Your SlideShare is downloading. ×
Interference of light 10
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×

Saving this for later?

Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime - even offline.

Text the download link to your phone

Standard text messaging rates apply

Interference of light 10

9,480
views

Published on

SACE Physics Section 3 Topic 2

SACE Physics Section 3 Topic 2

Published in: Education, Technology, Business

1 Comment
4 Likes
Statistics
Notes
No Downloads
Views
Total Views
9,480
On Slideshare
0
From Embeds
0
Number of Embeds
3
Actions
Shares
0
Downloads
0
Comments
1
Likes
4
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. INTERFERENCE OF LIGHT 12 SACE PHYSICS-STAGE 2 SECTION 3 TOPIC 2 PRINCE ALFRED COLLEGE
  • 2. DIFFRACTION
    • Diffraction is the bending of a wave as it passes near an obstacle while remaining in the same medium.
    • This phenomena occurs in water, sound and light waves.
  • 3. DIFFRACTION
    • When light or water pass through a wide slit we observe that the waves don’t spread out very much.
    • Diffraction - Changing Gap Width
  • 4. DIFFRACTION
    • However, as the slit becomes more narrow, the waves spread out more.
  • 5. DIFFRACTION
    • As the slit width approaches the wavelength of the water wave, the diffraction is very noticeable.
    • Diffraction - Changing Slit Width (again)
  • 6. DIFFRACTION
    • If light passes through an opening that is wide compared to the wavelength, the diffraction effect is small. The result is that we see a sharp shadow.
  • 7. DIFRACTION
    • If the opening width approaches the wavelength of the light, the diffraction produced becomes more pronounced and the shadow becomes more fuzzy around the edges. The light is diffracted by the thin slit.
  • 8. DIFFRA C TION
    • This can also happen when radio waves pass between two large objects.
    • Consider, two large buildings in the centre of Adelaide.
    • FM Waves have a wavelength of 2-3 m. They therefore pass through the space between buildings with little diffraction.
  • 9. DIFFRACTION
    • AM Radio waves have a much longer wavelength (200-300 metres). It is at least as large as the gap between the buildings. They therefore show considerable diffraction
    • This allows you to get good AM reception while the FM reception has “shadow zones” in cities.
  • 10. DIFFRACTION
    • This also explains why you get such poor FM reception in the hills.
    • Other examples of diffraction include sound diffracting around a building and
    • Ocean waves diffracting around a jetty.
  • 11. DIFFRACTION
    • Streetlights viewed through a dirty windscreen at night will show a diffraction pattern in the form of a cross.
    • This is due to the fine dust particles having a very small gap between them.
  • 12. INTERFERENCE
    • INTERFERENCE - occurs when two or more waves pass through the same space. CONSTRUCTIVE INTERFERENCE - two waves that have the same phase relationship and frequency.
    • Example: Two waves that have overlapping crests leads to a doubling of the amplitude of the wave.
    • Constructive Wave Interference
  • 13. INTERFERENCE
    • DESTRUCTIVE INTERFERENCE - two waves are out of step.
    • Example: One waves crest meets another waves trough. The waves will cancel each other out.
    • Destructive Interference Explanation
    • Wave Interference Example 1
    • Wave Interference Example 2
  • 14. INTERFERENCE
    • A constant phase relationship is maintained in both constructive and destructive interference when the frequency of both waves is the same.
    • We therefore say that the two sources that produced the waves are mutually coherent.
  • 15. INTERFERENCE
    • Mutual Coherence - two waves that are…
      • Always constructive (in phase)
      • Always destructive (one half wave out of phase)
      • At the same points.
      • Maintain a constant phase relationship.
  • 16. WATER WAVE INTERFERENCE
    • Here is an example of two water waves interfering with each other.
    • Two Source Interference
    • The two dippers are striking the water at the same time with the same frequency at S1 and S2.
  • 17. WATER WAVE INTERFERENCE
    • The Law of Superposition applies (sum of all waves added at a particular point).
