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Yr 12 Physics 16 - Superposition Of Waves > Interference > Flashcards

Interference Flashcards

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1
Q

when is a pattern of maxima and minima produced if waves from different sources overlap

A
  • if the waves combine constructively or destructively

- at fixed positions relative to the sources

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2
Q

what is this effect called

A

interference

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3
Q

what is the path difference of two waves

A
  • the number of wavelengths they are ‘away’ from each other

- measured in radians

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4
Q

what is a misconception about path differences

A
  • that they are the number of wavelengths out of phase with each other
  • two waves could have a path difference of 6 pi meaning there is q distance of three wavelengths between them
  • however they are in phase because they are perfectly 3 wavelengths apart
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5
Q

when do positions of maximum amplitude occur when waves interact and why

A
  • when the path difference is 0
  • or a whole number of wavelengths
  • because the waves are always in phase so constructive superposition occurs
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6
Q

when would the amplitude of a wave be zero when waves interact and why

A
  • when the path difference is an odd half wavelength
  • because the waves would be pi radians out of phase
  • meaning destructive superposition takes place
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7
Q

what three conditions need to be met in order for a stable interference pattern to occur

A
  • the waves are of the same type
  • the sources are coherent
  • the wave have similar amplitude at the point of superposition
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8
Q

what does it mean for sources to be coherent

A
  • they have the same frequency

- and maintain a constant phase relationship

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9
Q

if you have two sources of ripples close to each other which creates ripples with crests and troughs, when would constructive interference take place

A

when a crests of one ripple interfere with the crests of the other

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10
Q

when would destuctive interference occur

A

when the crests of one ripple interfere with the troughs of the other

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11
Q

in the double slit experiment where you have two waves entering two slits and interfering with each other, what would be points n or Px (P0 for example) be noting

A

the maxima of the new wave that has been formed from the interference between the two waves

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12
Q

where would those points therefore be if maxima were being formed

A

they would be point where the waves were constructively interfering

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13
Q

when the two waves constructively interfere, several maxima are formed next to each other, starting from the centre one (P0) to the left and right ones. if P0 is the central maximum, what is the path difference between the two waves at this point and why

A
  • it is 0
  • because if it is the central maximum it means both waves have had to travel equally far distances to reach that point
  • and as the waves have to have the same frequency, it means that there is no difference in the number of wavelengths it took to reach the point P0
  • essentially the waves are always in phase in this case
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14
Q

what would path difference for the first maximum and second maximum (P1 AND P2) on either side of the central maximum be and why

A
  • P1 would have a path difference of pi
  • P2 would have a path difference of 2 pi
  • this is because at P1 and P2 the waves are still in phase
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15
Q

what would be done in order to calculate the wavelengths of a type of wave using the double slit experiment

A
  • place a sheet of paper on a vertical surface
  • mark the position of the aerial at the highest available maximum
  • pencil the position of the slits onto the paper and measure the distance between each slit from the maximum
  • use a formula to calculate the path difference
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16
Q

what is the formula for calculating the path difference in this case

A

path difference = S2 * Pn - S1 * Pn = n * lambda

17
Q

what is n in the formula

A
  • the order of maximum from the centre

- the same as Px where x is n

18
Q

why is it hard to observe interference effects with light sources

A
  • because of their short wavelengths

- and the difficulty in providing coherent sources

19
Q

what did thomas young devise in order to produce coherent light

A
  • he passed monochromatic light through a fine slit

- then used a double slit arrangement

20
Q

what is monochromatic light

A

light with a single wavelength

21
Q

what did young observe when he looked through a microscope to see how the light reacted and what did this prove

A
  • he saw dark and bright fringes

- which proved that light had characteristics of a wave

22
Q

what formula is used to calculate the wavelength of light from the double slit experiment

A
  • lambda = Sx / D
  • S = slit width
  • x = fringe separation
  • D = distance from the slits to the microscope eyepiece
23
Q

what is a safety precaution when doing the double slit experiment with lasers

A

the laser should be a class 2 low power laser

24
Q

what do interferometers do

A
  • they use patterns created by the recombination of a laser beam that has been split
  • by doing this small changes in the path difference are detected by a shift in the fringe pattern
25
Q

what could the setup for an interferometer look like

A
  • you could have a laser beam being shot at a mirror angles at 45 degrees
  • where there is a reflecting surface past it and above it with a detector under it
  • so that when the laser beam is split as some is reflected to the top reflecting surface and refracted through to the other one
  • they will meet back at the mirror and both be reflected or refracted to the detector
  • making the recombined rays
26
Q

how are CDs built

A
  • they have a spiral groove with a width less than 2 micrometers
  • these are cut from the edge to the middle of a highly reflective silvered surface
27
Q

how do laser beams then interact with CDs when they are shun on them

A
  • the data in the form of binary is recorded onto the disc as millions of small bumps within the grooves
  • so when laser beams hit the surface, some waves are reflected before others
28
Q

what is the usual path difference that the bump heights in CDs cause

A

lamba / 4 (pi / 2) of the wavelength of the laser

29
Q

what happens to the reflected beams

A
  • they are reflected to a photodiode

- where the light is converted into an electrical signal

30
Q

how do these reflections of waves form a 0 in binary

A
  • because the bump height is a quarter of the wavelength
  • waves that are reflected from the top of the bump and waves from the bottom would be half a wavelength out of phase
  • (as the wave hitting the bottom has to travel an extra quarter wavelength to the bottom and then the top)
  • this antiphase causes the waves to interfere destructively
  • meaning the output of the photodiode is 0 in binary
31
Q

how do the reflection of the waves form a 1 in binary

A
  • when the entire beam is reflected from the upper surface or the gap between the bumps
  • this means the intensity of the reflected beam is strong
  • causing a high electrical output to be generated by the photodiode, making the 1 in binary

Yr 12 Physics 16 - Superposition Of Waves (4 decks)

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