An Experimental Problem of a Competition Discussed in a Secondary - - PowerPoint PPT Presentation

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An Experimental Problem of a Competition Discussed in a Secondary - - PowerPoint PPT Presentation

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An Experimental Problem of a Competition Discussed in a Secondary School Workshop Pter Vank Institute of Physics, Budapest University of Technology and Economics, Budapest, Hungary Inventing a new experimental problem is not easy. How can

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SLIDE 1

An Experimental Problem of a Competition Discussed in a Secondary School Workshop

Péter Vankó Institute of Physics, Budapest University of Technology and Economics, Budapest, Hungary

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Inventing a new experimental problem is not easy.

How can be used the apparatus later?

  • to discuss the problem in a

school workshop

  • as a problem of a local competition
  • as an measuring exercise for inquiring students
  • to prepare students for other competitions
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SLIDE 3

An experimental problem

in the competition

  • limited time
  • working alone
  • pocket calculator,

ruler, graph paper

  • limited knowledge

(without books) in the workshop

  • more time
  • working together
  • data analysis by PC

software's

  • background

information

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SLIDE 4

The problem

There are two optical structures to investigate by semiconductor laser. Both of them are multiple slits, i.e. a few parallel and identical transparent slits

  • n a dark background separated by

the same distance. From the diffraction pattern determine the distance, the number and the width

  • f the slits in both optical structures.
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SLIDE 5

The apparatus

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SLIDE 6

The slit structures B A

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SLIDE 7

The difficulties of the measurement

  • The careful adjusting of the arrangement
  • Reading detector position with half a mm

accuracy

  • Changing the range of the voltmeter

during the measurement

  • Measuring the background light intensity
  • Measuring the light intensity across the

„black” background of the slit structures

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SLIDE 8

The experimental setup

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SLIDE 9

Measured data A

2 4 6 8 10

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5 10 15 20 25 y (mm) U (V)

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SLIDE 10

Measured data B

2 4 6 8 10

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5 10 15 20 25 y (mm) U (V)

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SLIDE 11

Magnified graph A

200 400 600 800 1000

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5 10 15 20 25 y (mm) U (mV)

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SLIDE 12

Magnified graph B

200 400 600 800 1000

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5 10 15 20 25 y (mm) U (mV)

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SLIDE 13

Relative intensity A

0,05 0,1 0,15 0,2 0,25 0,3

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5 10 15 20 25 y (mm) I rel

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SLIDE 14

Relative intensity B

0,05 0,1 0,15 0,2 0,25 0,3

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5 10 15 20 25 y (mm) I rel

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SLIDE 15

The light intensity across the black background of the slit structures

1 2 3 4 5 6 7 8 9 10

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5 10 15 20 25 y (mm) U (V)

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SLIDE 16

Corrected graph A

0,05 0,1 0,15 0,2 0,25 0,3

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5 10 15 20 25 y (mm) I rel

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SLIDE 17

Corrected graph B

0,05 0,1 0,15 0,2 0,25 0,3

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5 10 15 20 25 y (mm) I rel

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SLIDE 18

Interpretation of the measured data

0,05 0,1 0,15 0,2 0,25 0,3

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5 10 15 20 25 y (mm) I rel 0,05 0,1 0,15 0,2 0,25 0,3

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5 10 15 20 25 y (mm) I rel

The distance of slits? (d) The number of slits? (n) The relative width of slits? (w/d)

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SLIDE 19

Diffraction on the multiple slit

slits diffracted beam x ⏐k⎪= 2π/λ y θ d w D n photodiode θ λ π θ ϕ x kx 2 sin ≈ = = ∆ kx θ θ D Dtg y ≈ =

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SLIDE 20

Determination of d

The position of the first maximum: (as for the double slit or for the grating) θ θ λ ≈ = sin d d D D D y λ θ θ ≈ ≈ = tan y = 9.75 ± 0.25 mm D = 1 ± 0.005 m λ = 650 ± 7 nm

d = 67 ± 3 µm for both structures

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SLIDE 21

Determination of n

The number of slits is related to the number of small maxima (or to the number of zeros) between two neighboring big maxima. To understand the relationship: use phasors. Phasors are (rotating) vectors expressing phase and amplitude of a quantity.

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SLIDE 22

Phasor representation of the E vector of the light

The phase difference between neighboring slits is

θ λ ϕ d π 2 =

ϕ

The light intensity is proportional to the square

  • f the (vectorial) sum of

the phasors.

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SLIDE 23

Zeros

n = 5

The sum of n identical phasors can be zero if

n m π 2 = ϕ

m < n is integer

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SLIDE 24

Determination of n

n-1 zeros and n-2 small maxima between two big maxima

0,05 0,1 0,15 0,2 0,25 0,3

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5 10 15 20 25 y (mm) I rel

n = 5 for structure A n = 4 for structure B

0,05 0,1 0,15 0,2 0,25 0,3

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5 10 15 20 25 y (mm) I rel

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SLIDE 25

Determination of w/d

Neither w/d ≈ 0 nor w/d ≈ 1 is possible.