    • Constructive Interference (crest meets crest or trough meets trough) - waves in phase and reinforcement occurs. Resultant wave has twice the amplitude.
    • Destructive Interference (crest meets trough) - waves half step out of phase and annulment occurs. Zero amplitude.
    • Superposition
  • 18. WATER WAVE INTERFERENCE
    • As the two waves have the same frequency they will have a constant phase relationship (coherence).
    • They therefore will have points that will always constructively interfere (ANTINODES) and destructively interfere (NODES).
  • 19. WATER WAVE INTERFERENCE
    • The nodes on the diagram are open circles.
    • The antinodes are shown as black dots.
  • 20. WATER WAVE INTERFERENCE
    • A two dimensional standing wave pattern will be seen.
    • If the dippers were out of phase by  radians (  /2), the 2D standing wave pattern would still be seen.
    • In both cases, the dippers maintain a constant phase relationship therefore they are mutually coherent sources.
  • 21. WATER WAVE INTERFERENCE
    • Because both waves are in phase, the central band will be constructive interference (antinodes). We call this the central order antinode. It will always be there.
  • 22. WATER WAVE INTERFERENCE
    • Confirm that P1, P2 and P3 on the diagram are points of destructive interference.
    • Confirm that P4 is a point of constructive interference.
    • Maxima & Minima
  • 23. WATER WAVE INTERFERENCE
    • For a point to be on a nodal line, the difference between its distance from one source and the distance from the other source must be an odd number of half wavelengths.
    • This is called the Geometric Path Difference (G.P.D.)
  • 24. WATER WAVE INTERFERENCE
      • For
  • 25. WATER WAVE INTERFERENCE
    • The G.P.D. is given by:
  • 26. WATER WAVE INTERFERENCE
    • For
  • 27. WATER WAVE INTERFERENCE
    • The G.P.D. is given by:
  • 28. WATER WAVE INTERFERENCE
    • In general, annulment will occur when…
    • m is an integer and the
    • G.P.D. = (m + ½)  m = 0, 1, 2 ……
  • 29. WATER WAVE INTERFERENCE
    • P4 is a point on an antinodal line
  • 30. WATER WAVE INTERFERENCE
    • The G.P.D. is given by:
  • 31. WATER WAVE INTERFERENCE
    • For any point on an antinodal line, the G.P.D. must be an even number of half -wavelengths.
    • Reinforcement (antinodes) will occur at
    • G.P.D. = m λ
    • m = 0, 1, 2…..
  • 32. WATER WAVE INTERFERENCE
    • What if we were to reverse the phase of one source?
    • A crest leaves S1 at the same time as a trough leaves S2.
    • Annulment… G.P.D. = m 
    • Reinforcement … G.P.D =
    (m + ½) 
  • 33. WATER WAVE INTERFERENCE
    • BANDWIDTH - the distance between consecutive nodes and antinodes.
    • If we decrease the wavelength, we decrease the bandwidth.
  • 34. WATER WAVE INTERFERENCE
    • What if we decreased the distance between the two sources (S1 and S2)?
    • The bandwidth would increase.
  • 35. WATER WAVE INTERFERENCE
    • This discussion has been based on water wave interference.
    • These principles also apply to other types of wave energy such as sound, microwave etc.
    • We will now focus for the rest of the chapter on how LIGHT energy disperses and produces interference patterns.
  • 36. LIGHT INTERFERENCE
    • In the case of light, it is the overlapping electromagnetic waves that produce the interference patterns.
    • It is difficult to observe interference patterns with light due its very small wavelength of around 4.7 x 10 -7 m
    • The bandwidth would be extremely small (anti nodes close together).
  • 37. LIGHT INTERFERENCE
    • Also, any interference pattern produced by an incandescent light bulb would not be sent out in a synchronized fashion.
    • It would not be a monochromatic light source due to the fact that there would be many different frequencies.
    • It would also not be coherent (same phase relationship) due to the fact that electrons...
  • 38. LIGHT INTERFERENCE
    • … give off light at different times within the atoms. This would mean that various crests and troughs would not be synchronized even if they were all of the same frequency.
