LIGHT INTERFERENCE 0.Particle-wave (or wave- corpuscle) duality. 1.Coherent light sources. Conditions for gaining and weakening of waves. Max & Min of. - презентация

1 LIGHT INTERFERENCE 0.Particle-wave (or wave- corpuscle) duality. 1.Coherent light sources. Conditions for gaining and weakening of waves. Max & Min of interference. 2.Light interference in parallel layers. Blooming of optical systems. 3. Michelson and Fabry-Perot INTERFEROMETERS.

2 0. Particle-wave (or wave-corpuscle) duality Light and other ELECTOMAGNETIC WAVES have dual nature: some OPTICAL PHENOMENAE (interference, diffraction, polarization) would be easier to explain on the basis of WAVE properties of light, while other phenomena (like photoelectric effect) more BEAM of PARTICLES calledPHOTONS.

3 LIGHT as ELECTROMAGNETIC WAVE (E, H) Plane polarized wave (linearly polarized wave)

4 May 7 th : DAY of RADIO (in Russia) Radio Waves discovered by Heinrich HERTZ are also ELECTROMAGNETIC WAVES (the only difference with the LIGHT is frequency range hence WAVELENGTH )!

5 · Energy of a Quantum h : 1 эВ3 эВ эВ 250 кэВ МэВ 1 km 1 m 1 mm 1 mcm 400 nm nm m ( Wavelength ) Radio Waves IR... visible light... UV X-rays Gamma-rays... Frequency (Hz): Red Org. Y Grn Blue Violet 10 SPECTRUM of ELECTROMAGNETIC WAVES

6 May 7 th : RADIO DAY in Russia Alexander POPOV MARCONI Heinrich HERTZ Nikola TESLA

7 May 7 th : DAY of RADIO (in Russia) also, Day of PHYSICS and PHYSICISTS (such as some of us)!

8 1. Light Interference Light interference is such superposition of the optical (electromagnetic) waves that results in a stable PATTERN showing DISTRIBUTION of the areas where they GAIN (thus possible to obtain max) or weaken (thus possible to obtain min) each other.

9 Summing up two waves… …coming from TWO sources, SOURCE (A) and SOURCE (B), to the point of observation M. Source B Source A M – point of observation x1x1 x2x2

10 Summing up two waves… …coming from source (A) and source (B) to the point of observation M: if the wave oscillations occur in the same plain with the same frequency, the resulting amplitude is given by eq. (1): where is the phase difference of the oscillations brought to the point of observation M by the two waves, WAVE1 and WAVE2 that come into superposition.

11 NO COHERENCE – NO INTERFERENCE: For ordinary light sources (natural like The Sun, stars, or artificial like electric bulbs/lamps) the mean value of averaged by the radiation produced by all atoms participating in the acts of irradiation is 0, therefore: THIS IS NOT the INTERFERENCE – but what is this?

12 if no coherence – then, INSTEAD of light inteference, we obtain simply the sum of illuminance (or intensities)! Due to the fact that the intensity of a wave is proportional to the square of its amplitude, the equation describing the summing of the waves from non-coherent sources is: It means that the intensity of the total radiation FOR THIS case equals simply the sum of the intensities of the superposing waves. E.g., the illuminance produced by two lamps equals the sum of the illuminance produced by each of the lamps.

13 Then what is the Interference? Not just the sum I 1 +I 2 but… (see examples):

14 To obtain the condition for the INTERFEREN- CE of LIGHT, the light sources (A) and (B) should be COHERENT: their phase difference should not change (not be varying) with time. Practically, COHERENT WAVES are obtained by SPLITTING" the waves that are coming from ONE AND THE SAME source therefore the constant phase difference between the sources is maintaned (Newton rings, soap bubbles, oil films on water suface…): COHERENCE

15 Newton Rings The following 2 beams are interfering: (1)the one reflected by the lower surface of the lens, and (2)another beam that is reflected by the surface of GLASS on which the lens is residing. If a lens or glass have defects which are normally not seen, on the interference pattern such defects would be clearly evident.

16 Interference maxima and minima: According to eq. (1), INTERFERENCE MAXIMA are obtained when =0 ( Cos = 1 ). In such case, INTERFERENCE MINIMA occur when Cos = - 1.

