ECE 5616Curtis

Nature of LightPart 2

• Fresnel Coefficients• From Helmholts equation see imaging conditions for

• Single lens• 4F system

• Diffraction ranges• Rayleigh Range• Diffraction limited resolution

• Interference – Newton’s Rings

ECE 5616Curtis

Specular vs diffuse reflections

ECE 5616Curtis

Derivation of Fresnel Coefficients

Continuity of E tangential at interface and θi=θr gives

Continuity of H tangential gives

Since E = vB, then nE ≈B this can be written as

Using ui = uo and the definition

Similar argument for the other polarization

Isotropic media

it

ti

i

i

ti

ii

t

t

nn

nn

rθ

μθ

μ

θμ

θμ

coscos

coscos

||

+

−=

ECE 5616Curtis

Fresnel CoefficientsAmplitude and phase of rays at a boundary

ECE 5616Curtis

Fresnel CoefficientsAmplitude and phase of rays at a boundary

ECE 5616Curtis

Uncoated Glass ReflectionNormal Incidence for intensity (square r)

R2 = ((n2-n1)/(n1+n2))2

Glass is 1.5 to 1.8 indexAir is 1

R2 = (.5/2.5)2 = 0.04 or 4%

Or

R2 = (.8/2.8)2 = 0.08 or 8%

This is why you must use AR coatings on lenses…

ECE 5616Curtis

Example of n1<n2

(For real n)

ECE 5616Curtis

Example n1>n2

ECE 5616Curtis

Phase of ReflectionInternal Reflection (nt>ni) External Reflection (nt<ni)

TM

TM

TE

TE

ECE 5616Curtis

Meaning of phase of reflection

TE TM

ECE 5616Curtis

IntensityDepends on area and velocity

tttiriiii EnEnEn θθθ coscoscos 222 +=22

coscos1 ⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=

i

t

ii

tt

i

r

EE

nn

EE

θθ

Area of reflected and incident are the same but area of transmitted is different with different angle

IiAcosθi = Ircosθi + ItAcosθt

Energy flow into an area A must equal energy flowing out of

Multiply by c and use expression for intensity results in

R =r2

{

T= t2

{

⎟⎟⎠

⎞⎜⎜⎝

⎛

ii

tt

nn

θθ

coscos

T+R=1

ECE 5616Curtis

Plots of R and T

ECE 5616Curtis

Relationship to Fresnel Equation

ECE 5616Curtis

Lens as a thin transparencyPositive thin lens transfer function is

P(x,y) x Phase Delay (function) of lens

Where P(x,y) is the pupil of the lens

Goodman, “Introduction to Fourier Optics”

Positive Lens

Lens Function (path length)

paraxial

Into phase

Use definition of f

ECE 5616Curtis

Fresnel Diffraction Eq & 1 lens ImagingImpulse response

Goodman, “Introduction to Fourier Optics”

Delta function at (ξ,η) goes to lens

After Lens

Using Fresnel equation to propagate a distance z2

Putting it all together….

ECE 5616Curtis

Fresnel Diffraction Eq & 1 lens ImagingImpulse response

Goodman, “Introduction to Fourier Optics”

1) Wave imaging condition of getting rid of quadratic phase turns into the Geometrical Optical Imaging condition

1/z1+1/z2-1/f =02) Small field curvature and only interested in intensity

3) Object is illuminated with spherical wave, OR small phase change in region of object and contributes to that image point

01

20 3

0

ECE 5616Curtis

Fresnel Diffraction Eq & 1 lens ImagingImpulse response

Goodman, “Introduction to Fourier Optics”

With all these phase factors assumed or hand waved away then can use the definition of M to rewrite remaining terms.

While these approximations work reasonably well under typical conditionsThis does not typically allow for these imaging systems to be cascaded.

M = -z2/z1, magnification of image

So, with imaging condition, the impulse response is the scaled FT of the lens aperture centered on coordinates u= Mξ, v=Mη

ECE 5616Curtis

Fourier Transform using Lens

Ut(x,y)

Amplitude distribution behind the lens

Plug into Fresnel equation and set z=f to find field at back focal plane

Plug in U’(x,y) and notice that quadratic phase factor cancels leaving FT with small field curvature (Petzal field curvature)

Notice fx = u/λf

Goodman, “Introduction to Fourier Optics”

ECE 5616Curtis

Definitions

ECE 5616Curtis

FT ExampleThin sinusoidal amplitude grating

Goodman, “Introduction to Fourier Optics”

FT of the 2 parts separately

Then use convolution theory (fortunately easy with delta functions)

ECE 5616Curtis

FT ExampleThin sinusoidal amplitude grating

Goodman, “Introduction to Fourier Optics”

Fraunhofer Diffraction Pattern

ECE 5616Curtis

QuestionSpatial frequency of pattern is 1000 lp/mm in x direction and is Fourier Transformed by lens of F=10mm and wavelength of 1 um. Where does this spatial frequency end up on the Fourier plane ?

fx = u/λf rewritten as u = fxλf = 1000(10-3)(10mm)

u = 10mm

Lens diameter would have to be larger – very fast lens !