    • The antinodal and nodal lines would be constantly shifting.
  • 39. LIGHT INTERFERENCE
    • A mutually coherent light source is necessary to see an interference pattern.
    • Thomas Young (1800’s) devised a method of doing this.
    • This is called the Young’s Double Slit Experiment.
  • 40. LIGHT INTERFERENCE
    • A single slit is illuminated with monochromatic (light of all the same frequency) light. It might be all blue light or all red light.
    • This slit becomes a point source and emits circular wavefronts.
    • The light emerging from the slit is coherent (only one crest or trough can get through at any given time).
  • 41. LIGHT INTERFERENCE
    • If the double slits are equidistant from the single slit, a crest (or trough) will reach the double slits simultaneously.
  • 42. LIGHT INTERFERENCE
    • The monochromatic light waves will be mutually coherent and in phase.
    • This will lead to a stable interference pattern as we saw with water waves.
  • 43. LIGHT INTERFERENCE
    • A sodium vapour lamp would be a good monochromatic light source if using the Young’s Slits Experiment.
    • If you are using a LASER there is no need for the single slit as LASER light is already monochromatic and coherent.
  • 44. LIGHT INTERFERENCE
    • What would you see from all of this?
    • If you were to shine the double slit towards a white screen you would see alternating bands of bright and dark fringes on the screen.
    • The bright bands would be antinodal lines, places where two crests (or troughs) were overlapping.
  • 45. LIGHT INTERFERENCE
    • The regions of darkness would be nodal lines, places where a crest and a trough were overlapping and causing cancellation.
    • Young's Double
    • Slit Experiment
  • 46. LIGHT INTERFERENCE
    • THE DERIVATION OF d sin θ = m λ
      • This is an essential derivation from the syllabus.
      • d = distance between the two slits.
      • L = distance between the slits and the screen.
      • y = the bandwidth (distance between consecutive antinodes or consecutive nodes).
      • Optical Path Difference (O.P.D.) = the extra difference that one of the slits is from the screen.
  • 47. LIGHT INTERFERENCE
  • 48. LIGHT INTERFERENCE
    • bisects the angle by the two waves that meet at . is at right angles to in order that we have which is the extra distance or the O.P.D. between the waves.
  • 49. LIGHT INTERFERENCE
    • Since is very small, the angle is approximately . This means that triangle
    • can be treated as a right angled triangle
    • sin = or = d sin
    • Remember that is the Optical Path Difference.
  • 50. LIGHT INTERFERENCE
    • Therefore, when there is reinforcement at
    • The O.P.D. = d sin
    • m = d sin
  • 51. LIGHT INTERFERENCE
    • When there is annulment at
    • The O.P.D. = d sin
    • (m + 1/2) = d sin
  • 52. DERIVING THE BANDWIDTH
    • (Not required in 2007)
    • For the bandwidth, y :
    • Consider which is a distance from the centre O and is a reinforcement position.
  • 53. DERIVING THE BANDWIDTH
    • In MP 1 O, tan =
    • In S 1 S 2 D, sin =
  • 54. DERIVING THE BANDWIDTH
    • ( S 2 D = m for reinforcement)
    • For a small , sin = tan
    • which then gives y 1 =
  • 55. DERIVING THE BANDWIDTH
    • Now, from the bandwidth equation, answer the following conceptual questions.
    • y 1 =
    • What happens to the bandwidth if you…
      • Decrease the wavelength (Red to blue light).
      • Decrease the distance from the slit to the screen
      • Decrease the slit separation.
  • 56. EVIDENCE FOR THE WAVE THEORY
    • Young’s experiment was important as at the time, light was thought to be particle in nature due to the influence of Sir Isaac Newton.
    • This result cannot be explained in terms of particles but only in terms of waves.
  • 57. EVIDENCE FOR THE WAVE THEORY
    • If it was particles, only one beam of particles would pass through the first slit and be blocked by the two slits as they are not in line.
    • Even if they did pass through, it would be expected that 2 bright spots would be seen on the screen.
