1. SACE Stage 2 Physics
Interference of Light

2. Diffraction Definition of Diffraction
Diffraction is the bending of light which results from limiting how much of a wavefront contributes to a reflected or transmitted wave.
When diffraction occurs, there is no change in the frequency, velocity or wavelength of the wave. The only changes which occur are changes in amplitude and direction.

3. Diffraction Consider plane waves approaching a single slit
 remains constant
v remains constant
f remains constant
Amplitude varies with position to the right of the slit.
geometric shadow zone

4. Diffraction
Amplitude is a measure of the variation in the energy of the wave, and so a variation in the energy transmitted at a point is observed when diffraction occurs.
Reflection from a small object and diffraction around the object (scattering).
Incident wavefronts
reflected wavefronts
Object
Diffracted wavefronts
Secondary wavelets produced by reflection

5. Very little diffraction is observed when  << d
Aperture Size
The amount of diffraction is significant if the wavelength of the wave is about the same size as the size of the diffracting object (or aperture).
1. Short wavelength compared with the slit size (d) ie  << d
Very little diffraction is observed when  << d d

6. Aperture Size
Huygens Principle - Every point on a wavefront may be considered as a source of secondary semi-circular wavelets. The new wavefront is formed along the common tangent to these secondary wavelets.
If the wavelength is small, we consider there to be a very large number of sources of secondary wavelets across the slit.
As the wave passes through the slit, most of the wavefront is partially blocked. Using Huygens principle, it can be seen that a new wave front is formed through the slit which is curved at the edges but straight in front of the slit.

7. Aperture Size 2. Long wavelength (   d )
Significant diffraction is observed.
The wavefront is now almost circular as the source of the new wave front is very short.

8. Coherent Wave Sources Phase:
Phase refers to the "relative position of the crests of a wave". Two waves with crests at the same point are "in phase".
If a crest from one wave and a trough from a second wave are present at a point, then the waves are "180o out of phase"
NB It is also possible for the waves emitted at the sources to be out of phase.

9. Coherent Wave Sources Coherence:
This refers to waves having a constant phase relationship. It is also necessary that the waves must travel at the same speed and have the same frequency (or the same wavelength).

10. Coherent Wave Sources Incandescent source
It is the oscillating electrons in an object that radiate light. Thus each electron is considered to be an individual light source. The type of light source is important in our society.
e.g. A torch is not a coherent source because there are a large number of electrons giving out light independently and not necessarily with a constant phase relationship
Constructive and destructive interference occurs and the waves "spread out".

11. Coherent Wave Sources Monochromatic light
The word has two parts- mono - meaning one and chromatic -referring to colour. Monochromatic literally means one colour.
The frequency of light determines the colour, hence monochromatic light has a single frequency.
Laser - is a source of monochromatic coherent light .
Laser light has a single frequency but also the same phase. Every single electron gives out light of the same frequency and in phase. These waves will show visible interference.

12. Principle of Superposition
When two waves coincide, their displacements are added vectorially. + =
This process is called reinforcement or constructive interference.
The two waves are said to be in phase.

13. Principle of Superposition+ =
This process is known as annulment or destructive interference.
The waves are said to be out of phase or have a phase difference of  or /2.

14. Principle of Superposition
Interference of Waves in 2 Dimensions
Interference is also easily seen for waves travelling in two dimensions. For example, the interference pattern of water waves from two coherent sources can be seen using a ripple tank.
Antinode – Minimal disturbance
Node – Maximum disturbance

15. Principle of Superposition
Necessary Conditions for Observing Interference with Light.
(a) The light from the different sources must overlap.
(b) The light must be of the same colour or frequency.
(c) The sources must be COHERENT (a constant phase relationship is needed).
(d) The amplitude of the waves must be approximately equal.
(e) External light must be excluded.

