Interference. Gambar: Museum Victoria Australia

Transcription

1 Interference Gambar: Museum Victoria Australia

2 Formulation of Interference Intensity Superposition between two waves (point sources) Two separate point sources S 1 (x 1 ) and S 2 (x 2 ) generate EM waves with field strength E at point P(x): E 1 (x,t) dan E 2 (x,t) respectively, S 1 (x 1 ) r 1 =x- x 1 S 2 (x 2 ) r 2 =x- x 2 P(x) The resultant field on P is the superposition of the two fields: E P (x,t)=e 1 (x,t)+e 2 (x,t)

3 Intensity of Interference Energy flux density (power/area) at point P will be determined by Poynting vector N. Generally the EM field has very high frequency (optical waves) at hz, hence the sensor effectively would only detect the average quantity. The average of energy flux density (irradiance or intensity) is: < Re(N)> < E 2 > I (intensity) thus I = <Re(E) Re(E)> (in εc unit) I = ½ < E 2 >

4 Far Field Approximation For P significantly far from S 1 and S 2, waves arriving at P can be approximated as plane waves. Thus, if waves from di S 1 and S 2 are harmonic waves, at P we will have:

5 Intensity of Interference Using the usual definition of intensity: The last term is the interference term which shows variation of intensity caused by the superposition.

6 Interference and Coherence Criteria When the waves amplitudes are orthogonal then Such that I 12 =0 If amplitudes are parallel : Stationary interference pattern will occur if and only if 1. ω 1 =ω 2 k 1 =k 2 with ϕ 1 -ϕ 2 = constant 2. If ϕ 1 -ϕ 2 is not constant, but time required to change by 1 rad (τ) for (ϕ 1 -ϕ 2 )>>T (observation time), as long as ω 1 =ω 2, then the wave is said to be in coherence.

7 Interference and Coherence Criteria In case of coherence waves : Clearly: Also when I 1 =I 2 =I 0 then, Which means that :

8 Interference and Coherence Condition If the waves are incoherent due to : - rapid Δϕ changes (τ << T obervation time) - or Δϕ randomly changes, even if ω 1 = ω 2 : Then the term: <Re (E 1 E* 2 )> = 0, bringing total intensity to just the simple sum of each intensity: I = I 1 + I 2 Which means: no interference pattern will be observed!!

9 Partial Coherence Case Define Cross-corelation function between two waves: T/2 1 Γ 12 τ =< E 1 t E 2 t + τ = lim E 1 t E 2 t + τ dt T T T/2 For total incoherence : Γ 12 =0 If τ > τ C (Characteristic value) then the relation between E 1 and E 2 is random, such as Γ 0 if τ < τ C the partial coherence is present. We define a normal correlation function γ 12 as: γ 12 τ = Γ 12 τ Γ 11 0 Γ 22 0 = Γ 12 τ 2 I 1 I 2 = γ 12 e iθ

10 Partial Coherence Case We have used the definition that Then Γ jj 0 = 2I j The absolute value γ 12 shows the degree of coherence: 0 γ Interference pattern is then given by: With the max and min values are :

11 Contrast and the degree of coherence Sharpness of interference pattern is determined by its fringes visibility, with a contrast parameter: Meaning the degree of coherence correlates to the contrast: For I 1 =I 2 we have V = γ 12 Classification: if 1.Perfectly coherent, γ 12 =1, thus if 2. Total incoherence γ 12 =0 so V=0, meaning no contrast 3. Partial coherence:

12 Interferometer: Wavefront Splitting Interferometer : tools to produce interference Two types : (1) based on the splitting of the wavefront (2) or amplitude splitting 1. Young s Interferometer A wavefront splitting S 1 θ r 1 r y S d S 2 θ θ r 2 L>>d

13 Far Field Approximation Far field approximation, L>>d, hence r 1 //r 2 //r and r 1 -r 2 =d sin θ Incident waves on S 1 and S 2 are from the same wavefront so those are coherent Each of S 1 and S 2 becomes a new wave source Then at P, the linearly polarized waves from S 1 and S 2 are: Superposition at P (for equal amplitudes E 10 = E 20 = E 0 )

14 Intensity Distribution Iintensity is given by : I = ½ < E p 2 > So that Where I 0 = E 0 2 2, Δr = d sinθ k = 2π λ

15 Max and Min Interference Assuming S 1 and S 2 are in-phase initially (Δφ = 0): I max = 4I 0 if and I min = 0 if Under this ideal condition, the maximum intensities are constant throughout (independent of θ). The same applies for the minimum

16 Interference from two parallel surfaces: amplitude splitting d n 0 n θ 1 θ 1 A θ 2 B D E 1 E 2 Geometrical Analysis: AD = AC cos(π/2-θ 1 ) = AC sin (θ 1 ) tan θ 2 = ½ AC/d AC = 2d tan (θ 2 ) thus AD = 2d tan(θ 2 ) sin(θ 1 ), C 2 beams of light reflected from the top surface (AD) as well as the bottom surface (ABC). For n>n 0 each optical path is, respectively, ϕ 1 = k 0 AD+π ϕ 2 = k ABC = k2d/cos(θ 2 )

17 Phase Difference : Optical Path Length phase difference between the two is then: Δφ = φ 2 φ 1 = 2kd cos θ 2 2k 0 d tanθ 2 sin θ 1 π Snell s law gives : k 0 sin θ 1 = k sin θ 2 Δφ = 2kd cos θ 2 2kd sin2 θ 2 cos θ 2 π Δφ = 2kd cos θ 2 1 sin 2 θ 2 π = 2kd cos θ 2 π Note we use k and 2 are in the second instead of the first (incidence light) medium, because it is shorter. We can always express the wave number k for the incidence light by using the Snell s law.

18 Comparison to Young s Interference Δφ = 2kd cos θ 2 π But k = ω v = n ω c = n k 0 = n 2π λ 0, where n : second medium Index of refraction and 0 is the vacuum wavelength! So Small angle Δφ = 4πnd cos θ λ 2 π Δφ = 4πnd 2 1 cos θ 0 λ 1 π 0 Intensity pattern resulting from the interference is similar to the one we get from Young s: With n:glass refractive index, λ 0 wavelength in air, θ incident angle with respect to the glass (small). Interference pattern for different θ value can be derived from the expression above

19 Maximum & Minimum I = 4I 0 sin 2 I min =0, when 2πnd cos θ λ 0 2πnd cos θ λ 0 = mπ d cos θ = mλ 0 2n, m = 0,1,2, Find the condition for I max!