  • 58. Try Example 1
    • Light of wavelength 5.0 x 10 -7 m is used in a Young’s double slit interferometer. The distance between the double slits and the screen is 25 cm. The centre of the third bright fringe is a distance of 5.0 mm from the middle of the central bright fringe in the pattern. What is the separation of the slits?
  • 59. Example 1 Solution
  • 60. Example 1 Solution
    • Use the bandwidth equation…
    • d = 7.5 x 10 -5 m
  • 61. Try Example 2
    • Light of a certain wavelength produces an interference pattern with bright fringes 2.0 mm apart. The screen is 50 cm from the two slits which are separated by 1.0 x 10 -1 mm. Determine the wavelength of light used.
  • 62. Example 2 Solution
    • Use the bandwidth formula…
    • 4 x 10 -7 m or 400nm (nanometres)
    • (Violet Light)
  • 63. LASERS
    • This application will be dealt with in more detail in Section 4, Topic 1.
    • In relatively recent times, light sources have been developed in which all the waves emitted by the source are the same wavelength and are all in phase.
  • 64. LASERS
    • The light from these sources is perfectly coherent and is produced by a process known as Light Amplification by Stimulated Emission of Radiation (LASER).
    • The full explanation of a laser can only be given with quantum theory.
  • 65. LASERS
    • The atoms of (for example) a He - Ne gas laser are optically stimulated to emit light all of the same frequency and in such a way that the waves emitted are in phase with each other.
  • 66. LASERS
    • The emitted light gains in intensity by being reflected many times through the active material which is being stimulated, and then being released from the laser through a partially reflecting end mirror.
    • The resultant beam is highly monochromatic, coherent, intense and unidirectional.
  • 67. LASERS
    • Often, the laser light appears to be speckled when shone onto a screen or wall (such as in one of the year 12 practicals).
    • Speckle is produced whenever a laser beam is reflected by a rough surface. It is due to the interference between light reflected in different directions from the surface.
  • 68. LASERS
    • This is interference due to reflection (several waves combining for amplification)
  • 69. SINGLE SLIT DIFFRACTION
    • The following work on single slit, double slit and multiple slit diffraction is for reference only. It is not part of the syllabus but it will help you to understand how diffraction gratings work.
  • 70. SINGLE SLIT DIFFRACTION
    • The pattern we should expect to see from a narrow slit illuminated with monochromatic light is a broad band of light whose intensity decreases further from the centre.
    • This does happen if the slit is one wavelength or less as shown below:
  • 71. SINGLE SLIT DIFFRACTION
  • 72. SINGLE SLIT DIFFRACTION
    • If the slit is greater than 2-3 wavelengths, annulment lines appear.
    • This occurs because there are several sources of secondary wavelets which allow for reinforcement and annulment as shown below:
  • 73. SINGLE SLIT DIFFRACTION
  • 74. SINGLE SLIT DIFFRACTION
    • The characteristics that the above diagram shows include:
    • The central bandwidth is twice the size of those on either side (i.e. 2 y )
    • The intensity of the maxima diminishes quickly with the increased distance from the centre of the screen.
  • 75. SINGLE SLIT DIFFRACTION
    • The central band has about 86% of the light energy.
    • If the slit width (D) decreases, then the bandwidth increases and the pattern spreads outwards, losing intensity since the amount of light must be spread out over a bigger area,
  • 76. SINGLE SLIT DIFFRACTION
    • If the distance from the slit to the screen ( L ) increases, the bandwidth increases and there is a reduction in intensity with the spreading area.
    • If the wavelength (  ) decreases, bandwidth decreases and the pattern becomes more intense.
  • 77. SINGLE SLIT DIFFRACTION
    • The relationship between the wavelength, the distance from the slit to the screen (L), the slit width (D) and the bandwidth (y) are summarised in the equation:
    • This is the equation for a SINGLE SLIT.
  • 78. SINGLE SLIT DIFFRACTION
    • Notice that this equation is different from the equation derived for a DOUBLE SLIT.
    • …which has the symbol, d (distance between the two slits).
