contactless calipper - english

Contactless Calliper

Reporter – Giga Khizanishvili

Giga Khizanishvili

დასრულებულია

Giga Khizanishvili

Hi my name is Giga Khizanishvili and today im going to present problem N15 - Contactless Calliper

01/05/2023Giga Khizanishvili 2

CONTACTLESS CALLIPER

Invent and construct an optical device that uses a laser pointer and allows contactless determination of thickness, refractive index, and other properties of a glass sheet.

Giga Khizanishvili



PLAN Problem Theoretical model Experiments The margin of error Compare of theory and experiments Conclusion

Giga Khizanishvili

There is my plan: at first i'll tell you about my problem. Then i will show you my theoretical moden and after that i will show you experiments which i have done. next i will measure the margin of error and after that i will compare my theoritical model and experiments. And to the and i'll have conclusion


𝛽 𝛽

𝛼 𝛼

THEORETICAL MODEL

d

𝑠

n2

n1

𝑥s=𝑥∗𝑠𝑖𝑛𝑎𝑠

Giga Khizanishvili



Refractive index 1,0002926

𝒔𝒊𝒏𝜶𝒔𝒊𝒏 𝜷=

𝒏𝟐𝒏𝟏

n1 (Air) 1n1

n2 𝒔𝒊𝒏𝜶=𝒔𝒊𝒏 𝜷∗𝒏


cos𝛼=𝑠𝑙

𝑐𝑡𝑔 𝛽=2𝑑𝑙 𝑙= 2𝑑

𝑐𝑡𝑔 𝛽

𝑐𝑜𝑠𝛼=𝑠∗𝑐𝑡𝑔 𝛽2𝑑

𝑑=𝑠∗𝑐𝑡𝑔 𝛽2∗𝑐𝑜𝑠𝑎 = 𝑠∗𝑐𝑜𝑠 𝛽

2∗𝑐𝑜𝑠𝛼∗𝑠𝑖𝑛 𝛽= 𝑠∗√1−𝑠𝑖𝑛2𝛽2∗𝑐𝑜𝑠𝛼∗𝑠𝑖𝑛 𝛽

⇒ 𝑠𝑖𝑛𝛽=𝑠𝑖𝑛𝑎𝑛

𝑑=𝑠∗𝑛∗√1− 𝑠𝑖𝑛2𝑎𝑛2

𝑠𝑖𝑛 2𝑎


dn

𝑠1

1 𝑎1𝑠1

𝑎1