Draw picture

ECE 5616Curtis

4F Lens System

Images with out phase terms – true image.

These systems can be cascaded.

Magnification is ratio of focal lengths M=f2/f1

Filtering or correlations can be done in FT plane

Shortest track length imaging system

U(x,y) FT(U(x,y) U’(-x,-y)

f1 f1 f2 f2

ECE 5616Curtis

Phase Contrast ViewingHow to see phase only object?

Reynolds, et al “Physical Optics Notebook: Tutorials in Fourier Optics’

U(x) FT(U(x)) U’(-x)

f1 f1 f2

f2

U(x) = e iϕ(x)

If phase is small (| ϕ(x) |2<<1) then object can be written as

Phase Object

U(x) =1+iϕ(x)At FT plane this is approximately

U(ξ) = δ(ξ) +iϕ~(x)

Now in order to see something in intensity we must filter this tryM(x) = e iπ/2 if |x|<ε, for ε small M(x) = 1 if |x|>ε

With this multiplied into U() and then FT again, the intensity becomes

|U’(x)|2 = |1 +ϕ(x)|2 ~ 1+2 ϕ(x)

Phase is visible on camera !!!

ECE 5616Curtis

Fourier Optics in 1 Equation

ECE 5616Curtis

Propagation and DiffractionRegions of Diffraction

Rayleigh Range – near field, far field boundary

ECE 5616Curtis

Diffraction vs propagation distance

ECE 5616Curtis

First Imaging LimitationSpatial Frequencies beyond TIR do not propagate and result in a band limited image with propagation. For example a 1D rect function.

ECE 5616Curtis

Near Field/ Far Field Boundary

ECE 5616Curtis

Near Field/ Far Field Boundary

ECE 5616Curtis

Diffraction Limited Resolution

ECE 5616Curtis

Linear Resolution of Points

Rayleigh Criterion

Sparrow Criterion

Rayleigh criterion for resolution is points separated by distance:

Depends on wavelength and largest angle of system (angular bandwidth)

)/#(22.161.0sin

61.0

max

FNAn

x λλθλ

===Δ

Note: this is for points, not lines

ECE 5616Curtis

Angular ResolutionFor telescopes or object in distance

wλα 22.1

=

Where α is in radians and w is the DIAMETER of the circular aperture

Notice for angular resolution the wavelength and the diameter are the key parameters. Note that the focal length does not directly determine the angular resolution.

ECE 5616Curtis

Questions for the day

t < d2/λ - Rayleight range for direct transmissiont < (λ/10)2/λ = λ/100∼ d/10t < d/10 – answer is D

ECE 5616Curtis

InterferenceCombinedWavefront

Wave 1Wave 2

InPhase out of phase

Visibility = Imax-Imin/(Imax+Imin)

ECE 5616Curtis

Interference pattern

DC terms Interference terms

S R

m

I (intensity) hologram index modulation

1

Modulation depth:

DC component of illuminationwasted media dynamic range

m = 1 when beam ratio is 1:1

SR

SR

IIII

m+

=2

ECE 5616Curtis

Interference patternS R

m

I (intensity) hologram index modulation

1

Spatial frequency of this pattern is

fx = 2 sinθ/λ

θθ

ECE 5616Curtis

InterferenceExample Newton’s Rings

Newton's rings is an interference pattern caused by the reflection of light between a spherical surface and an adjacent flat surface. When viewed with monochromatic light it appears as a series of concentric, alternating light and dark rings centered at the point of contact between the two surfaces. When viewed with white light, it forms a concentric ring pattern of rainbow colors because the different wavelengths of light interfere at different thicknesses of the air layer between the surfaces. The light rings are caused by constructive interference between the light rays reflected from both surfaces, while the dark rings are caused by destructive interference. Also, the outer rings are spaced more closely than the inner ones. Moving outwards from one dark ring to the next, for example, increases the path difference by the same amount λ, corresponding to the same increase of thickness of the air layer λ/2. Since the slope of the lens surface increases outwards, separation of the rings gets smaller for the outer rings.

Wikipedia

ECE 5616Curtis

InterferenceExample Newton’s Rings

Wikipedia

The radius of the Nth Newton's bright ring is given by

rn = ((N-1/2)λR)1/2

where N is the bright ring number, R is the radius of curvature of the lens the light is passing through, and λ is the wavelength of the light passing through the glass.

ECE 5616Curtis

Homework

Available at the website under homework

http://ecee.colorado.edu/~ecen4616http://ecee.colorado.edu/~ecen5616

Due in 2 weeks