nature of light part 2 - university of colorado boulderecee.colorado.edu/~ecen5616/webmaterial/02...

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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

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Specular vs diffuse reflections

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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

||

+

−=

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Fresnel CoefficientsAmplitude and phase of rays at a boundary

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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…

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Example of n1<n2

(For real n)

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Example n1>n2

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

TM

TM

TE

TE

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Meaning of phase of reflection

TE TM

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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

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Plots of R and T

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Relationship to Fresnel Equation

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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

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Fresnel Diffraction Eq & 1 lens ImagingImpulse response


Delta function at (ξ,η) goes to lens

After Lens

Using Fresnel equation to propagate a distance z2

Putting it all together….

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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

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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

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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


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Definitions

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FT ExampleThin sinusoidal amplitude grating


FT of the 2 parts separately

Then use convolution theory (fortunately easy with delta functions)

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FT ExampleThin sinusoidal amplitude grating


Fraunhofer Diffraction Pattern

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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

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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

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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 !!!

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Fourier Optics in 1 Equation

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Propagation and DiffractionRegions of Diffraction

Rayleigh Range – near field, far field boundary

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Diffraction vs propagation distance

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

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Near Field/ Far Field Boundary

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Diffraction Limited Resolution

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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

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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.

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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

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InterferenceCombinedWavefront

Wave 1Wave 2

InPhase out of phase

Visibility = Imax-Imin/(Imax+Imin)

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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

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Interference patternS R

m

I (intensity) hologram index modulation

1

Spatial frequency of this pattern is

fx = 2 sinθ/λ

θθ

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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

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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.

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Homework

Available at the website under homework

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

Due in 2 weeks

nature of light part 2 - university of colorado boulderecee.colorado.edu/~ecen5616/webmaterial/02...

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