0,2 0,4 0,6 0,8 1 0,5 1 1,5 2 2,5

The maxima would have the same (similar) intensity.

0,2 0,4 0,6 0,8 1 0,5 1 1,5 2 2,5

It would behave as a single slit with d’ = nd.

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SLIDE 26

E1

2 2 2

w n E I = = r

Phasor diagram of the multiple slit

n = 5 w/d = 5/8 θ = 0

E1 w d E1 = C⋅ w = w (by C = 1)

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SLIDE 27

E1 R ϕ ψ ψ/2 E1 E ϕ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = w R E λ θ θ λ ψ π sin π 2 sin 2

1

Phasor diagram of the multiple slit

n = 5 w/d = 5/8 θ > 0

θ λ ⋅ = π 2 1 R d R d ⋅ ⋅ = = θ λ ϕ π 2 w R w ⋅ ⋅ = = θ λ ψ π 2

i

E E ∑ = r r

2

E I r =

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SLIDE 28

Phasor diagram of the multiple slit

n = 5 w/d = 5/8 0 < θ ≤ λ /d

θ ⋅ d /λ = 1/10 1/5 3/10 2/5 1/2 3/5 7/10 4/5 9/10 1 E1 / E0 = 0.99 0.97 0.94 0.9 0.85 0.78 0.71 0.64 0.56 0.47 I / I0 = 0.41 0 0.05 0 0.03 0 0.03 0 0.13 0.22

0,2 0,4 0,6 0,8 1

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SLIDE 29

Determination of w/d

The intensity of the first big maximum:

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5 10 15 20 25 y (mm) I rel

Irel ≈ 0.25 ⇒ w/d ≈ 0.6 Irel ≈ 0.15 ⇒ w/d ≈ 0.7

0,05 0,1 0,15 0,2 0,25 0,3

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5 10 15 20 25 y (mm) I rel

⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = d w w d I I Irel π sin π

2 2

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SLIDE 30

R

Determination of w/d

Investigation of the second big maximum:

E1 ϕ =2π R ψ = 5/4 π E1 ψ = 5/2 π ϕ =4π

w/d = 0.625

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SLIDE 31

Determination of w/d

Investigation of the second big maximum:

ϕ =2π E1 = 0 R ϕ =4π ψ = 2π E1 =2/π⋅E0 R ψ = π

w/d = 0.5

0,05 0,1 0,15 0,2 0,25 0,3

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5 10 15 20 25 y (mm) I rel 0,05 0,1 0,15 0,2 0,25 0,3

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5 10 15 20 25 y (mm) I rel

w/d ≈ 0.5

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SLIDE 32

Determination of w/d

Considered both argumentations

w/d ≈ 0.55 ± 0.05

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SLIDE 33

The concept of the workshop

  • An afternoon event - free from the syllabus
  • For inquiring students and teachers
  • The participation is voluntary
  • Interdisciplinary approach
  • Enough time
  • for mathematical and physical background
  • for nice experiments and measurements
  • to analyze measured data by PC software’s
  • to discuss details and the consequences
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SLIDE 34

Mathematical background

Fourier-series and Fourier-integral (introduction without exactness) Diffraction of light on the multiple slit as a physical realization

( ) ( )

2

d f i exp ) ( x x kx k

∫

∞ ∞ −

= I

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SLIDE 35

E1 R ϕ ψ ψ/2 E1 E ϕ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = w R E λ θ θ λ ψ π sin π 2 sin 2

1

Phasor diagram of the multiple slit

n = 5 w/d = 5/8 θ > 0

θ λ ⋅ = π 2 1 R d R d ⋅ ⋅ = = θ λ ϕ π 2 w R w ⋅ ⋅ = = θ λ ψ π 2

i

E E ∑ = r r

2

E I r =

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SLIDE 36

⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = w R E λ θ θ λ ψ π sin π 2 sin 2

1

The calculation of Irel(y)

i

E E ∑ = r r E1 E ϕ

( ) ( ) ( ) ( )

[ ]

2 2 2 1 2

1 sin ... 2 sin sin 1 cos ... 2 cos cos 1 ϕ ϕ ϕ ϕ ϕ ϕ − + + + + − + + + + = = n n E E I r d θ λ ϕ π 2 =

2 2

w n I I I I rel = = θ θ D D y ≈ = tan

Microsoft Excel worksheet

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SLIDE 37

Conclusions

Playing with the apparatus and the simulation: a better understanding of diffraction The apparatus, measured data and simulations can be used for physics lessons The most important effect: the free atmosphere and the interdisciplinary approach of the workshop can arouse some participants’ interest in physics

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SLIDE 38

Thank You for Your attention

Downloads:

The measured data (Excel) The simulation (Excel) The presentation (PowerPoint) http://goliat.eik.bme.hu/~vanko/wfphc/wfphc.htm