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INTERFERENCE Interference patterns are a direct result of superpositioning. Antinodal and nodal lines are produced. These patterns can be enhanced using.

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Presentation on theme: "INTERFERENCE Interference patterns are a direct result of superpositioning. Antinodal and nodal lines are produced. These patterns can be enhanced using."— Presentation transcript:

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3 INTERFERENCE Interference patterns are a direct result of superpositioning. Antinodal and nodal lines are produced. These patterns can be enhanced using diffraction gratings, where all waves pass through each other from multiple point sources. We also learnt that the path difference for a point on a an antinodal line is always a factor of a wavelength,, whereas for a nodal line is half a wavelength, ½. Antinodal line path difference = n Nodal line path difference = n½ Where n = order 0, 1, 2, 3, ……. Interference patterns are a direct result of superpositioning. Antinodal and nodal lines are produced. These patterns can be enhanced using diffraction gratings, where all waves pass through each other from multiple point sources. We also learnt that the path difference for a point on a an antinodal line is always a factor of a wavelength,, whereas for a nodal line is half a wavelength, ½. Antinodal line path difference = n Nodal line path difference = n½ Where n = order 0, 1, 2, 3, …….

4  Red: crest meets crest Or trough meets trough. Constructive interference  Blue: crest meets a trough and they cancel out. Destructive interference

5  Can be used to calculate the path difference.  Whole numbers: antinodal lines  Half numbers: nodal lines

6 6 wavelength s 5 wavelengths

7 S1S1 S2S2 S 1 and S 2 are two coherent sources All points on a wavefront are in phase with one another All points on a wavefront are in phase with one another Waves interfere constructively where wavefronts meet. = antinodal lines Waves interfere constructively where wavefronts meet. = antinodal lines Along the nodal lines, destructive interference occurs. Here antiphase wavefronts meet. Along the nodal lines, destructive interference occurs. Here antiphase wavefronts meet. Wave Intensity (Fringes) Wave Intensity (Fringes) n = order number

8 Double slit Screen Monochromatic light, wavelength Young’s Double Slits A series of dark and bright fringes on the screen. A series of dark and bright fringes on the screen.

9 Young’s Double Slit Experiment THIS RELIES INITIALLY ON LIGHT DIFFRACTING THROUGH EACH SLIT. Where the diffracted light overlaps, interference occurs THIS RELIES INITIALLY ON LIGHT DIFFRACTING THROUGH EACH SLIT. Where the diffracted light overlaps, interference occurs Double slit Double slit screen Light INTERFERENCE Diffraction Some fringes may be missing where there is a minimum in the diffraction pattern

10 A B P Wave trains AP & BP have travelled the same distance (same number of ’s) Wave trains AP & BP have travelled the same distance (same number of ’s) Assuming the sources are coherent Hence waves arrive in-phase CONSTRUCTIVE INTERFERENCE (Bright fringe) Hence waves arrive in-phase CONSTRUCTIVE INTERFERENCE (Bright fringe)

11 L L Screen Slits d d = slit separation x = fringe separation

12 Normal light sources emit photons at random, so they are not coherent. LASER LASERS EMIT COHERENT LIGHT

13 Example 5: Monochromatic light from a point source illuminates two parallel, narrow slits. The centres of the slit openings are 0.80mm apart. An interference pattern forms on screen placed 2.0m away. The distance between two adjacent dark fringes is 1.2mm. Calculate the wavelength,, of the light used. Example 5: Monochromatic light from a point source illuminates two parallel, narrow slits. The centres of the slit openings are 0.80mm apart. An interference pattern forms on screen placed 2.0m away. The distance between two adjacent dark fringes is 1.2mm. Calculate the wavelength,, of the light used.

14 Example 5: Monochromatic light from a point source illuminates two parallel, narrow slits. The centres of the slit openings are 0.80mm apart. An interference pattern forms on screen placed 2.0m away. The distance between two adjacent dark fringes is 1.2mm. Calculate the wavelength,, of the light used. Example 5: Monochromatic light from a point source illuminates two parallel, narrow slits. The centres of the slit openings are 0.80mm apart. An interference pattern forms on screen placed 2.0m away. The distance between two adjacent dark fringes is 1.2mm. Calculate the wavelength,, of the light used. SOLUTION: The distance to the screen (2.0m) is large compared with the fringe spacing (1.2mm). The approximation formula can be used. n = dx/L[n = 1 because the fringe spacing is being calculated] = (8.0 x x 1.2 x ) / 2.0 = 4.8 x m SOLUTION: The distance to the screen (2.0m) is large compared with the fringe spacing (1.2mm). The approximation formula can be used. n = dx/L[n = 1 because the fringe spacing is being calculated] = (8.0 x x 1.2 x ) / 2.0 = 4.8 x m

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16  Decide which points are Constructive interference and which are Destructive interference?

17  In phase  Out of phase By 180 deg (half a wavelength)

18  Quantum Physics. Quantum Physics. 

19  Double slit animation. Double slit animation.  edinger/two-slit2.html edinger/two-slit2.html

20 A student uses a laser and a double-slit apparatus to project a two-point source light interference pattern onto a whiteboard located 5.87 meters away. The distance measured between the central bright band and the fourth bright band is 8.21 cm. The slits are separated by a distance of mm. What would be the measured wavelength of light?

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23 PD= m λ Two point sources, 3.0 cm apart, are generating periodic waves in phase. A point on the third antinodal line of the wave pattern is 10 cm from one source and 8.0 cm from the other source. Determine the wavelength of the waves. Two point sources are generating periodic waves in phase. The wavelength of the waves is 3.0 cm. A point on a nodal line is 25 cm from one source and 20.5 cm from the other source. Determine the nodal line number.

24 The Diffraction Grating: This is a piece of glass with tiny slits made in it to produce small point sources. A formula can be used to relate to the interference pattern produced by a particular diffraction grating. dsin  = n (Where n = 0, 1, 2, 3 …….) Often N, the number of slits per metre, or slits per centimetre is given. The slit spacing d is related to N by: d = 1/N

25 Grating Monochromatic light  C  For light diffracted from adjacent slits to add constructively, the path difference = AC must be a whole number of wavelengths. AC = AB sin  and AB is the grating element = d Hence d sin  n d = grating element A B

26 DIFFRACTION GRATING WITH WHITE LIGHTHence in any order red light will be more diffracted than blue.A spectrum will result White Central maximum, n = 0 First Order maximum, n = 1 First Order maximum, n = 1 Second Order maximum, n = 2 Second Order maximum, n = 2 Several spectra will be seen, the number depending upon the value of d Grating screen

27 n=0 n=2n=1n=3 grating Note that higher orders, as with 2 and 3 here, can overlap Note that higher orders, as with 2 and 3 here, can overlap Note that in the spectrum produced by a prism, it is the blue light which is most deviated

28 Example: Light from a laser passes through a diffraction grating of 2000 lines per cm. The diagram below shows the measurement made. laser 0 order 2 nd order 0.5m Grating  2m Calculate the wavelength of the light. SOLUTION:Slit spacing d = 1/N= 1/200000= 5.00 x m sin  = 0.5/2 =  = dsin  /n = (5.00 x x 0.250) / 2 = 6.25 x m

29  /ipmj/java/slitdiffr/index.html /ipmj/java/slitdiffr/index.html


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