This preview shows pages 1–6. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1 Interference and diffraction • Interference and diffraction phenomena are both results of the superposition of EM waves from sources that are coherent to each other • Interference: combined effects from discrete sources • Diffraction: combined effects from a continuous distribution of sources Interference from two coherent sources  Young’s experiment : At point Q , the distance traveled for EM wave from source A is different from that from source B , ∆ : path difference. phase difference δ = π 2 period travel of time extra × = T 1 c 2 ⋅ ⋅ ∆ π = λ ∆ π ⋅ 2 • Phase difference of 0, 2 π , 4 π , … are equivalent as far as interference is concerned 2 Assuming sources A and B gives out EM waves of the same intensity, so that E T = E A + E B = E cos ω t + E cos( ω t+ δ ) Using complex notation, T E ~ = E 0 e i ω t + E 0 e i( ω + δ ) = E 0 e i ω t (1 + e i δ ) = E 0 e i ω t e i δ/2 (ei δ/2 + e i δ/2 ) ∴ E T = 2 E cos ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 2 δ cos ( ω t + 2 δ ) Since I T ∝ < E T 2 > and I ∝ < E 2 > , ∴ I T = 4 I 0 cos 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 2 δ = 0 when δ = π , 3 π , 5 π … = 4I when δ = 0, 2 π , 4 π … If the screen is located far from the sources, ∆ ≅ d sin θ ≅ d tan θ = L dx ∴ I T = 4 I 0 cos 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ L dx λ π = 0 when x = d 2 L λ , d 2 L 3 λ , d 2 L 5 λ … = 4I when x = 0, d L λ , d L 2 λ … ⇒ The wellknown Young’s interference bright and dark fringes on the screen 3 • What happen if the two sources are not completely coherent with respect to each other? • A wave is perfectly coherent if the phase has a definite value by any delay on the wave itself • A perfectly coherent wave must be monochromatic , i.e. a perfect sinusoidal wave with no beginning and ending • An EM wave with a finite duration must consist of a band of frequency components e.g. A wave with finite duration of 2 ∆ t : E(t) = t i e ω − ∆ t < t < ∆ t = otherwise This can be written as an integral over the frequency: ( ) ∫ ∞ ∞ − − = ω ω π ω d e E t E t i ) ( ~ 2 1 The spectral distribution E( ω ) is given by the Fourier transform of E(t) : ( ) ∫ ∫ ∆ ∆ − − ∞ ∞ − = = t t t i t i dt e dt e t E E ) ( ) ( ~ ω ω ω ω = t 2 t t sin ∆ ∆ ω ω ∆ ω ω ⋅ − − ) ( ] ) [( 4 The power spectrum I ( ω ) is proportional to the square of ( ) ω E ~ : ( ) ( ) 2 ~ ω ω E I ∝ The power spectrum of our e.g. look like this: Since ] ) [( t sin ∆ ω ω − = 0 when ω = ω ± t ∆ π ∴ The bandwidth of this spectral distribution, ∆ω , is given by ∆ω∆ t = π • This ‘uncertainty principle’ tell us that a finite wave duration requires a nonzero spread of frequency • Atoms emit light by jumping down from an excited state. Finite excited state lifetime ⇒ finite pulse duration ⇒ frequency spread • A collection of uncorrelated atoms gives an incoherent source 5 Coherent length l c : average length of the light wave within...
View
Full
Document
 Spring '11
 WKLIU
 Diffraction

Click to edit the document details