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Interference of Light Waves

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1 Interference of Light Waves
Chapter 3 Interference of Light Waves

2 To observe interference in light wave, conditions are:
The sources must be coherent that is, they must maintain a constant phase with respect to each other. The sources must be monochromatic that is of a single wavelength. The superposition principle must apply.

3 Huygen’s Principle, Wavefronts and Coherence
Examples of coherence are: Laser light Small spot on tungsten filament Wavefront k Most light is incoherent: Two separate light bulbs Two headlight beams on a car Sun is basically incoherent

4 Interference is the combination of two or more waves to form a composite wave, based on the principle of superposition

5 Out of Phase by 180 degrees or p radians or l/2
In between In Phase

6 Young’s Double Slit Experiment,
The narrow slits, S1 and S2 act as sources of waves The waves emerging from the slits originate from the same wave front and therefore are always in phase

7 Resulting Interference Pattern
The light from the two slits form a visible pattern on a screen The pattern consists of a series of bright and dark parallel bands called fringes Constructive interference occurs where a bright fringe appears Destructive interference results in a dark fringe

8 Fringe Pattern The fringe pattern formed from a Young’s Double Slit Experiment would look like this The bright areas represent constructive interference The dark areas represent destructive interference

9 Interference Patterns
Constructive interference occurs at the center point The two waves travel the same distance Therefore, they arrive in phase

10 The upper wave has to travel farther than the lower wave
The upper wave travels one wavelength farther Therefore, the waves arrive in phase A bright fringe occurs

11 The upper wave travels one-half of a wavelength farther than the lower wave
The trough of the bottom wave overlaps the crest of the upper wave This is destructive interference A dark fringe occurs

12 Interference Equations
The path difference, δ, is found from the tan triangle δ = r2 – r1 = d sin θ This assumes the paths are parallel Not exactly parallel, but a very good approximation since L is much greater than d

13 For a bright fringe, produced by constructive interference, the path difference must be either zero or some integral multiple of the wavelength δ = d sin θbright = m λ m = 0, ±1, ±2, … m is called the order number When m = 0, it is the zeroth order maximum When m = ±1, it is called the first order maximum

14 The positions of the fringes can be measured vertically from the zeroth order maximum
y = L tan θ  L sin θ Assumptions L>>d d>>λ Approximation θ is small and therefore the approximation tan θ  sin θ can be used

15 When destructive interference occurs, a dark fringe is observed
This needs a path difference of an odd half wavelength δ = d sin θdark = (m + ½) λ m = 0, ±1, ±2, …

16 Interference Equations, final
For bright fringes For dark fringes

17 Intensity distribution of the double-slit interference pattern
The total electric field intensity at point P on the screen is the vector superposition of the two waves.

18 We suppose the two slits represent coherent sources of sinusoidal waves. Hence,
the waves have the same angular frequency ω and a constant phase difference φ .

19 Assuming the two waves have the same amplitude Eo , the electric field intensities at P due to each wave separately is : E1 = Eo sin ω t ,and E2 = Eo sin ( ω t + φ )  Although the waves are in phase at the slits, their phase difference φ at P depends on the path difference δ = r2 - r1 = d sin θ .

20 Because a path difference of λ (constructive interference) corresponds to a phase difference of 2 π rad, a path difference of λ/2 (destructive interference) corresponds to a phase difference of π rad, we obtain the ratio:

21 Ep = 2 Eo cos (φ/2) sin (ω t + φ/2)
Using the superposition principle we can obtain the resultant electric field at P Ep = E1 + E2 = Eo [sin ω t + sin (ω t + φ)] sin A+sin B = 2 sin[(A + B) / 2] cos[(A – B)/2] Taking A = ω t + φ , and B = ω t, Ep = 2 Eo cos (φ/2) sin (ω t + φ/2) the electric field at P has the same frequency as the light at the slits, but its amplitude is multiplied by the factor: 2 cos (φ /2).

22 the intensity of a wave is proportional to the square of the resultant electric field at that point.
I α Ep2 = 4 Eo2 cos2 (φ /2) sin2 (ω t + φ/2) the time average value of sin2 (ω t + φ/2) over one cycle is 1/2, Iav = Io cos2 ( φ / 2) where lo is the maximum possible time average light intensity

23 Iav = Io cos2 ( π d sin θ /λ )   since sinθ ≈ y/ L for small values of θ Iav = Io cos2 ( π yd /λ L)   Constructive interference, which produces intensity maxima, occurs when the quantity ( π yd / λ L) is an integral multiple of π , corresponding to : y = ( λ L / d) m. This is consistent with the equation of ybright .