𝐹𝑖𝑟𝑠𝑡 𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡


dn

𝑠2

𝑎 2 𝑎 2𝑠2

𝑎1

𝑎 2

𝑆𝑒𝑐𝑜𝑛𝑑 𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡


𝑑=s1∗n∗√1− 𝑠𝑖𝑛2𝑎1𝑛2

𝑠𝑖𝑛2𝑎1 𝑑=s2∗n∗√1− 𝑠𝑖𝑛2𝑎 2𝑛2

𝑠𝑖𝑛2𝑎2

s1∗n∗√1− 𝑠𝑖𝑛2𝑎 1𝑛2𝑠𝑖𝑛2𝑎1 =

s2∗n∗√1− 𝑠𝑖𝑛2𝑎2𝑛2𝑠𝑖𝑛2𝑎2

𝒏=√𝐬𝟏𝟐∗ 𝒔𝒊𝒏𝟐𝒂𝟏∗ 𝒔𝒊𝒏𝟐𝟐𝒂𝟐−𝐬𝟐𝟐∗𝒔𝒊𝒏𝟐𝒂𝟐∗𝒔𝒊𝒏𝟐𝟐𝒂𝟏𝐬𝟏𝟐∗ 𝒔𝒊𝒏𝟐𝟐𝒂𝟐−𝐬𝟐𝟐∗𝒔𝒊𝒏𝟐𝟐𝒂𝟏


F𝑎

𝑎𝑥

𝑦

𝑦=𝑥∗𝑠𝑖𝑛𝑎

𝑠

𝐹𝑑

cos (90−𝑎)−𝑥∗𝑐𝑜𝑠𝑎

2∗𝐹∗ cos (90−𝑎)𝑑−𝑥∗ 𝑠𝑖𝑛2𝑎 = 𝑠

𝑥∗𝑐𝑜𝑠𝑎

𝑠=2∗ F∗ cos (90−𝑎)∗𝑥∗𝑐𝑜𝑠𝑎𝑑−𝑥∗ 𝑠𝑖𝑛2𝑎

𝑑

𝑠=𝐹∗ 𝑥∗𝑠𝑖𝑛2𝑎𝑑−𝑥 ∗𝑠𝑖𝑛2𝑎

¿ 𝑠𝑦


THE MARGIN OF ERROR

0,03%

The margin of error of the refractive index of the air


The margin of the error of the angle (

THE MARGIN OF ERROR

𝑙

h=132±0,5cm

𝑥

𝑎

𝑥=126±0.5 cm

h

𝑡𝑔𝑎 (𝑚𝑎𝑥 )= 126+0.5132−0.5=0,96 1965

𝑡𝑔𝑎 (𝑚𝑖𝑛)=126−0.5132+0.5 =0.9 4749

Giga Khizanishvili




𝑙

𝑥

𝑎

h

𝑎𝑟𝑐𝑡𝑔 (0,961965 )=43.88 °

𝑡𝑔𝑎 (𝑚𝑖𝑛)=126−0.5132+0.5 =0.94749

𝑎𝑟𝑐𝑡𝑔 (0.94749 )=43.45 °

𝑎=𝑎 (𝑚𝑎𝑥 )+𝑎 (𝑚𝑖𝑛)

2= 43.88+43.45

2=43.665 ° ±0. 215 °

∆𝑎𝑎 =

0.215 °43.665 ° ≈0.005≈0.5%

𝑡𝑔𝑎 (𝑚𝑎𝑥 )= 126+0.5132−0.5=0,96 1965

THE MARGIN OF ERROR

Giga Khizanishvili

დასრულებულია.



𝑙

𝑥

𝑎

h

h=130±0,5𝑐𝑚

𝑥=81±0,5𝑐𝑚

𝑎=𝑎 (𝑚𝑎𝑥 )+𝑎 (𝑚𝑖𝑛)

2=32.44 °+31 ,4 °

2=31.92° ±0. 26 °

∆𝑎𝑎 =

0. 26 °31.95 ° ≈0.008 ≈0.8%

𝑎 (𝑚𝑎𝑥 )=32.18 °

𝑎 (𝑚𝑖𝑛 )=31 ,66 °

THE MARGIN OF ERROR

Giga Khizanishvili

დასრულებულია.


