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Lecture Notes on Wave (03/05/14) 2.71/2.710 Introduction to Optics –Nick Fang

Outline: A. B. Frequency Domain (Fourier transform) C. EM waves in Cartesian coordinates D. Energy Flow and Poynting Vector E. Connection to geometrical optics F. Eikonal Equations: Path of in an Inhomogeneous Medium

A. Electromagnetism and Maxwell Equations, Differential Forms:

- In real space, time-dependent fields: D (displacement field) D dS dS + V

A Coulomb

(Coulomb’s Law, electric field) (1)

dS there are no B magnetic charges

V

A (2) (Gauss’ Law, )

1 Lecture Notes on Wave Optics (03/05/14) 2.71/2.710 Introduction to Optics –Nick Fang

E dl B(t) up or down

A C

𝐸 − 푡 (3) (Faraday’s Law)

H H dl dl D I current capacitor A CA C

푡 (4) (Ampere-Maxwell’s Law)

Note: J, q are sources of EM radiation and E, D, H, B are induced fields.

B. From time domain to frequency domain (Fourier Transform):

Continuous wave laser light field understudy are often mono-chromatic. These problems are mapped in the Maxwell equations by expanding complex time signals to a series of time harmonic components (often referred to as “single” light): ⃗ ( ) ∞ ⃗⃗ ( ) ( ) e.g. 𝐸 푟 , 푡 ∫−∞ 푬 푟 , 𝜔 푒푥푝 −푖𝜔푡 푑𝜔 (5) Advantage:

휕 ∞ 휕 ∞ 𝐸⃗ (푟 , 푡) ∫ ⃗푬⃗ (푟 , 𝜔) 푒푥푝(−푖𝜔푡)푑𝜔 ∫ [−풊흎⃗푬⃗ (⃗풓 , 흎)]푒푥푝(−푖𝜔푡)푑𝜔 (6) 휕푡 −∞ 휕푡 −∞

2 Lecture Notes on Wave Optics (03/05/14) 2.71/2.710 Introduction to Optics –Nick Fang

휕 So, we can replace all time derivatives by −푖𝜔 in frequency domain: 휕푡

(Faraday’s Law:) (7)

(Ampere-Maxwell’s Law:) (8)

The other two equations remain unaltered.

There are total of 12 unknowns (E, H, D, B) but so far we only obtained 8 equations from the Maxwell equations (2 vector form x3 + 2 scalar forms) so more information needed to understand the complete wave behavior!

Generally we may start to construct the response of a material by applying a excitation field E or H in vacuum. Therefore it is more typical to consider the E, H field as input and D, B fields as output. In common optical materials, we may enjoy the following simplification of local (i.e. independent of neighbors) and linear relationship:

푫⃗⃗ (𝜔) 휀(𝜔)휀0푬⃗⃗ (𝜔) (9)

푩⃗⃗ (𝜔) 휇(𝜔)휇0푯⃗⃗⃗ (𝜔) (10)

The so called (electric) permittivity 휀(𝜔) and (magnetic) permeability 휇(𝜔)are unitless parameters that depend on the frequency of the input field. In the case of anisotropic medium, both 휀(𝜔) and 휇(𝜔)become 3x3 dimension tensor.

Now we have 6 more equations from material response, we can include them together with Maxwell equations to obtain a complete solution of optical fields with proper boundary condition.

C. Maxwell’s Equations in Cartesian Coordinates:

To solve Maxwell equations in Cartesian coordinates, we need to practice on the vector operators accordingly. Most unfamiliar one is probably the curl of a vector 𝐴 . In Cartesian coordinates it is often written as a determinant:

3 Lecture Notes on Wave Optics (03/05/14) 2.71/2.710 Introduction to Optics –Nick Fang

풙̂ 풚̂ 풛̂ ⃗⃗ 휕 휕 휕 훁 푨 | 휕푥 휕푦 휕푧 | (11) 𝐴풙 𝐴풚 𝐴풛

In this fashion, we may write the Faraday’s law in frequency domain,

, with 3 components in Cartesian coordinates: 휕퐸 휕퐸 ( 푧 − 푦) 푖𝜔 (12) 휕푦 휕푧 푥 휕퐸 휕퐸 ( 푧 − 푥) 푖𝜔 (13) 휕푥 휕푧 푦 휕퐸 휕퐸 ( 푦 − 푥) 푖𝜔 (14) 휕푥 휕푦 푧