17 Summing up two waves with constant phase difference Let us add oscillations of electric field (described by Ē vector) brought to some single point M by TWO WAVES separated from the source of WAVE1 by distance х 1 and from source of the WAVE2 by the distance х 2 :

18 Often BEAM1 and BEAM2 travel in different media thus their wave velocities may be different:

19 BEAM1 and BEAM2 sometimes travel to M in different MEDIA (n 1 & n 2 ) …coming from TWO sources, SOURCE (A) and SOURCE (B) to the point of observation M. Source B Source A M – point of observation x 1 (n 1 ) x 2 (n 2 )

20 BEAM1 and BEAM2 ARRIVE to M in PHASE => MAX …because one wave is strengthening the other. (They support one another). This SUM (GAIN) produces the MAXIMUMs. Source B Source A M – point of observation x1x1 x2x2

21 BEAM1 and BEAM2 ARRIVE to M in COUNTER-PHASE => MIN …because one wave is weakening the other. (They fight one another). This DIFFERENCE (subtraction) produces the MINIMUMs. Source B Source A M – point of observation x1x1 x2x2

22 Then, the PHASE DIFFERENCE between WAVE1 and WAVE2 arriving to M point from sources (A) and (B)

23 Path-Length Difference When analyzing phenomena of light INTERFERENCE, it is more convenient to utilize PATH- LENGTH DIFFERENCE rather than PHASE DIFFERENCE.

24 The PRODUCT of the PATH-LENGTH by absolute refraction index n of the respective MEDIA is called the OPTICAL PATH (x n); the difference of the two OPTICAL PATH- LENGTH is OPTICAL PATH DIFERENCE (SI: m). OPTICAL PATH DIFFERENCE

25 Phase difference and OPTICAL PATH DIFFERENCE obey the follow. relationship: or Relation between PHASE DIFFERENCE and OPTICAL PATH DIFFERENCE:

26 condition for a MINIMUM : = k -where k=0, 1, 2, … CONDITIONS for MAX GAIN and MIN (weakening) at waves interference in terms of PHASE DIFFERENCE : condition for a MAXIMUM : =2 k

27 condition for a MINIMUM : where k=0, 1, 2, … condition for a MAXIMUM : CONDITIONS for MAX GAIN and MIN (weakening) at waves interference in terms of optical path-length difference :

28 Interference MAXIMAE will be observed in such points where optical path-length difference equals integer WAVELENGTHs (i.e. EVEN number of half- wavelengths), while the MINIMAE where the optical path-length difference equals ODD number of half- WAVELENGTHs.

29 Interference of the two beams originating from two slots in a screen (distance between A & B is D). Slots SOURCE (A) & SOURCE (B) are the secondary sources of irradiation. Compare the distances from A & B to the point of observation M: Source B Source A M – point of observation x1x1 x2x2 observation screen D

30 angle Maximums of interference: D Sin = k (k = 0, ±1, ±2…). Source B Source A M – point of observation x1x1 x2x2 observation screen D Opaque (non- transparent) screen with slots A & B

31 2. Light Interference in a LAYER between Parallel Surfaces If a beam falls on a surface of a FLAT PLAIN BULK: Figure 1

32 To find the optical path-length difference between BEAMS 2 & 3, we put the normal from B point onto the direction of the BEAM 2:

33 When BEAM 2 reflects from optically MORE DENSE MEDIA, it results in the optical path-length difference SHIFT (OFFSET) of half wavelength. From Fig. 1 it follows:

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35 Optical Path-Length Difference would equal:

36 because then.

37 Using the condition for a MAXIMUM, we obtain:,.

38 Using the condition for the MINIMUM, we obtain:

39 Applications: Soap bubbles; LENS BLOOMING; INTERFEROMETERS.

40 Blooming of OPTICAL LENSES The idea of BLOOMING OPTICS is to arrange a condition for LIGHT INTERFERENCE MINIMUM for REFLECTION from the outer surface of a LENS (which is achieved with the aid of a thin METAL OXYDE FILM such as LaO). *It is possible to obtain such reflection MIN only for a single WAVELENGTH (GREEN is the medium in the visible light spectrum), =________?

41 3. Such PARALLEL LAYER (made at very high PRESISION) is in fact a FABRY-PEROT Interferometer! In FABRY-PEROT INTERFEROMETERS, the perpendicular beam is used: L=N stand =N ( n /2) where n = / n L n

42 Find N: if L=0.01 m, =535 nm (green light): L=N ( n / 2) L

43 …thus the quality of the PARALLEL surfaces is estimated for FARY-PEROT INTERFEROMETERS at a level of /10 or so (very difficult thus very expensive!).

44 In a MONOCHROMATIC light ( =const), Newton Rings would be alternating LIGHT and DARK. In a WHITE light (where one can find with various wavelength from 400 nm up to 780 nm (red), the colours of Newton Rings would be alternating LIGHT and DARK, as in the case of the SOAP BUBBLES :

45 3. INTERFEROMETERS (continued) INTERFEROMETERS are instruments based on the phenomenon of INTERFERENCE of waves. INTERFEROMETERS are used for HIGH PRECISION measurements of the WAVELENGTH, small distances, REFRACTION INDICES n (hence water and air purity), quality (smoothness) of surfaces that require high precision machining (such as optical lenses, mirrors, and prisms).

46 Michelson Interferometer

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48 Linnik Interferometer

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