  • 79. DOUBLE SLIT DIFFRACTION
    • The pattern produced from monochromatic light also has distinctive characteristics.
    • Each slit produces a diffraction pattern and because the slits are so close, they coincide on the screen.
    • From Young’s double slit, the bandwidth is given by where d is the distance
    • between the centres of the slits.
  • 80. DOUBLE SLIT DIFFRACTION
    • The bandwidth for the diffraction fringes is
    • , where D is the width of each slit.
    • This is the same as the bandwidth for a single slit diffraction pattern, and the central diffraction band is twice the width of the side bands.
  • 81. DOUBLE SLIT DIFFRACTION
    • Since D is much smaller than d , the diffraction bands are larger than those in the interference pattern. The result is an interference pattern with fringes missing at regular intervals.
  • 82. DOUBLE SLIT DIFFRACTION
    • Revisiting the single slit in Young’s experiment, we would expect to get a diffraction pattern from the single slit.
    • Over this you would also get an interference pattern from the two slits.
  • 83. DOUBLE SLIT DIFFRACTION
  • 84. DOUBLE SLIT DIFFRACTION
    • We get not only an interference pattern but also a diffraction pattern.
    • This is why we see a gradual dimming of light intensity as we move away from the central order antinode when we shine a LASER on a wall.
  • 85. DOUBLE SLIT DIFFRACTION
    • Using width of the slits = 10 -4 m
    • Distance from single to double slits =10 -1 m
    • Light of wavelength = 5 x 10 -7 m
    • The bandwidth becomes:
  • 86. DOUBLE SLIT DIFFRACTION
    • Thus the central band (2 y ) is 1.0 x 10 -3 m.
    • The double slits are separated by a distance of 3 x 10 -4 m, which is well within the intense bright diffraction band produced by the single slit.
    • Using a much narrower or broader band single slit may reduce the intensity of the light reaching the double slit.
  • 87. MULTIPLE SLIT DIFFRACTION
    • The characteristics of multiple slits can be summarised as below:
    • Interference fringes are superimposed on the single slit diffraction pattern only when there are least two slits enabling the interference of waves to occur.
  • 88. MULTIPLE SLIT DIFFRACTION
    • As the number of slits increases, the diffraction pattern spreads and the intensity of each reinforcement diminishes.
    • As the number of slits increases, each fringe becomes narrower. This results in sharper reinforcement bands.
  • 89. MULTIPLE SLIT DIFFRACTION
    • The central bandwidth is no longer twice that of fringe bands. The bandwidth equation can no longer be used.
    • When there are a large number of slits, very sharp reinforcement lines occur with non-uniform spacing between them.
  • 90. THE DIFFRACTION GRATING
    • A large number of equally spaced parallel slits which can also be called an ‘interference grating’. There are two types:
    • TRANSMISSION : Large number of equally spaced scratches (6000 per cm is common) are inscribed mechanically into a transparent material such as glass. Each scratch becomes an opaque line and the space in between becomes a slit.
  • 91. THE DIFFRACTION GRATING
    • REFLECTION : Regularly spaced grooves in a medium such as metallic or glass surface. You can see the diffraction pattern on the surface of a CD or DVD.
  • 92. THE DIFFRACTION GRATING
    • Coherent light is directed on the diffraction grating by passing light through a collimator (light gatherer). The pattern can be seen through the telescope. The whole arrangement is called a spectrometer.
  • 93. THE DIFFRACTION GRATING
    • The pattern is similar to Young’s double slit pattern. The slits are narrow enough so that diffraction by each of them spreads light over a very wide angle on to a screen. Interference occurs with light waves from all slits.
  • 94. THE DIFFRACTION GRATING
    • We use the Spectroscope to find the wavelengths of the different colours.
    • For reinforcement
    • Each wavelength of light (or different colour) will have a unique angle that it is diffracted to on the screen.
    • Diffraction Grating - Monochromatic Light
  • 95. THE DIFFRACTION GRATING
    • When m = 0, the central reinforcement line is produced and is called the zero order maximum.
    • First order maxima occur when m = 1 and second order maxima when m = 2.
