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Measurement of Film Thickness An optical measurement of thin films is fairly easy.  Film Thickness is measured using a Spectrophotometer. In order to use this technique the optical thickness of the films has to be within the wavelength range of your spectrophotometer . Spectrophotometer has reference slide as an ordinary glass plate. In the other holder the coated film is loaded and the interference pattern is obtained by passing  light through the film, at selected wavelengths with the help of a grating.
n = n – i k d Film Substrate S  s  = 0 T T 1  = 1 n 0  = 1 n 0  = 1 Thin Film on a thick finite  transparent substrate ,[object Object],[object Object],For the film: Refractive index n = n –i k, k = extinction coefficient k =   /4  Substrate thickness d s  >>> d film thickness  For the substrate: Refractive index = S,   S  = 0 d s
Materials to be coated : SnO 2  ,   ZnO Substrate Film Air d t Refractive index  = 1 Refractive index of film = n Refractive index of substrate = s Refractive index  = 1 Figure showing the directions of incident and transmitted light in different regions
   (nm) If the thickness d is uniform interference effects gives this spectrum 500 900 Transparent Medium weak Absorption
The basic equation for  Interference :  2 n d = m     m is an integer  for constructive interference (maxima) and half integer for destructive interference (minima) Wave lengths of different regions: Ultraviolet rays ( UV rays )  :  below 300 nm Visible rays   :  400nm – 800 nm Infrared  ( IR Rays )   : above 800 nm We use 300 – 900 nm range
Weak Medium Strong Transmission is For Glass has n    1.5 Transmission    92 %, For Pure Quartz   93.5 % in visible region In the weak transmission region,    falls close to optical band gap of the material showing strong absorption α  is high and T is low
The Reference Spectrum for glass is shown below T %  UV  Visible  IR In this spectrum there occurs no Interference 92 %
In the available Spectrophotometer in the laboratory, the possible wave length is 300 – 900 nm E g  for ZnS = 3.5 eV  E g  for SnO 2  = 3.2 eV  Evaporated in the Mo boat Because its melting point is lower For higher MP- tungsten is used Optical band gap values are given below:
Reflection coefficient = R Transmission coefficient = T T is a function of ( n, s, d,   ,    ) Here n = refractive index of the film s = refractive index  of the substrate d = thickness of the film    =  absorption coefficient     = wavelength of the light being used In other words T= T (n, x),  x(   ) is the absorbance = exp (-   d)
T max  = T M T min  = T m
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],A = 16n 2 s B = (n+1) 3 (n+s 2 ) C = (n-1) 3 (n-s 2 ) D=(n-1) 3 (n-s 2 )    = 4  nd/  
In the Transparent region ,    = 0 and x = 1  T M  = [ 2S / ( S 2  + 1 ) ] T m  = [ 4n 2 S / ( n 4  + n 2 (S 2  + 1) + S 2 ) ] Where n is given as n= [ M + ( M 2 -S 2  ) 1/2  ] 1/2 Here M is calculted using the relation, M = [ ( 2S / T M  ) –  {( S 2  + 1 ) / 2 }] Thus n can be calculated for each of the wavelengths Analysis of the spectrum:
In the region of weak absorption,       0 Here n is the refractive index of the film N = [ 2S( T M  - T m ) / T M  T m  ] + {( S 2  + 1 ) / 2 }] where N is a mathematical simplification parameter Once T M  at a given    are known then N and n can be calculated.  Take T M  or T m  alone and then calculate d, the thickness of the film using the below relation d = [ (  m(   1   2  ) / 2 ( n 2    1  – n 1    2   ) ] n 1 , n 2  are the refractive indices at two adjacent maxima
In the Absorptive region , In this region,       0 and x < 1  The relations used for calculating different parameters are given below T M  = [ 2S / ( S 2  + 1 ) ]( 1 / T m  ) – ( 1 / T M  ) =  ( 2 C / A ) n = [ N + ( N 2 -S 2  ) 1/2  ] 1/2 Where N = [ 2S( T M  - T m ) / T M  T m  ] + {( S 2  + 1 ) / 2 }] Thus n (    ) can be calculated for each of the wavelengths
d  =  (   m   1   2  ) / 2 ( n 2    1  – n 1    2  )  m = 1 for successive maxima d  =  (   1   2  ) / 2 ( n 2    1  – n 1    2  ) for maxima and d  =  (   1   2  ) / 4 ( n 2    1  – n 1    2  ) for minimum and  successive maximum
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

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6.b.measurement of film thickness

  • 1. Measurement of Film Thickness An optical measurement of thin films is fairly easy.  Film Thickness is measured using a Spectrophotometer. In order to use this technique the optical thickness of the films has to be within the wavelength range of your spectrophotometer . Spectrophotometer has reference slide as an ordinary glass plate. In the other holder the coated film is loaded and the interference pattern is obtained by passing light through the film, at selected wavelengths with the help of a grating.