16. Two Slit Interference
The Double Slit Experiment (Young's Interferometer) demonstrates interference of waves travelling in two dimensions using light waves.
source of
monochromatic
light
single
slit
double
light arrives
here in phase
screen
Central
bright band
Single slit small enough to diffract light. This causes the same wavefront of light to hit both slits,
ie. The single slit acts as a source of Huygen's double slits. A plane parallel wavefront at the Secondary Wavelets, producing a new wave.
The double slits also diffract the
light, again acting as sources of
COHERENT Light
secondary wavelets. Producing a new wave front.

17. Two Slit Interference
The dark bands on the screen occur where the light from the two slits is out of phase at that point on the screen. For these points, the path difference of light from the two sources is an odd number of half wavelengths.
 = (m + )
There is a bright band at the centre of the pattern on the screen. This is the point where the path difference is zero. At this point the light from the two sources meets in phase. ie
 = 0

18. Two Slit Interference
Other bright bands occur where the path difference is a whole number of wavelengths. At these points the light is also in phase.
 = m 

19. Young’s Double Slit Experiment
Derivation of the positions for maxima using Young's Double Slit
Assume L>>d
S1P is nearly parallel to S2P.
 is very small O S1 S2 R P  L A
S1P = RP and
S2R is the path difference ()
d=separation of the slits

20. Young’s Double Slit Experiment
Now angle  S1OA = 90o
Hence angle S1OP = 90o - 
As  is small, the lines S1P, S2P and OP are nearly parallel
Hence S1R is very nearly perpendicular to OP
So, to a good approximation,  S2S1R = POA =  q S 1 O 2 R A d
Using triangle S1S2R, sin  = or

21. Young’s Double Slit Experiment
For REINFORCEMENT to occur, need a whole number of wavelengths,  = m
ie d sin  = m, m = 0, 1, 2,
For ANNULMENT (Destructive Interference)  = (m ) 
ie d sin  = (m + 1/2) , m = 0, 1, 2, .....

22. Fringe Separation ("bandwidth") of the Interference Pattern (Dy)
Consider two consecutive bright bands on the screen: m
m+1 y ym
ym+1 d L

23. Fringe Separation ("bandwidth") of the Interference Pattern (Dy)
Fringe separation, Dy = ym+1 - ym
but tanm+1 =
and tan m =
Dy = L. tanm+1 - L. tanm
But, assume tan ≈ sin ≈  as  is very small,
Dy = L. sin m+1 - L. sin m
Now sin m+1 =
and sin m =

24. Fringe Separation ("bandwidth") of the Interference Pattern (Dy)
Now, we are considering bright bands (reinforcement) for which
d sin = m
but sin = tan =
ie sin = -
For two consecutive bright fringes
Dy = -
We call Dy the fringe separation.

25. The Speckle Effect of Laser Light
When laser light reflects of a slightly irregular surface, it appears to be grainy. The reflected area appears to be made of bright and dark spots.
This is of course, not the case, the reflected area is evenly illuminated but because of the irregularity on the surface, the laser light reaching your eye is not consistent but consists of different path lengths that interfere to give the appearance of the spots.
This effect can be used to determine an eye condition known as Myopia (near sightedness). When observing the spots and moving your head slowly to the left, if the spots appear to move in the opposite direction, then you are near sighted otherwise you vision is normal or you are far sighted.

26. Examples
A young’s double slit interferometer is illuminated with green light of wavelength 546nm. The double slits are 0.100mm apart and the screen is 20.0cm away form the slits. Find,
(1) The angular position of the first minimum on the screen.
The angular position of the fifth minimum on the screen.
The distance, on the screen, between the first maximum and the fifth maximum.

27. Examples (1) The angular position of the first minimum on the screen.
At the first minimum, path difference =

28. Examples The angular position of the fifth minimum on the screen.
At the fifth maximum, path difference = 5

29. Examples
The distance, on the screen, between the first maximum and the fifth maximum.
The fringe separation is given by,
Distance between first and fifth maxima is 4x fringe seperation
distance = 4.37mm

30. Examples
Monochromatic light from a single slit illuminates two, narrow parallel slits. The centres of the two slits are 0.800mm apart. An interference pattern forms on a screen 50.0cm away. The fringe separation on the screen is 0.304mm. Find the wavelength  of the light.