𝑑=s∗n∗√1− 𝑠𝑖𝑛2𝑎𝑛2

𝑠𝑖𝑛2𝑎

∆ 𝑑𝑑 =4.8%⇒

∆𝑎1𝑎1 =0.5%

∆𝑎 2𝑎2 =0.8%

THE ERROR OF THE GLASS SHEET’S THICKNESS


DI ELECTRICAL PERMEABILITY𝑐=

1√𝜇0∗𝜀 0

𝑣=1

√𝜇𝜇0∗𝜀 0𝜀

𝑐=𝑣∗𝑛 𝑐𝑛=𝑣⇒

𝑛∗√𝜇 0∗𝜀 0=√𝜇𝜇0∗𝜀 0 𝜀

𝑛=√𝜀 𝑛2=𝜀⇒


DI ELECTRICAL PERMEABILITY OF A GLASS SHEET

I Exp

II Exp

III Exp

IV exp

V Exp

Vi asfas

0 0.5 1 1.5 2 2.5

2.37

2.28

2.19

2.28

2.2

0.0001

Di electrical permeability


EXPERIMENTS


DECLINATION OF LASER

Angle of laser’s declination - 𝛼1 = 43,7 Angle of laser’s declination - 𝛼2= 32


REFRACTIVE INDEX OF A GLASS SHEET

𝑠1=2.8𝑚𝑚


𝑠2=2.4𝑚𝑚𝑎1=43.7 °𝑎 2=32 °

⇒ 𝑛=1.53


THICKNESS OF GLASS SHEET


𝑠𝑖𝑛2𝑎

𝑛=1.53𝑠𝑖𝑛𝑎=0.69𝑠𝑖𝑛2𝑎=0.99


F𝑎

𝑎𝑥

𝑦

𝑠

𝑑

𝑑1

𝑑2

𝑑1=953 cm

𝑑2=𝐹∗𝑠𝑖𝑛𝑎

𝑑=𝑑 1+𝑑 2

𝑑=953+15∗0.69

𝑑=953+10.35

𝑑=963cm

01/05/2023 23


Giga Khizanishvili

𝑛=1.55𝑠𝑖𝑛𝑎=0.69

𝑠𝑖𝑛2𝑎=0.9989

𝑠=𝐹∗ 𝑥∗𝑠𝑖𝑛2𝑎𝑑−𝑥 ∗𝑠𝑖𝑛2𝑎

𝐹=15𝑐𝑚𝑑=963𝑐𝑚


𝑠𝑖𝑛2𝑎

𝑠𝑖𝑛2𝑎=0.99

𝑥=¿

𝑠=15∗0.999∗6.5172.35∗2 =2.8𝑚𝑚⇒𝑥=36𝑐𝑚

𝑥=36𝑐𝑚



𝑛=1.53𝑠𝑖𝑛𝑎=0.69𝑠𝑖𝑛2𝑎=0.9989


𝑠𝑖𝑛2𝑎

𝑠=2.8𝑚𝑚𝑛=1.53

𝑠𝑖𝑛2𝑎=0.999𝑠𝑖𝑛𝑎=0.69

𝑑=2.8∗1.55∗√1− 0.476792.4

0.999=4.39∗√0.8=3.83𝑚𝑚


COMPARE OF THEORY AND EXPERIMENTS

𝑑=3.83𝑚𝑚𝑛=1.53 𝐶𝑎𝑙𝑙𝑖𝑝𝑒𝑟



n

d

0 1 2 3 4 5 6

1.5

4.96

1.54

5.09

d = 4.96mm

Experiment Theory



n

d

0 0.5 1 1.5 2 2.5 3

1.5

2.39

1.51

2.44

d = 2.39mm

Experiment Theory



n

d

0 1 2 3 4 5 6 7

1.5

6.21

1.48

6.14

d = 6.21mm

Experiment Theory



n

d

0 0.5 1 1.5 2 2.5 3 3.5

1.5

3.24

1.51

3.28

d = 3.24mm

Experiment Theory



n

d

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

1.5

3.91

1.485

3.85

d = 3.91mm

Experiment Theory


CONCLUSION We came up with a mechanism that measures the thickness of glass sheet The thickness of a glass sheet depends upon We have derived a formula for the refractive index of glass Experimental Errors we are measured Experiments have been conducted Thickness of glass sheet and refractive index have been determined Theoretical model was compared to experiments We have solved the problem


USED LITERATURE

Baldwin, G. C.; Solem, J. C. (1997). "Recoilless gamma-ray lasers". Reviews of Modern Physics 69 (4): 1085–1117.

Vogel, Werner. Physics of Glass. Wiley, 1985. Bundschuh, Bernhard; Himmel, Jörg: Optische Informationsübertragung.

Oldenbourg Verlag, München, Wien 2003 Baldwin, G. C.; Solem, J. C. (1982). "Is the time ripe? Or must we wait so

long for breakthroughs?". Laser Focus 18

Giga Khizanishvili

Here are literature which i have use


https://www.google.com/patents/US4902902 http://russianpatents.com/patent/242/2429447.html http://physics.aidio.net/index.php/18-teoria/73-geometriuli-optika-ziritadi-kanonebi http://www.google.com/patents/US7414740 http://vlab.amrita.edu/?sub=1&brch=189&sim=1519&cnt=1 https://www.google.com/patents/US5636027

USED WEB-SPACE

Giga Khizanishvili

Here are web-space which i have use

01/05/2023Giga Khizanishvili

Thanks For Your

Attention 34


Thanks For Your Attention


Added Slides

01/05/2023

LASERსინათლის გაძლიერება იძულე-ბული გამოსხივების მეშვეობითმოწყობილობა, რომელიც ენერგიას გარდაქმნის მიმარ-თულ სინათლის სხივად.