Likewise, we arrive at the rest of Maxwell’s equations: 휕퐻 휕퐻 ( 푧 − 푦) −푖𝜔 (15) 휕푦 휕푧 푥 휕퐻 휕퐻 ( 푧 − 푥) −푖𝜔 (16) 휕푥 휕푧 푦 휕퐻 휕퐻 ( 푦 − 푥) −푖𝜔 (17) 휕푥 휕푦 푧 together with 휕 휕 휕 (18) 휕푥 푥 휕푦 푦 휕푧 푧 And

휕 휕 휕 (19) 휕푥 푥 휕푦 푦 휕푧 푧

- Example: EM wave in 1-D homogeneous medium (e.g. an expanded 흏 흏 laser beam in +z direction, ퟎ, ퟎ) 흏풙 흏풚 We can now further simplify from the above equations in Cartesian coordinates: 휕퐸 휕퐸 ( 푧 − 푦) 푖𝜔 (20) 휕푦 휕푧 푥 휕퐸 휕퐸 ( 푧 − 푥) 푖𝜔 (21) 휕푥 휕푧 푦 And 휕퐻 휕퐻 ( 푧 − 푦) −푖𝜔 (22) 휕푦 휕푧 푥 휕퐻 휕퐻 ( 푧 − 푥) −푖𝜔 (23) 휕푥 휕푧 푦

From the above we found only 4 non-trivial equations. In the isotropic case, they can be further divided into 2 independent sub-groups (two Polarizations!):

4 Lecture Notes on Wave Optics (03/05/14) 2.71/2.710 Introduction to Optics –Nick Fang

(Ex, Hy only) 휕 푖𝜔휇휇 − 𝐸 (24) 0 푦 휕푧 푥 휕 and 푖𝜔휀휀 𝐸 (25) 0 푧 휕푧 푦

Or (Ey, Hx only) 휕 푖𝜔휇휇 − 𝐸 (26) 0 푥 휕푧 푦 휕 And 푖𝜔휀휀 𝐸 (27) 0 푦 휕푧 푥

Observations: - We see that in such medium is purely transverse, i.e. only components of E, H field that are orthogonal to propagation direction (+z) survived in the wave field. 흏 - Taking the derivative again on any of these equations, we obtain wave 흏풛 equation such as: 휕2 휕 𝐸 −푖𝜔휇휇 −푖𝜔휇휇 (푖𝜔휀휀 )𝐸 (28) 휕푧2 푥 0 휕푧 푦 0 0 푥 2 2 휕 2 휀휇휔 2 𝐸푥 (𝜔 휇0휀0)휀휇𝐸푥 ( 2 ) 𝐸푥 (29) 휕푧 푐0

2 1 since the c0 in vacuum satisfy ( c0 ) 휀0휇0 Therefore the index of is found as: 푛(𝜔) √휀(𝜔)휇(𝜔) (30)

D. The Poynting Vector ⃗ 2 The Poynting vector, 푺 휀0푐0 𝐸⃗ ⃗ , is used to power per unit area in the direction of propagation.

5 Lecture Notes on Wave Optics (03/05/14) 2.71/2.710 Introduction to Optics –Nick Fang

o Justification:

U = Energy density Energy passing through area A in time t: A

So the energy per unit time per unit area:

c t

- The Irradiance (often called the Intensity) Visible light wave oscillates in 1014-1015Hz. Since we don’t have a detector that responds in such a high speed yet, for convenience we take the average power per unit area, as the irradiance.