31. Examples d = 0.800mm = 8.00 x 10-4m L = 50.0cm = 5.00 x 10-1m
y = 0.304mm = 3.04 x 10-4m

32. Examples
Laser light illuminates two narrow parallel slits and the interference pattern producedis observed on a wall 4.00m away. The distance between the to slits is 0.500mm and the angular position of the fourth maximum is 0.30o. Find the wavelength  of the light.

33. Examples
d = 0.500mm = 5.00 x 10-4m L = 4.00m,  = 30o

34. Examples
Two narrow parallel slits are illuminated by light of two wavelengths. 1 = 600nm and 2 is unknown. On the interference patterns produced on th screen, the 4th bright fringe for 1 coincides with the 6th bright fringe of 2. Find 2,

35. Examples
As the 4th and 6th bright fringes coincide, they must both lie at the same angular position.
 At this position dsin = 41 and dsin = 62

36. Transmission Gratings
Single Slit Diffraction with Light
We deal only with the situation where both the source and the slit are a relatively large distance from the screen. Hence the rays involved are effectively parallel. *
effectively parallel rays
laser
source
screen
slit

37. Transmission Gratings
The pattern is more usefully represented using an intensity (brightness) diagram.
Intensity I2 I1
Secondary or subsidiary maxima Io
Central maximum x 2x
minima (I = 0)

38. Transmission Gratings
Intensity
(1st Order) = of the intensity of the central maximum
The width of the central maximum happens to be twice the width of the other maxima.
The width of the diffraction pattern depends on the wavelength of the light incident on the slit.

39. The Diffraction Grating
Diffraction Grating is effectively a very large number of narrow slits arranged so that they are parallel to each other. It is made by scratching lines on glass (transmission grating).
eg. Transmission Grating
Transparent Slits
scratches d
d is the spacing between the slits.

40. The Diffraction Grating
Each slit is effectively a point source.
Slits
Scratches d

41. The Diffraction Grating
Central Maximum (  = 0o )
Assumption of a large distance to the screen compared with slit separation.
ie Assume the light rays are parallel.
Light from each slit travels an equal distance to the screen. Hence the waves reinforce to produce a bright image. (Path difference is zero)

42. The Diffraction Grating
Other points on the Screen 
90 -  d X Z Y 
Effectively parallel rays
From Young’s Double Slit Experiment, it can be seen that,

43. The Diffraction Grating
Pattern observed with a Diffraction Grating
Intensity Maxima occur when: d sin  = N
Where: d is the distance between slits
 angle from the Normal to the grating
N an integer
 the wavelength of the light
A number of bright reinforcement lines will be seen on the screen corresponding to integer multiples of the wavelength. They are referred to as "orders".

44. The Diffraction Grating
2nd Order
1st Order
0th Order
N=1
Laser light
N=0
N=1
grating
N=2
screen

45. The Diffraction Grating
Sharp bright maxima are seen with wide dark minima between them.
Bright Sharp maxima
Wide dark minima
N = 0
N = 1
N = 2

46. The Diffraction Grating
Number of Images Observed
The spacing of the "slits" on the grating determines how many "orders" of the diffraction pattern are seen.
Consider what happens to the path difference as the angle of the rays leaving the grating increases.

47. The Diffraction Grating
a)  = 0, path difference  = 0
b) As the angle  becomes larger,  increases.    
(c) The largest possible value for  occurs when
 = d sin  = d sin 90 = d
Now, for maxima,  = N
ie: The largest possible order for a given grating is given by  = d = N

48. The Diffraction Grating
Explanation of intense peaks:
Constructive interference occurs when =m.
Path difference

49. The Diffraction Grating
Explanation of the regions of negligible intensity between the intense peaks:
Consider a position on the screen where the path difference is not = m
Consider 2 adjacent slits emitting a wave such that the path difference is 1/100th of a wavelength.
In this case then the next consecutive slit produces a wave with a path difference of 2/100th of  relative to the first slit
The next consecutive slit is 3/100th , etc until the slit with a PD = 50/100th  relative to the first slit results in annulment with the wave from the first slit considered.

