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Problem 15.

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Problem 15. Optical Tunnelling. Problem. Take two glass prisms separated by a small gap. Investigate under what conditions light incident at angles greater than the critical angle is not totally internally reflected. Experiment. Two measurement ranges: Centimeter waves – accurate measuring - PowerPoint PPT Presentation
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Problem 15. Problem 15. Optical Optical Tunnelling Tunnelling
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Page 1: Problem 15.

Problem 15.Problem 15.

Optical Optical TunnellingTunnelling

Page 2: Problem 15.

ProblemProblem

Take two glass prisms separated by a Take two glass prisms separated by a small gap. Investigate under what small gap. Investigate under what conditions light incident at angles conditions light incident at angles greater than the critical angle is not greater than the critical angle is not totally internally reflected.totally internally reflected.

Page 3: Problem 15.

Experiment Experiment • Two measurement ranges:Two measurement ranges:

• Centimeter waves – accurate Centimeter waves – accurate measuringmeasuring

• Visile light – obtaining the effectVisile light – obtaining the effect

• Parameters:Parameters:• Waveelngth of the light usedWaveelngth of the light used• Refraction index of prisms and Refraction index of prisms and

medium in the gapmedium in the gap• Polarization Polarization • Distance between prismsDistance between prisms

Page 4: Problem 15.

1. Centimeter waves 1. Centimeter waves

• Wavelength: 3 cmWavelength: 3 cm• Polarization: linear, electrical field Polarization: linear, electrical field

perpendicular to plane of propagationperpendicular to plane of propagation• Prism refraction index: 1.5 (paraffin)Prism refraction index: 1.5 (paraffin)• Measurements:Measurements:

• Intensity of tunneled wavesIntensity of tunneled waves• Intenzity of reflected wavesIntenzity of reflected waves

in dependence on prism distancein dependence on prism distance

• Measured – voltage in the detectorMeasured – voltage in the detector• Voltage – proportional to field!Voltage – proportional to field!

Page 5: Problem 15.

Centimeter waves Centimeter waves cont.cont.

• Apparatus schematic:Apparatus schematic:

2,78Izvor

Senzor

multimetermultimeter

Translation Translation systemsystem

Radiation Radiation sourcesource

detectordetector

Page 6: Problem 15.

Centimeter waves Centimeter waves cont.cont.

Page 7: Problem 15.

Centimeter waves Centimeter waves cont.cont.

distance [cm]

0 2 4 6 8 10 12

volta

ge [V

]

0,00

0,02

0,04

0,06

0,08

0,10

Tunelled field:Tunelled field:

d

eUU

0

Page 8: Problem 15.

2. Visible light2. Visible light• Wavelenght: 780 nmWavelenght: 780 nm• Polarization: linearPolarization: linear

• electric field perpendicular to plane electric field perpendicular to plane of propagationof propagation

• Prism index of refraction: 1.48 Prism index of refraction: 1.48 (measured)(measured)

• Measurements:Measurements:• Intensity of tunneled wavesIntensity of tunneled waves• Intensity of reflected wavesIntensity of reflected waves

• Intensity measurement: photodiodeIntensity measurement: photodiode

• Voltage – proportional to square of field!Voltage – proportional to square of field!

Page 9: Problem 15.

2. Visible light2. Visible light

• Measurement in timeMeasurement in time• A slow motor (0.5 r/min) moves A slow motor (0.5 r/min) moves

the translatorthe translator• Voltage sampling at the diode Voltage sampling at the diode

every 1/50 of a secondevery 1/50 of a second• The signal grows in timeThe signal grows in time• Change of prism distance:Change of prism distance:

vtd v – translator speed

t – elapsed time

Page 10: Problem 15.

Visible light Visible light cont.cont.

• Apparatus schematic:Apparatus schematic:

3,15

3,15

Slow motor + Slow motor + translationtranslationlaserlaser

prismsprisms

oscilloscopeoscilloscope

Data Data receiving receiving computercomputer

Page 11: Problem 15.
Page 12: Problem 15.

Visible light Visible light cont.cont.

time [s]

0,6 0,7 0,8 0,9 1,0 1,1

inte

nsity

[arb

itrar

y un

its]

0,11

0,12

0,13

0,14

0,15

0,16

0,17

vt

eUU

0

Page 13: Problem 15.

Explanation Explanation

Electromagnetic waves in dielectrics – Electromagnetic waves in dielectrics – the resultant of interference of the the resultant of interference of the initial wave and all scattered wavesinitial wave and all scattered waves

• Huygens principle:Huygens principle:

Every atom ˝through˝ which light Every atom ˝through˝ which light passes is a source of light identical to passes is a source of light identical to the incident lightthe incident light

Page 14: Problem 15.