24 Intensity distribution versus d sin θ or the double-slit pattern when the screen is far from the two slits (L » d).

25 Interference in Thin Films
Interference effects are commonly observed in thin films Examples are soap bubbles and oil on water The interference is due to the interaction of the waves reflected from both surfaces of the film

26 An electromagnetic wave traveling from a medium of index of refraction n1 toward a medium of index of refraction n2 undergoes a 180° phase change on reflection when n2 > n1 There is no phase change in the reflected wave if n2 < n1 The wavelength of light λn in a medium with index of refraction n is λn = λ/n where λ is the wavelength of light in vacuum

27 Ray 1 undergoes a phase change of 180° with respect to the incident ray
Ray 2, which is reflected from the lower surface, undergoes no phase change with respect to the incident wave

28 Ray 2 also travels an additional distance of 2t before the waves recombine
For constructive interference 2nt = (m + ½ ) λ m = 0, 1, 2 … This takes into account both the difference in optical path length for the two rays and the 180° phase change For destruction interference 2 n t = m λ m = 0, 1, 2 …

29 Two factors influence interference
Possible phase reversals on reflection Differences in travel distance The conditions are valid if the medium above the top surface is the same as the medium below the bottom surface If the thin film is between two different media, one of lower index than the film and one of higher index, the conditions for constructive and destructive interference are reversed

30 Be sure to include two effects when analyzing the interference pattern from a thin film
Path length Phase change

31 Newton’s Rings Another method for viewing interference is to place a planoconvex lens on top of a flat glass surface The air film between the glass surfaces varies in thickness from zero at the point of contact to some thickness t A pattern of light and dark rings is observed This rings are called Newton’s Rings The particle model of light could not explain the origin of the rings Newton’s Rings can be used to test optical lenses

32 The interference effect is due to the combination of ray l, reflected from the flat plate, with ray 2, reflected from the lower part of the lens. Imperfection in lens

33 Ray 1 undergoes a phase change of 180o upon reflection, because it is reflected from a medium of higher refractive index, whereas ray 2 undergoes no phase change the conditions for constructive and destructive interference are given, with n = 1 because the film is air. Point O is dark, because ray 1, reflected from the flat surface, undergoes a 180o phase change with respect to ray 2

34 Expressions for the radii of the bright and dark bands can be obtained in terms of the radius of curvature R and wavelength λ. For example, the dark rings have radii given by: r ≈ √ m λ / n . By measuring the radii of the rings, the wavelength can be obtained, provided R is known. Conversely, if the wavelength is accurately known, this effect can be used to obtain R.

35 The Michelson Interferometer
A beam of light provided by a monochromatic source is split into two rays by a partially silvered mirror, M, inclined at 45o to the incident light beam. One ray is reflected vertically upward toward mirror M2, while the second ray is transmitted horizontally through M toward mirror M2.

36 Hence, the two rays travel separate paths L1 and L2.
After reflecting from M1 and M2, the two rays eventually recombine to produce an interference pattern, which can be viewed through a telescope. The interference condition for the two rays is determined by their path length differences. When the two mirrors are exactly perpendicular to each other, the interference pattern is a target pattern of bright and dark circular fringes, similar to Newton’s rings.

37 As M1 is moved, the fringe pattern collapses or expands, depending on the direction in which M1 is moved. For example, if a dark circle appears at the center of the target pattern (corresponding to destructive interference) and M1 is then moved a distance λ /4 toward M0, the path difference changes by λ /2. What was a dark circle at the center now becomes a bright circle.

38 As M1 is moved an additional distance &/4
toward M0, the bright circle becomes a dark circle again. Thus, the fringe pattern shifts by one-half fringe each time M1 is moved a distance λ /4. The wavelength of light is then measured by counting the number of fringe shifts for a given displacement of M1. If the wavelength is accurately known, mirror displacements can be measured to within a fraction of the wavelength.

39 Anti-Reflecting Coatings
We can apply a thin film of a material with the proper thickness so that it will cause destructive interference of reflected light. Thus all the light must be transmitted through the film into the lens.

40 Problem Solving Strategy with Thin Films
Identify the thin film causing the interference Determine the indices of refraction in the film and the media on either side of it Determine the number of phase reversals: zero, one or two

41 The interference is constructive if the path difference is an integral multiple of λ and destructive if the path difference is an odd half multiple of λ The conditions are reversed if one of the waves undergoes a phase change on reflection

42 Problem Solving with Thin Films
Equation 1 phase reversal 0 or 2 phase reversals 2nt = (m + ½) l constructive destructive 2nt = m l

43 Example An example of different indices of refraction
A coating on a solar cell There are two phase changes

44 CD’s and DVD’s Data is stored digitally
A series of ones and zeros read by laser light reflected from the disk Strong reflections correspond to constructive interference These reflections are chosen to represent zeros Weak reflections correspond to destructive interference These reflections are chosen to represent ones

45 CD’s and Thin Film Interference
A CD has multiple tracks The tracks consist of a sequence of pits of varying length formed in a reflecting information layer The pits appear as bumps to the laser beam The laser beam shines on the metallic layer through a clear plastic coating

46 Reading a CD As the disk rotates, the laser reflects off the sequence of bumps and lower areas into a photodector The photodector converts the fluctuating reflected light intensity into an electrical string of zeros and ones The pit depth is made equal to one-quarter of the wavelength of the light

47 When the laser beam hits a rising or falling bump edge, part of the beam reflects from the top of the bump and part from the lower adjacent area This ensures destructive interference and very low intensity when the reflected beams combine at the detector The bump edges are read as ones The flat bump tops and intervening flat plains are read as zeros

48 DVD’s DVD’s use shorter wavelength lasers
The track separation, pit depth and minimum pit length are all smaller Therefore, the DVD can store about 30 times more information than a CD

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