  • 96. THE DIFFRACTION GRATING
    • The diffraction pattern for a grating is different to that from a double slit pattern.
    • The bright maxima are much narrower and brighter for a grating.
  • 97. THE DIFFRACTION GRATING
    • For a grating, the waves from two adjacent slits will not be significantly out of phase but those from a slit maybe 500 away, may be exactly out of phase. Nearly all the light will cancel out in pairs this way. The more lines, the sharper the peaks will be.
    • Double Slit Multiple Slits
  • 98. THE DIFFRACTION GRATING
    • This is a much more accurate way of measuring the wavelength of light than using two slits.
  • 99. PRODUCING SPECTRUMS
    • PRODUCING SPECTRUMS FROM WHITE LIGHT USING A SPECTROSCOPE.
    • If the light is not monochromatic, the angles at which the wavelengths produce their m th order maxima are different.
  • 100. PRODUCING SPECTRUMS
    • If what light strikes a grating:
    • The central maximum (m=0) will be a sharp white light since d sin  = 0 for all wavelengths at the zero order maxima
  • 101. PRODUCING SPECTRUMS
    • On either side after the central white maxima and an area of darkness, violet reinforces first as it has the smallest wavelength, then the other colours through to red.
    • A clear first order (m = 1) continuous spectrum can be seen.
  • 102. PRODUCING SPECTRUMS
    • Higher order spectra become spread further and become less intense.
    • The pattern will overlap from the third order onwards and the bandwidth formula used in the double slit cannot be used.
  • 103. PRODUCING SPECTRUMS
  • 104. PRODUCING SPECTRUMS
    • A small wavelength equals a small sin
    • Blue light makes a small angle.
  • 105. EXAMPLE 3
    • Light from a sodium vapour lamp (m) falls on a grating with 5000 lines per centimetre. Find the angular positions of all the reinforcement positions observed. The wavelength of Sodium vapour (orange light) is 5.98 x 10 -7 metres.
  • 106. EXAMPLE 3 SOLUTION
  • 107. EXAMPLE 3 SOLUTION
  • 108. EXAMPLE 4
    • White light containing wavelengths from 400 to 750 nm strikes a grating containing 4000 lines per centimetre. Show that the violet of the third order (m=3) overlaps the red of the second order (m=2) spectrum.
    • Violet light has a wavelength of 400nm.
    • Red light has a wavelength of 750 nm.
  • 109. EXAMPLE 4 SOLUTION
  • 110. EXAMPLE 4 SOLUTION
  • 111. EXAMPLE 5
    • What spacing is required in a reflection grating so that the first order maximum obtained with X-rays of wavelength
    • 1x10 -10 m appears at 30 0 from the normal? The X-rays are incident normally on the grating.
  • 112. EXAMPLE 5 SOLUTIONS
  • 113. APPLICATION -COMPACT DISCS (CD’S)
    • A CD is a fairly simple piece of plastic about 1.2 mm thick.
    • The CD consists of a moulded piece of plastic that is impressed with microscopic bumps arranged as a single, continuous spiral track of data.
    • A thin, reflective aluminium layer is placed onto the top of the disc, to cover the bumps.
  • 114. APPLICATION -COMPACT DISCS (CD’S)
    • A thin acrylic layer is sprayed over the aluminum to protect it.
    • The label is then printed onto the acrylic.
  • 115. APPLICATION -COMPACT DISCS (CD’S)
    • A CD has a single spiral track of data circling from the inside of the disc to the outside.
    • The data track is incredibly small. It is about 0.5 microns wide, with 1.6 microns separating one track from the next.
  • 116. APPLICATION -COMPACT DISCS (CD’S)
  • 117. APPLICATION -COMPACT DISCS (CD’S)
    • Due to the extreme thinness, the total length of the track squeezed onto this small disc is about 8 km.
    • The information on the disc is read by shining a laser beam from the underside of the compact disc.
    • Thus the laser is seeing the “bumps”, not the “pits”.