50. The Diffraction Grating
Then the slit immediately after this has a PD = 51/100th  relative to the wave produced by the first slit which annuls with the wave from the second slit considered.
This process continues so that with a large number of slits, there is a corresponding slit resulting in annulment.
Hence any position other than the case producing a path difference of m results in a position of negligible intensity.

51. The Diffraction Grating
Example
A laser of wavelength 6.3 x 10-7m is passed through a diffraction grating of 100 lines/mm. What is the maximum number of reinforcements seen on the screen?

52. The Diffraction Grating
Example
100 lines/mm = lines/m
slit seperation = 1  (no. of lines/m) = 1 x 10-5m
Therefore there are 15 orders of reinforcement, therefore there are (2x15+1)=31 maximums.

53. The Diffraction Grating
White Light Pattern from a Diffraction Grating
White light consists of varying wavelengths. The angle at which reinforcement occurs depends on the wavelength of the light.
Hence bright orders for each colour occur in different positions The result of this is that white light is dispersed (spread or separated) into its various colours for each order. (ie: spectra are seen for each order).

54. The Diffraction Grating
Red
N = 2
Second Order Spectrum
Blue
Red
N = 1
First Order Spectrum
Blue
white light
N = 0 All colours
Have maximum here
White
grating
screen
NOTE: Red has the largest wavelength and so has the largest angle for its maxima

55. The Diffraction Grating
The Properties of a Grating Useful in Spectroscopy
Spectroscopy is used to make absolute measurements of wavelength. Diffraction gratings are used in such instruments since the spacing between lines d can be measured accurately using microscopes.
A diffraction grating produces very sharp lines of maxima only ie. pure lines. Particularly for the first order spectra of visible light where there is no overlap or mixtures of colours (frequencies).
The resolving power of a grating compared to a prism (the other device used to split light into its constituent frequencies) is greater.

56. Application: Compact Discs
Converting and Analogue waveform to a Digital Signal
A microphone can be used to convert an analogue signal to a digital signal by using an Analogue to Digital Converter (ADC).
The ADC does this by measuring the amplitude of the waveform 44,100 times every second (this is a sampling rate of 44.1 kHz). The amplitude is then converted into a binary number between -32,768 to 32,768. This stream of binary numbers is then recorded onto the compact disc.
When the disc is read a DAC recreates the original waveform.

57. Application: Compact Discs
Encoding the Digital Data on a Compact Disc
A compact disc consists of one long spiralling track that’s starts in the centre of the CD and spirals out. The digital data is encoded onto the disc by pressing bumps of varying length onto this track. The intervening space between the bumps are known as the ‘land’.

58. Application: Compact Discs
As the disc spins and the laser tracks along the spiral track, it encounters the boundary between a bump and the land. The CD player interprets this as a ‘1’.
A sequence of a long land or bump is interpreted as a row of zero’s.

59. Application: Compact Discs
Reading the Digital Data From The Disc
As the beam of laser light moves along the track it is reflected of the land and the bumps. When the laser goes from the land to a bump, a change in the intensity of the light that the receiver picks up occurs and this is interpreted as a ‘1’.
The wavelength used is 503nm when travelling in the polycarbonate plastic coating of the disc. The height of a bump is approximately ¼ of this.

60. Application: Compact Discs
When the receiver picks up the light from the bump, it also receives light reflected from the land either side of the bump.
As the bump is approx ¼ the height of a wavelength, the path difference between the light reflecting of the top of a bump and the land is approx ½ a wavelength there is a drop in the intensity of the light received as there is destructive interference occurring.

61. Application: Compact Discs
Keeping the laser on track
For the laser to remain on track, it is first passed through a diffraction grating,

62. Application: Compact Discs
The main central beam is focussed on the track and the two first order beams are placed either side on the land as shown.
When the laser is on track, the two side beams are reflecting the same amount of light intensity of the land.

63. Application: Compact Discs
When the laser begins to stray, the reflected intensity of the two side beams are now different.
The CD player now knows to correct the allignment and adjusts the laser’s position.