4


n2 > n1

n1

n2

სრული შინაგანი არეკვლა


s12∗(𝑛2−𝑠𝑖𝑛2𝑎1)𝑠𝑖𝑛22𝑎1 =

s22∗(𝑛2−𝑠𝑖𝑛2𝑎2)𝑠𝑖𝑛22𝑎2

s12∗𝑛2−s12∗ 𝑠𝑖𝑛2𝑎1𝑠𝑖𝑛22𝑎1 =

s 22∗𝑛2−s 22∗𝑠𝑖𝑛2𝑎 2𝑠𝑖𝑛22𝑎2

s12∗𝑛2∗𝑠𝑖𝑛 22𝑎 2−s12∗𝑠𝑖𝑛2𝑎 1∗𝑠𝑖𝑛22𝑎 2=s22∗𝑛2∗𝑠𝑖𝑛22𝑎1−s 22∗𝑠𝑖𝑛2𝑎2∗𝑠𝑖𝑛22𝑎1

s12∗𝑛2∗𝑠𝑖𝑛 22𝑎 2−s 22∗𝑛2∗𝑠𝑖𝑛 22𝑎1=s12∗𝑠𝑖𝑛2𝑎1∗ 𝑠𝑖𝑛22𝑎2−s 22∗𝑠𝑖𝑛2𝑎2∗𝑠𝑖𝑛22𝑎1

𝑛2∗(s 12∗𝑠𝑖𝑛22𝑎2−s22∗𝑠𝑖𝑛22𝑎1)=s12∗𝑠𝑖𝑛2𝑎1∗𝑠𝑖𝑛22𝑎2−s22∗ 𝑠𝑖𝑛2𝑎2∗𝑠𝑖𝑛22𝑎1

𝑛2= s12∗𝑠𝑖𝑛2𝑎1∗𝑠𝑖𝑛22𝑎2−s22∗ 𝑠𝑖𝑛2𝑎2∗𝑠𝑖𝑛22𝑎1s12∗𝑠𝑖𝑛22𝑎2−s22∗𝑠𝑖𝑛22𝑎1



𝑑=s1∗n∗√1− 𝑠𝑖𝑛2𝑎1𝑛2

𝑠𝑖𝑛2𝑎1 𝑑=s2∗n∗√1− 𝑠𝑖𝑛2𝑎 2𝑛2

𝑠𝑖𝑛2𝑎2

s1∗n∗√1− 𝑠𝑖𝑛2𝑎 1𝑛2𝑠𝑖𝑛2𝑎1 =

s2∗n∗√1− 𝑠𝑖𝑛2𝑎2𝑛2𝑠𝑖𝑛2𝑎2

s12∗n2∗ 𝑛2−𝑠𝑖𝑛2𝑎1

𝑛2𝑠𝑖𝑛22𝑎1 =

s 22∗n2∗𝑛2−𝑠𝑖𝑛2𝑎 2

𝑛2𝑠𝑖𝑛22𝑎2

s12∗(𝑛2−𝑠𝑖𝑛2𝑎1)𝑠𝑖𝑛22𝑎1 =

s22∗(𝑛2−𝑠𝑖𝑛2𝑎2)𝑠𝑖𝑛22𝑎2

01/05/2023 41Giga Khizanishvili

𝒅=𝐬∗𝐧∗√𝟏− 𝒔𝒊𝒏𝟐𝒂

𝒏𝟐𝒔𝒊𝒏𝟐𝒂


𝑑=

s∗√ 𝑠12∗𝑠𝑖𝑛2𝑎1∗𝑠𝑖𝑛22𝑎2−𝑠22∗𝑠𝑖𝑛2𝑎2∗𝑠𝑖𝑛22𝑎1𝑠12∗𝑠𝑖𝑛22𝑎2−𝑠22∗𝑠𝑖𝑛 22𝑎 1 ∗√1− 𝑠𝑖𝑛2𝑎𝑠12∗𝑠𝑖𝑛2𝑎1∗ 𝑠𝑖𝑛22𝑎2−𝑠 22∗𝑠𝑖𝑛2𝑎2∗𝑠𝑖𝑛22𝑎1