  • 118. APPLICATION -COMPACT DISCS (CD’S)
  • 119. APPLICATION -COMPACT DISCS (CD’S)
    • The diagram below gives you some idea of how small a CD “bump” is compared to a human hair.
  • 120. APPLICATION -COMPACT DISCS (CD’S)
    • The sequence of bumps, the length of the bumps and the length of the spaces between the bumps provides the information that the CD player decodes.
  • 121. APPLICATION -COMPACT DISCS (CD’S)
    • The laser that is used is an infrared laser emitting light at a wavelength of 780 nm.
    • The laser passes through the plastic and is reflected off the aluminum coating on the bumps and the land between them.
  • 122. APPLICATION -COMPACT DISCS (CD’S)
    • A very important point is that the height of the “bumps” is approximately one quarter the wavelength of the laser light.
    • When the laser light is passing over the “land”, all of the light is reflected off and it travels back to photoelectric cell.
    • The photoelectric cell then produces an electric current.
    • CD Diagram
  • 123. APPLICATION -COMPACT DISCS (CD’S)
    • This electric current then goes on to generate sound in a loudspeaker (see loudspeaker application).
    • Now lets look at what happens when the laser light approaches a “bump”.
    • When the light reaches a bump, half of the light is reflected off the “bump” and half of the light is reflected off the “land”.
  • 124. APPLICATION -COMPACT DISCS (CD’S)
  • 125. APPLICATION -COMPACT DISCS (CD’S)
    • Because the bump is ¼ of a wavelength in height, the light being reflected off the land travels one half a wavelength further .
    • The light reaching the photoelectric cell coming from the “land” and the “bump” is out of phase . This leads to partial cancellation and a decrease in intensity.
    • This leads to decreased current being produced.
  • 126. APPLICATION -COMPACT DISCS (CD’S)
    • As the laser moves along the track the intensity of the light falling on the photoelectric cell changes every time it comes into approaches or leaves a bump.
    • It is this change in intensity which causes the fluctuation in electric current, which causes the movement of the loudspeaker and ultimately the fluctuation in sound.
  • 127. APPLICATION -COMPACT DISCS (CD’S)
    • USING INTERFERENCE TO KEEP A LASER ON TRACK
    • The musical data on the CD is read from the inside out.
    • The CD spins above the laser.
    • After one revolution, the laser must move to the outside exactly 1.6 microns to remain on track.
  • 128. APPLICATION -COMPACT DISCS (CD’S)
    • This requires a very precise tracking mechanism and an accurate correction mechanism to move the laser back on track if it should stray off the line.
    • The tracking correction is achieved by first passing the laser beam through a diffraction grating, before it reaches the CD.
  • 129. APPLICATION -COMPACT DISCS (CD’S)
    • When the monochromatic light passes through the diffraction grating a central beam and a first order diffracted beam will land on the CD.
  • 130. APPLICATION -COMPACT DISCS (CD’S)
    • The central beam is focused on the track of the CD and passes over the bumps while the two first order diffracted beams are focused on the land on either side of the bumps.
    • One diffracted beam is slightly ahead of the other.
  • 131. APPLICATION -COMPACT DISCS (CD’S)
    • The laser beam is tracking correctly when the central beam is varying in intensity from 35% to 100% and the two diffracted beams have a constant intensity of 100%.
  • 132. APPLICATION -COMPACT DISCS (CD’S)
    • If the laser beam should stray to one side of its correct position, then the variation in intensity of the main beam is now reduced.
    • The leading tracking beam will also have a variation in the intensity because some of it is passing over the bumps.
    • The tracking mechanism “senses” that it must adjust the position of the laser down in order to put it back on track.
  • 133. APPLICATION -COMPACT DISCS (CD’S)
  • 134. APPLICATION -COMPACT DISCS (CD’S)
    • If the laser beam were to stray to the other side of its correct position, then the variation in intensity of the main beam is again reduced.
    • The TRAILING beam will now have a reduction in intensity.
    • The tracking mechanism “senses” that it must adjust its position up in order to get back on track.
  • 135. APPLICATION -COMPACT DISCS (CD’S)