Nature of Light Part 2

Transcription

1 Nature of Light Part 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

2 Specular vs diffuse reflections

3 Derivation of Fresnel Coefficients Continuity of E tangential at interface and θ i =θ r gives Isotropic media Continuity of H tangential gives Since E = vb, then ne B this can be written as Using u i = u o and the definition r = nt cosθi μt ni cosθi + μ i ni cosθt μi nt cosθi μ t Similar argument for the other polarization

4 Fresnel Coefficients Amplitude and phase of rays at a boundary

5 Fresnel Coefficients Amplitude and phase of rays at a boundary

6 Uncoated Glass Reflection Normal Incidence for intensity (square r) R 2 = ((n 2 -n 1 )/(n 1 +n 2 )) 2 Glass is 1.5 to 1.8 index Air is 1 R 2 = (.5/2.5) 2 = 0.04 or 4% Or R 2 = (.8/2.8) 2 = 0.08 or 8% This is why you must use AR coatings on lenses

7 Example of n 1 <n 2 (For real n)

8 Example n 1 >n 2

9 Phase of Reflection Internal Reflection (n t >n i ) External Reflection (n t <n i ) TM TE TE TM

10 Meaning of phase of reflection TE TM

11 Intensity Depends on area and velocity t t t i r i i i i E n n E E n θ θ θ cos cos cos = 2 2 cos cos 1 + = i t i i t t i r E E n n E E θ θ Area of reflected and incident are the same but area of transmitted is different with different angle I i Acosθ i = I r cosθ i + I t Acosθ t Energy flow into an area A must equal energy flowing out of Multiply by c and use expression for intensity results in R =r 2 { T= t 2 { i i t t n n θ θ cos cos T+R=1

12 Plots of R and T

13 Relationship to Fresnel Equation

14 Lens as a thin transparency Positive thin lens transfer function is P(x,y) x Phase Delay (function) of lens Where P(x,y) is the pupil of the lens Lens Function (path length) Positive Lens Into phase paraxial Use definition of f Goodman, Introduction to Fourier Optics

15 Fresnel Diffraction Eq & 1 lens Imaging Impulse response Delta function at (ξ,η) goes to lens After Lens Using Fresnel equation to propagate a distance z 2 Putting it all together. Goodman, Introduction to Fourier Optics

16 Fresnel Diffraction Eq & 1 lens Imaging Impulse response ) Wave imaging condition of getting rid of quadratic phase turns into the Geometrical Optical Imaging condition 1/z 1 +1/z 2-1/f =0 2) 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 Goodman, Introduction to Fourier Optics

17 Fresnel Diffraction Eq & 1 lens Imaging Impulse response With all these phase factors assumed or hand waved away then can use the definition of M to rewrite remaining terms. M = -z 2 /z 1, 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 While these approximations work reasonably well under typical conditions This does not typically allow for these imaging systems to be cascaded. Goodman, Introduction to Fourier Optics

18 Fourier Transform using Lens Amplitude distribution behind the lens U t (x,y) 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 f x = u/λf Goodman, Introduction to Fourier Optics

19 Definitions

20 FT Example Thin sinusoidal amplitude grating FT of the 2 parts separately Then use convolution theory (fortunately easy with delta functions) Goodman, Introduction to Fourier Optics

21 FT Example Thin sinusoidal amplitude grating Fraunhofer Diffraction Pattern Goodman, Introduction to Fourier Optics

22 Question Spatial 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? f x = u/λf rewritten as u = f x λf = 1000(10-3 )(10mm) u = 10mm Lens diameter would have to be larger very fast lens! Draw picture

23 4F Lens System U(x,y) FT(U(x,y) U (-x,-y) f 1 f 1 f 2 f 2 Images with out phase terms true image. These systems can be cascaded. Magnification is ratio of focal lengths M=f 2 /f 1 Filtering or correlations can be done in FT plane Shortest track length imaging system

24 Phase Contrast Viewing How to see phase only object? U(x) FT(U(x)) U (-x) f 1 f 1 f f 2 2 Phase Object U(x) = e iϕ(x) If phase is small ( ϕ(x) 2 <<1) then object can be written as 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 try M(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!!! Reynolds, et al Physical Optics Notebook: Tutorials in Fourier Optics

25 Fourier Optics in 1 Equation

26 Propagation and Diffraction Regions of Diffraction Rayleigh Range near field, far field boundary

27 Diffraction vs propagation distance

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

29 Near Field/ Far Field Boundary

30 Near Field/ Far Field Boundary

31 Diffraction Limited Resolution

32 Linear Resolution of Points Sparrow Criterion Note: this is for points, not lines Rayleigh Criterion Rayleigh criterion for resolution is points separated by distance: Δx = 0.61λ nsinθ max 0.61λ = = 1.22λ( F /#) NA Depends on wavelength and largest angle of system (angular bandwidth)

33 Angular Resolution For telescopes or object in distance α = 1.22 w λ 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.

34 Questions for the day t < d 2 /λ - Rayleight range for direct transmission t < (λ/10) 2 /λ = λ/100 d/10 t < d/10 answer is D

35 Interference Combined Wavefront Wave 1 Wave 2 InPhase out of phase Visibility = I max -I min /(I max +I min )

36 Interference pattern DC terms Interference terms S R I (intensity) 1 hologram index modulation m Modulation depth: m = 2 I m = 1 when beam ratio is 1:1 DC component of illumination wasted media dynamic range R I R I + I S S

37 Interference pattern θ S θ R I (intensity) 1 hologram index modulation m Spatial frequency of this pattern is f x = 2 sinθ/λ

38 Interference Example 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

39 Interference Example Newton s Rings The radius of the Nth Newton's bright ring is given by r n = ((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. Wikipedia

40 Homework Available at the website under homework Due in 2 weeks