

⎡

⎣⎢⎢

⎤

⎦⎥⎥+ exp − jk

x − a( )2 + y22z

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪Factoring the common terms

U r( ) = A0zexp − jkz( )exp − jk

x2 + a2 + y2

2z⎡

⎣⎢

⎤

⎦⎥ exp − jk

2ax + y2

2z⎡

⎣⎢

⎤

⎦⎥ + exp jk

2ax + y2

2z⎡

⎣⎢

⎤

⎦⎥

⎧⎨⎩⎪

⎫⎬⎭⎪

Using the definition of cosine in terms of exponentials

cosφ =e jφ + e− jφ

2we obtain

U r( ) = 2 A0zexp − jkz( )exp − jk

x2 + a2 + y2

2z⎡

⎣⎢

⎤

⎦⎥cos k

2ax + y2

2z⎡

⎣⎢

⎤

⎦⎥

The intensity along the y = 0 line at z = d

I(x,0,d) = U(x,0,d) 2 = 4A0

2

d 2cos2 k

axd

⎡⎣⎢

⎤⎦⎥

Employing the trigonometric identity2cos2φ = 1+ cos2φand the result for the intensity of a single wave I0 we obtain

I(x,0,d) = 2I0 1+ cos2kaxd

⎛⎝⎜

⎞⎠⎟

Defining

θ ≈2ad

Finally we obtain

I(x,0,d) ≈ 2I0 1+ cos kxθ( )( ) = 2I0 1+ cos 2π xθλ⎛⎝⎜

⎞⎠⎟

Exercise 3.2-2

(a) It will be convenient to define the following

ζ = zf−1 and ζ ' = z '

f−1 and ζ0 =

z0

fSo that from equation 3.2-9 we have

r =z0

z − f=ζ0

ζ and Mr =

fz − f

= 1ζ

and finally

M =Mr

1+ r2=

1ζ

1+ ζ0ζ

⎛⎝⎜

⎞⎠⎟

2= 1

ζ 2 +ζ 20

Equation 3.2-6z '− f( ) = M 2 z − f( )

then becomes

ζ ' = ζζ 2 +ζ 20

which can also be written asz 'f−1= z / f −1

z / f −1( )2 + z0 / f( )2

(b) To find the maximum of z’ (or equivalently ζ’), we set the derivative equal to zerodζ 'dζ

=ζ 2 +ζ 20 −ζ2ζ

ζ 2 +ζ 20( )2=

ζ 20 −ζ2

ζ 2 +ζ 20( )2= 0

and find that

ζ = ζ0 zf−1= z0

fwhich simplifies toz = f + z0

(c) Givenλ = 1 µm z0 = 1 cm f = 50 cmwe find

z = 51 cm M ≈ 35.4 z ' ≈ 1300 cm W0 =λz0

π≈ 56.4 µm

Prove that the parallel component of the electric field is constant across a dielectric boundary (Figure 5.1-1) using Maxwell’s equations and Stokes’ Theorem.

t̂

n̂

E1

E2

C

S

!

l

Integrate Maxwell’s equation

∇×E = − ∂B

∂tover a surface S, shown in the figure above

∇×E ⋅dS

S∫ = − ∂B∂t

⋅dSS∫

Use Stokes’ theorem

∇× F ⋅dS

S∫ = F ⋅dlC∫

to convert the surface integral of the curl into a line integral, where the path of the line integral, C, bounds the surface, S, as required by the theorem

E ⋅dl

C∫ = − ∂B∂t

⋅dSS∫

Evaluating the path integral we assume the entire path is sufficiently small that the electric field has the constant value, E1, on one side of the boundary and E2 on the other. Letting the length of the short sides, ε, approach zero

limε→0

E ⋅dlC∫ = E1 ⋅ t̂ −E2 ⋅ t̂( )l

Evaluating the surface integral of the time derivative of B, we note that this derivative is finite while the surface area approaches zero with ε, so that the integral approaches zero

limε→0

∂B∂t

⋅dSS∫ = 0

Thus we conclude that

E1 ⋅ t̂ = E2 ⋅ t̂