A wave through two slits Screen P=d sin d In Phase, i.e. Maxima when P = d sin = n Out of Phase, i.e. Minima when P = d sin = (n+1/2)
A wave through two slits In Phase, i.e. Maxima when P = d sin = n Out of Phase, i.e. Minima when P = d sin = (n+1/2) + +
The Intensity What is the intensity at P? The only term with a t dependence is sin 2 ( ).That term averages to ½. If we had only had one slit, the intensity would have been, So we can rewrite the total intensity as, with
The Intensity We can rewrite intensity at point P in terms of distance y Using this relation, we can rewrite expression for the intensity at point P as function of y Constructive interference occurs at where m=+/-1, +/-2 …
Phasor Addition of Waves Consider a sinusoidal wave whose electric field component is Consider second sinusoidal wave The projection of sum of two phasors E P is equal to E0E0 E 1 (t) tt E 2 (t) E0E0 E P (t) ERER /2 E0E0 tt E 1 (t) t+ E0E0 E 2 (t)
Phasor Diagrams for Two Coherent Sources E R =2E 0 E0E0 E0E0 E0E0 E0E0 ERER 45 0 E0E0 E0E0 ERER 90 0 E R =0 E0E0 E0E0 E0E0 E0E0 ERER 270 0 E R =2E 0 E0E0 E0E0
SUMMARY 2 slits interference pattern (Young’s experiment) How would pattern be changed if we add one or more slits ? (assuming the same slit separation ) 3 slits, 4 slits, 5 slits, etc.
Change of Phase Due to Reflection Lloyd’s mirror P2P2 P1P1 S I L Mirror The reflected ray (red) can be considered as an original from the image source at point I. Thus we can think of an arrangement S and I as a double-slit source separated by the distance between points S and I. An interference pattern for this experimental setting is really observed ….. but dark and bright fringes are reversed in order This mean that the sources S and I are different in phase by 180 0 An electromagnetic wave undergoes a phase change by 180 0 upon reflecting from the medium that has a higher index of refraction than that one in which the wave is traveling.
Change of Phase Due to Reflection 180 0 phase change n1n1 n2n2 n 1 <n 2 n1n1 n2n2 no phase change n 1 >n 2
Interference in Thin Films Air Film t 1 2 180 0 phase change no phase change A wave traveling from air toward film undergoes 180 0 phase change upon reflection. The wavelength of light n in the medium with refraction index n is The ray 1 is 180 0 out of phase with ray 2 which is equivalent to a path difference n /2. The ray 2 also travels extra distance 2t. Constructive interference Destructive interference
Chapter 33 – Act 1 Estimate minimum thickness of a soap-bubble film (n=1.33) that results in constructive interference in the reflected light if the film is Illuminated by light with =600nm. A) 113nm B) 250nm C) 339nm
Problem Consider the double-slit arrangement shown in Figure below, where the slit separation is d and the slit to screen distance is L. A sheet of transparent plastic having an index of refraction n and thickness t is placed over the upper slit. As a result, the central maximum of the interference pattern moves upward a distance y’. Find y’ where will the central maximum be now ?
Solution Corresponding path length difference: Phase difference for going though plastic sheet: Angle of central max is approx: Thus the distance y’ is: gives
Phase Change upon Reflection from a Surface/Interface Reflection from Optically Denser Medium (larger n) Reflection from Optically Lighter Medium (smaller n) by analogy to reflection of traveling wave in mechanics 180 o Phase ChangeNo Phase Change
Examples : constructive: 2t = (m +1/2) n destructive: 2t = m n constructive: 2t = m n destructive: 2t = (m +1/2) n
Application Reducing Reflection in Optical Instruments