Substituting a sinusoidal light wave into the expression for the Poynting vector yields 1 〈⃗푺 〉 휀 푐2〈𝐸⃗ ⃗ 〉 휀 푐2𝐸 (31) 0 0 2 0 0 0 0

E. High Frequency Limit, connection to Geometric Optics:

How can we obtain Geometric optics picture such as tracing from wave equations? Now let’s go back to real space and frequency domain (in an isotropic medium but with spatially varying permittivity 휀(푥, 푧), for example).

휕2 휕2 휔2 2 𝐸푥 2 𝐸푥 휀(푥, 푧) ( 2 ) 𝐸푥 (32) 휕푥 휕푧 푐0

Now we decompose the field E(r, ) into two

forms: a fast oscillating component exp(ik0),

k0 𝜔/푐0 and a slowly varying E0(r) as illustrated in the textbox.

6 Lecture Notes on Wave Optics (03/05/14) 2.71/2.710 Introduction to Optics –Nick Fang

With this tentative solution, we can rewrite the wave equation:

휕 휕 휕 𝐸 ( 𝐸 (푥, 푧)) exp(푖푘Φ(푥, 푧)) 𝐸 (푥, 푧) [ exp(푖푘Φ(푥, 푧))] (33) 휕푥 푥 휕푥 0 0 휕푥

휕 휕 휕Φ(푥,푧) 𝐸 ( 𝐸 (푥, 푧)) exp(푖푘Φ(푥, 푧)) 𝐸 (푥, 푧) [푖푘 ] exp(푖푘Φ(푥, 푧)) (34) 휕푥 푥 휕푥 0 0 휕푥

휕 휕 휕Φ(푥,푧) 𝐸 [ 𝐸 (푥, 푧) 푖푘𝐸 (푥, 푧) ] exp(푖푘Φ(푥, 푧)) (35) 휕푥 푥 휕푥 0 0 휕푥

휕2 휕2 휕2 휕 휕 𝐸 [ 𝐸 (푥, 푧) 푖푘𝐸 (푥, 푧) Φ(푥, 푧) 2푖푘 𝐸 (푥, 푧) Φ(푥, 푧) − 휕푥2 푥 휕푥2 0 0 휕푥2 휕푥 0 휕푥 2 휕 푘2𝐸 (푥, 푧) ( Φ(푥, 푧)) ] exp(푖푘Φ(푥, 푧)) (36) 0 휕푥

Similarly you can find the derivative along the z direction. So our wave equation becomes:

2 2 휕 휕 푘2 [휀(푥, 푧) − ( Φ(푥, 푧)) − ( Φ(푥, 푧)) ] 𝐸 (푥, 푧) 휕푥 휕푧 0

휕2 휕2 휕 휕 휕 휕 [ 𝐸 (푥, 푧) 𝐸 (푥, 푧)] 2푖푘 [ 𝐸 (푥, 푧) Φ(푥, 푧) 𝐸 (푥, 푧) Φ(푥, 푧)] 휕푥2 0 휕푧2 0 휕푥 0 휕푥 휕푧 0 휕푧 휕2 휕2 푖푘𝐸 (푥, 푧) [ Φ(푥, 푧) Φ(푥, 푧)] (37) 0 휕푥2 휕푧2

Furthermore, if the envelope of field varies slowly with E 1 휕 1 휕 0 ∇Φ wavelength (e.g. 𝐸0 ≪ 1, 𝐸0 ≪ 1) then only 푘 휕푥 푘 휕푧 H0 the first term is important.

2 2 3 휕Φ 휕Φ  ( ) ( ) 휀(푥, 푧) 푛2(푥, 푧) (38) 2 휕푥 휕푧  1 This is the well-known Eikonal equation, Φ being the Geometrical relationship eikonal (derived from a Greek word, meaning image). of E, H, and ∇Φ

7 Lecture Notes on Wave Optics (03/05/14) 2.71/2.710 Introduction to Optics –Nick Fang

F. Path of Light in an Inhomogeneous Medium

A. Example 1: 1D problems (Gradient index waveguides, Effects)

The best known example of this kind is probably the Mirage effect in desert or near a seashore, and we heard of the explanation such as the increases with density (and hence decreases with temperature at a given altitude). With the picture in mind, now can we predict more accurately the ray path and image forming processes?