Explanation Explanation cont.cont.

• At total reflection – the reflected ray is At total reflection – the reflected ray is the only interference maximumthe only interference maximum

• Behind the reflection plane – Behind the reflection plane – destructive interference, but only destructive interference, but only far far awayaway from the plane from the plane

• Close to the plane (distances of the Close to the plane (distances of the order of the wavelength) the waves order of the wavelength) the waves haven’t interfered completely and a haven’t interfered completely and a decaying field existsdecaying field exists

Page 15: Problem 15.

Explanation Explanation cont.cont.

• That field decays fast due to That field decays fast due to interferenceinterference

• If a prism is put into the field – a new If a prism is put into the field – a new interference maximum can be formed interference maximum can be formed in the prismin the prism

• A new, tunnelled wave is formed in A new, tunnelled wave is formed in the prismthe prism

• The energy of the reflected wave The energy of the reflected wave becomes smallerbecomes smaller

Page 16: Problem 15.

Maxwell equations Maxwell equations

PE 0

1

t

B

E

0B

ttc

EP

B0

2 1

E – electric field

B – magnetic field induction

P – polarization

c – speed of light in a vacuum

ε0 – vacuum permittivity

Page 17: Problem 15.

Plane wave solutions Plane wave solutions

Electrical field:Electrical field:

krEE tie 0

E0 – amplitude

ω – frequency

t – time

k – wave vector

r - radiusvector

Magnetic field:Magnetic field:

EkB 1

Page 18: Problem 15.

Geometry of the Geometry of the problem problem

x

y

k1

krkt'

E1

Er Et'

d

φ

φ

Incident waveIncident wave

Reflected Reflected wavewave

Tunnelled Tunnelled wavewave

Prism 1Prism 1 Prism 2Prism 2

nn11nn00 nn00

Page 19: Problem 15.

Boundary conditionsBoundary conditions

• If the electric field is perpendicular If the electric field is perpendicular to the wave vector plane:to the wave vector plane:

0010 tr EEE

00101 ttxrrxx EkEkEk

E10, Er0, Et0– amplitudes of the incident, reflected and transmitted wavesk1x, krx, ktx– x components

of the wave vectors of the incident, reflected and transmitted waves

Page 20: Problem 15.

Boundary conditions Boundary conditions cont.cont.

• For the wave vectors:For the wave vectors:

yty kk 1

21

21

2

0

12ytx kk

n

nk

k1y, kty– y components of the incident and transmitted wave vectors

n0 – prism index of refraction

n1 – medium between prisms index of refraction

Page 21: Problem 15.

Solution Solution

• If the incident angle is greater than If the incident angle is greater than the reflection angle, Snell’s law givesthe reflection angle, Snell’s law gives

• xx – component of the wave vector is a – component of the wave vector is a pure imaginary => the wave pure imaginary => the wave propagates along the planepropagates along the plane

=> the => the amplitude decays amplitude decays exponentiallyexponentially

φ – incident angle1sin22

1

0

0

11

n

n

n

nikktx

Page 22: Problem 15.

Solution Solution cont.cont.

• The field in the second prism:The field in the second prism:

d

n

ndn

t eEeEE

10

1sin2

10

22

1

01

'

d – prism distance

λ – vacuum wavelegth of incident light

Θ – decay coefficient

Page 23: Problem 15.

Comparation – Comparation – decay coefficientdecay coefficient

• Centimeter waves:Centimeter waves:

• Optical range:Optical range:

Experimental 2.5±0.1

Theoretical 2.22

Experimental 1.1±0.1

Theoretical 1.94

Page 24: Problem 15.

ComparationComparation cont.cont.

• Agreement is relatively goodAgreement is relatively good

• Error causes:Error causes:

• Imprecise prism refraction Imprecise prism refraction index valuesindex values

• In optical range: In optical range:

• Prism surface defects and Prism surface defects and dustdust

• Motor precision ...Motor precision ...

Page 25: Problem 15.

Conclusion Conclusion

• We have obtained, measured and We have obtained, measured and modelled optical tunnellingmodelled optical tunnelling

• It may be said:It may be said:

The only condition for light incident on a The only condition for light incident on a prism plane with an angle greater than the prism plane with an angle greater than the critical angle not reflecting completely is to critical angle not reflecting completely is to put another prism plane next to the original put another prism plane next to the original plane to a distance of the order of the plane to a distance of the order of the wavelength usedwavelength used


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