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

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

Inventing a new experimental problem is not easy. How can be used the apparatus later? •to prepare students for other competitions •as an measuring exercise for inquiring students • as a problem of a local competition • to discuss the problem in a school workshop

An experimental problem in the competition in the workshop • limited time • more time • working alone • working together • pocket calculator, • data analysis by PC ruler, graph paper software's • limited knowledge • background (without books) information

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 on a dark background separated by the same distance. From the diffraction pattern determine the distance, the number and the width of the slits in both optical structures.

The apparatus

The slit structures B A

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

The experimental setup

y (mm) 25 20 15 10 Measured data A 5 0 -5 -10 -15 -20 U (V) -25 10 8 6 4 2 0

y (mm) 25 20 15 10 Measured data B 5 0 -5 -10 -15 U (V) -20 -25 10 8 6 4 2 0

Magnified graph A U (mV) 1000 800 600 400 200 y (mm) 0 -25 -20 -15 -10 -5 0 5 10 15 20 25

Magnified graph B U (mV) 1000 800 600 400 200 y (mm) 0 -25 -20 -15 -10 -5 0 5 10 15 20 25

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

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

The light intensity across the black background of the slit structures 10 U (V) 9 8 7 6 5 4 3 2 y (mm) 1 0 -25 -20 -15 -10 -5 0 5 10 15 20 25

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

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

Interpretation of the measured data The distance of slits? ( d ) The number of slits? ( n ) The relative width of slits? ( w / d ) 0,3 0,3 I rel I rel 0,25 0,25 0,2 0,2 0,15 0,15 0,1 0,1 0,05 0,05 y (mm) y (mm) 0 0 -25 -20 -15 -10 -5 0 5 10 15 20 25 -25 -20 -15 -10 -5 0 5 10 15 20 25

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

Determination of d The position of the first maximum: λ λ = sin θ ≈ θ = θ ≈ θ ≈ tan y D D D d d (as for the double slit or for the grating) y = 9.75 ± 0.25 mm D = 1 ± 0.005 m λ = 650 ± 7 nm d = 67 ± 3 µ m for both structures

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.

Phasor representation of the E vector of the light The phase difference between neighboring slits is 2 π ϕ = θ d λ The light intensity is proportional to the square of the (vectorial) sum of ϕ the phasors.

Zeros The sum of n identical phasors can be zero if m π 2 ϕ = m < n is integer n n = 5

Determination of n n -1 zeros and n -2 small maxima between two big maxima n = 5 for structure A n = 4 for structure B 0,3 0,3 I rel I rel 0,25 0,25 0,2 0,2 0,15 0,15 0,1 0,1 0,05 0,05 y (mm) y (mm) 0 0 -25 -20 -15 -10 -5 0 5 10 15 20 25 -25 -20 -15 -10 -5 0 5 10 15 20 25

Determination of w / d Neither w / d ≈ 0 nor w / d ≈ 1 is possible. 1 1 0,8 0,8 0,6 0,6 0,4 0,4 0,2 0,2 0 0 0 0,5 1 1,5 2 2,5 0 0,5 1 1,5 2 2,5 The maxima would have the It would behave as a single same (similar) intensity. slit with d’ = nd .

Phasor diagram of the multiple slit n = 5 w / d = 5/8 θ = 0 d E 1 = C ⋅ w = w w (by C = 1) E 1 r 2 = = 2 2 I E n w 0 E 1

Phasor diagram of the multiple slit n = 5 w / d = 5/8 θ > 0 1 2 π = λ ⋅ θ R 2 π d ϕ = = ⋅ θ ⋅ d λ R 2 π w ψ = = ⋅ θ ⋅ w λ R ψ λ θ ⎛ ⎞ ⎛ ⎞ π ψ = = 2 sin ⎜ ⎟ sin ⎜ ⎟ E R w E ϕ 1 θ λ ⎝ 2 ⎠ π ⎝ ⎠ ϕ R r r E ∑ = E E 1 E 1 ψ /2 i r 2 = I E

Phasor diagram of the multiple slit n = 5 w / d = 5/8 0 < θ ≤ λ / d 1 θ ⋅ d / λ = 1/10 1/5 3/10 2/5 1/2 3/5 7/10 4/5 9/10 1 E 1 / E 0 = 0.99 0.97 0.94 0.9 0.85 0.78 0.71 0.64 0.56 0.47 0,8 I / I 0 = 0.41 0 0.05 0 0.03 0 0.03 0 0.13 0.22 0,6 0,4 0,2 0

Determination of w / d The intensity of the first big maximum: 2 ⎛ ⎞ ⎛ ⎞ π I d w = = 2 ⎜ ⎟ ⎜ ⎟ sin I rel ⎝ π ⎠ ⎝ ⎠ I w d 0 I rel ≈ 0.25 ⇒ w / d ≈ 0.6 I rel ≈ 0.15 ⇒ w / d ≈ 0.7 0,3 0,3 I rel I rel 0,25 0,25 0,2 0,2 0,15 0,15 0,1 0,1 0,05 0,05 y (mm) y (mm) 0 0 -25 -20 -15 -10 -5 0 5 10 15 20 25 -25 -20 -15 -10 -5 0 5 10 15 20 25

Determination of w / d Investigation of the second big maximum: w / d = 0.625 ϕ =2 π E 1 ψ = 5/4 π ϕ =4 π ψ = 5/2 π R E 1 R

Determination of w / d Investigation of the second big maximum: 0,3 0,3 I rel I rel E 1 =2/ π⋅ E 0 w / d = 0.5 0,25 0,25 ϕ =2 π 0,2 0,2 w / d ≈ 0.5 0,15 0,15 ψ = π 0,1 0,1 ϕ =4 π 0,05 0,05 y (mm) y (mm) ψ = 2 π R 0 0 R -25 -25 -20 -20 -15 -15 -10 -10 -5 -5 0 0 5 5 10 10 15 15 20 20 25 25 E 1 = 0

Determination of w / d Considered both argumentations w / d ≈ 0.55 ± 0.05

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

Mathematical background Fourier-series and Fourier-integral (introduction without exactness) Diffraction of light on the multiple slit as a physical realization 2 ∞ ( ) ( ) ∫ = ( ) exp i f d I k kx x x − ∞

Phasor diagram of the multiple slit n = 5 w / d = 5/8 θ > 0 1 2 π = λ ⋅ θ R 2 π d ϕ = = ⋅ θ ⋅ d λ R 2 π w ψ = = ⋅ θ ⋅ w λ R ψ λ θ ⎛ ⎞ ⎛ ⎞ π ψ = = 2 sin ⎜ ⎟ sin ⎜ ⎟ E R w E ϕ 1 θ λ ⎝ 2 ⎠ π ⎝ ⎠ ϕ R r r E ∑ = E E 1 E 1 ψ /2 i r 2 = I E

The calculation of I rel ( y ) ψ λ θ ⎛ ⎞ ⎛ ⎞ π = = 2 sin ⎜ ⎟ sin ⎜ ⎟ E R w 1 θ λ ⎝ 2 ⎠ π ⎝ ⎠ r r E ∑ = E E i 2 π ϕ = θ d ϕ λ E 1 [ ] r ( ( ) ) ( ( ) ) 2 2 2 = = 2 + ϕ + ϕ + + − ϕ + ϕ + ϕ + + − ϕ 1 cos cos 2 ... cos 1 sin sin 2 ... sin 1 I E E n n 1 I I = = I rel 2 2 I n w Microsoft Excel worksheet 0 = θ ≈ θ tan y D D

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

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