Starting from the Eikonal equation and we assume 푛2(푥, 푧) is only a function of x, then we find:

휕Φ 2 휕Φ 2 ( ) ( ) 푛2(푥) (39) 휕푥 휕푧

Since there is the index in independent of z, we may assume the slope of phase change in z direction is linear:

휕Φ ( )= C(const) (40) 휕푧

This allows us to find 휕Φ √푛2(푥) − 𝐶2 (41) 휕푥

From Fermat’s principle, we can visualize that direction of rays follow the gradient of phase front:

푑푟 푛 ∇Φ (42) 푑푙

8 Lecture Notes on Wave Optics (03/05/14) 2.71/2.710 Introduction to Optics –Nick Fang

푑푧 z-direction: 푛(푥) C (43) 푑푙

푑푥 2 x-direction: 푛(푥) √푛2(푥) − 𝐶 (44) 푑푙

Therefore, the light path (x, z) is determined by: 푑푧 C (45) 푑푥 √푛2(푥)−퐶2

Hence 푥 C 푧 − 푧0 ∫ 푑푥 (46) 푥0 √푛2(푥)−퐶2

Without loss of generality, we may assume a quadratic index profile along the x direction, such as found in gradient index optical fibers or rods:

2 2 2 푛 (푥) 푛0(1 − 훼푥 ) (47)

푥 C 푧 − 푧0 ∫ 푑푥 (48) 푥0 2 2 2 √푛 (1−훼푥 )−퐶

To find the integral explicitly we may take the following transformation of the variable x:

2 2 푛0−퐶 푥 √ 2 푠푖푛휃 (49) 푛0훼 Therefore,

휃 C 푧 − 푧0 ∫ 푑휃 (50) 휃0 푛0√훼

C 푧 푧0 (휃 − 휃0) (51) 푛0√훼

Or more commonly, 2 2 푛0훼 √푛0훼 푥√ 2 2 푠푖푛휃 푠푖푛 (휃0 2 (푧 − 푧0)) (52) 푛0−퐶 퐶

As you can see in this example, ray propagation in the gradient index waveguide follows a 2휋퐶 sinusoid pattern! The periodicity is determined by a constant . 푛0√훼

9 Lecture Notes on Wave Optics (03/05/14) 2.71/2.710 Introduction to Optics –Nick Fang

Index of x refraction n(x) dx dz

z

Observation: the constant C is related to the original “launching” angle of the optical ray. To check that we start by:

푑푧 C | (53) 푑푥 2 2 푥=푥0 √푛 (푥0)−퐶 If we assume C=푛(푥0)푐표푠훽, then 푑푧 cosβ | 푐표푡훽 (54) 푑푥 푠푖푛훽 푥=푥0

B. Other popular examples: Luneberg

The Luneberg lens is inhomogeneous sphere that brings a collimated beam of light to a focal point at the rear surface of the sphere. For a sphere of radius R with the origin at the center, the gradient index function can be written as:

푟2 푛 √2 − , 푟 ≤ 푅 푛(푟) { 0 푅2 (55) 푛0 푟 > 푅 Such lens was mathemateically conceived during the 2nd world war by R. K. Luneberg, (see: R. K. Luneberg, Mathematical Theory of Optics (Brown University, Providence, Rhode Island, 1944), pp. 189-213.) The applications of such Luneberg lens was quickly demonstrated in microwave frequencies, and later for optical communications as well as in . Recently, such device gained new interests in in phased array communications, in illumination systems, as well as concentrators in solar energy harvesting and in imaging objectives.

10 Lecture Notes on Wave Optics (03/05/14) 2.71/2.710 Introduction to Optics –Nick Fang

Left: Picture of an Optical Luneberg Lens (a ball 60 mm in diameter) used as spherical retro-reflector on Meteor-3M spacecraft. (Nasa.gov) Right: Ray Schematics of Luneberg Lens with a radially varying index of refraction. All parallel rays (red solid curves) coming from the left-hand side of the Luneberg lens will to a point on the edge of the sphere.

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