𝑠12∗𝑠𝑖𝑛22𝑎2−𝑠22∗𝑠𝑖𝑛 22𝑎 1𝑠𝑖𝑛2𝑎


𝒏=√ 𝒙𝟏𝟐∗𝒔𝒊𝒏𝟐𝒂𝟏∗ 𝒔𝒊𝒏𝟐𝒂𝟏∗𝟒∗𝒔𝒊𝒏𝟐𝒂𝟐∗𝒄𝒐𝒔𝟐𝒂𝟐− 𝒙𝟐𝟐∗𝒔𝒊𝒏𝟐𝒂𝟐∗ 𝒔𝒊𝒏𝟐𝒂𝟐∗𝟒∗𝒔𝒊𝒏𝟐𝒂𝟏∗𝒄𝒐𝒔𝟐𝒂𝟏𝒙𝟏𝟐∗𝒔𝒊𝒏𝟐𝒂𝟏∗𝟒∗ 𝒔𝒊𝒏𝟐𝒂𝟐∗𝒄𝒐𝒔𝟐𝒂𝟐−𝒙𝟐𝟐∗𝒔𝒊𝒏𝟐𝒂𝟐∗𝟒∗ 𝒔𝒊𝒏𝟐𝒂𝟏∗ 𝒔𝒊𝒏𝟐𝒂𝟏

𝒏=√𝟒∗𝒔𝒊𝒏𝟐𝒂𝟏∗𝒔𝒊𝒏𝟐𝒂𝟐(𝒙 𝟏𝟐∗ 𝒔𝒊𝒏¿¿𝟐𝒂𝟏∗𝒄𝒐𝒔𝟐𝒂𝟐− 𝒙𝟐𝟐∗ 𝒔𝒊𝒏𝟐𝒂𝟐∗𝒄𝒐𝒔𝟐𝒂𝟏)𝟒∗ 𝒔𝒊𝒏𝟐𝒂𝟏∗𝒔𝒊𝒏𝟐𝒂𝟐(𝒙𝟏𝟐∗𝒄𝒐𝒔𝟐𝒂𝟐−𝒙 𝟐𝟐∗𝒄𝒐𝒔𝟐𝒂𝟏)

¿

𝒏=√ 𝒙𝟏𝟐∗𝒔𝒊𝒏𝟐𝒂𝟏∗𝒄𝒐𝒔𝟐𝒂𝟐− 𝒙𝟐𝟐∗ 𝒔𝒊𝒏𝟐𝒂𝟐∗𝒄𝒐𝒔𝟐𝒂𝟏𝒙𝟏𝟐∗𝒄𝒐𝒔𝟐𝒂𝟐− 𝒙𝟐𝟐∗𝒄𝒐𝒔𝟐𝒂𝟏

s=𝑥∗𝑠𝑖𝑛𝑎⇒


დახრილი სიბრტყის ცდომილება

𝛼𝑜𝑠

𝑙

𝑙= 𝑠𝑐𝑜𝑠𝑎

∆ 𝑠𝑠 =

𝑠𝑠∗𝑐𝑜𝑠𝑎 −1

∆ 𝑠𝑠 =

𝑙𝑠 −1

∆ 𝑠𝑠 =

1𝑐𝑜𝑠𝑎−1

𝑎=2.5 ° 𝑐𝑜𝑠𝑎=0.999048221582

∆ 𝑠𝑠 =

10.999048221582 −1

∆ 𝑠𝑠 =1.0009526851631775−1

∆ 𝑠𝑠 =0.0009526851631775

∆ 𝑠𝑠 ≈ 0 .01%

Giga Khizanishvili


contactless calipper - english

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