  • 2.
  • 3. Materials to be coated : SnO 2 , ZnO Substrate Film Air d t Refractive index = 1 Refractive index of film = n Refractive index of substrate = s Refractive index = 1 Figure showing the directions of incident and transmitted light in different regions
  • 4. (nm) If the thickness d is uniform interference effects gives this spectrum 500 900 Transparent Medium weak Absorption
  • 5. The basic equation for Interference : 2 n d = m  m is an integer for constructive interference (maxima) and half integer for destructive interference (minima) Wave lengths of different regions: Ultraviolet rays ( UV rays ) : below 300 nm Visible rays : 400nm – 800 nm Infrared ( IR Rays ) : above 800 nm We use 300 – 900 nm range
  • 6. Weak Medium Strong Transmission is For Glass has n  1.5 Transmission  92 %, For Pure Quartz  93.5 % in visible region In the weak transmission region,  falls close to optical band gap of the material showing strong absorption α is high and T is low
  • 7. The Reference Spectrum for glass is shown below T %  UV Visible IR In this spectrum there occurs no Interference 92 %
  • 8. In the available Spectrophotometer in the laboratory, the possible wave length is 300 – 900 nm E g for ZnS = 3.5 eV E g for SnO 2 = 3.2 eV Evaporated in the Mo boat Because its melting point is lower For higher MP- tungsten is used Optical band gap values are given below:
  • 9. Reflection coefficient = R Transmission coefficient = T T is a function of ( n, s, d,  ,  ) Here n = refractive index of the film s = refractive index of the substrate d = thickness of the film  = absorption coefficient  = wavelength of the light being used In other words T= T (n, x), x(  ) is the absorbance = exp (-  d)
  • 10. T max = T M T min = T m
  • 11.
  • 12. In the Transparent region ,  = 0 and x = 1 T M = [ 2S / ( S 2 + 1 ) ] T m = [ 4n 2 S / ( n 4 + n 2 (S 2 + 1) + S 2 ) ] Where n is given as n= [ M + ( M 2 -S 2 ) 1/2 ] 1/2 Here M is calculted using the relation, M = [ ( 2S / T M ) – {( S 2 + 1 ) / 2 }] Thus n can be calculated for each of the wavelengths Analysis of the spectrum:
  • 13. In the region of weak absorption,   0 Here n is the refractive index of the film N = [ 2S( T M - T m ) / T M T m ] + {( S 2 + 1 ) / 2 }] where N is a mathematical simplification parameter Once T M at a given  are known then N and n can be calculated. Take T M or T m alone and then calculate d, the thickness of the film using the below relation d = [ (  m(  1  2 ) / 2 ( n 2  1 – n 1  2 ) ] n 1 , n 2 are the refractive indices at two adjacent maxima
  • 14. In the Absorptive region , In this region,   0 and x < 1 The relations used for calculating different parameters are given below T M = [ 2S / ( S 2 + 1 ) ]( 1 / T m ) – ( 1 / T M ) = ( 2 C / A ) n = [ N + ( N 2 -S 2 ) 1/2 ] 1/2 Where N = [ 2S( T M - T m ) / T M T m ] + {( S 2 + 1 ) / 2 }] Thus n (  ) can be calculated for each of the wavelengths
  • 15. d = (  m  1  2 ) / 2 ( n 2  1 – n 1  2 )  m = 1 for successive maxima d = (  1  2 ) / 2 ( n 2  1 – n 1  2 ) for maxima and d = (  1  2 ) / 4 ( n 2  1 – n 1  2 ) for minimum and successive maximum
  • 16.
  • 17.