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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter2:ELECTROMAGNETIC PLANE WAVES"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2.6.5:pg-69"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
" \n",
"import math \n",
"#(a) Program to find gold-film surface resistance \n",
" \n",
" \n",
"t=80*(10**(-10)) #Film Thickness\n",
"o=4.1*(10**7) #Bulk conductivity \n",
"p=570*(10**(-10)) #Electron mean free path \n",
"of=((3*t*o)/(4*p))*(0.4228 + math.log(p/t)) #the gold-film conductivity is of=(3*t*o/4*p)*(0.4228 + ln(p/t)) \n",
"\n",
"Rs=1/(t*of) #the gold-film surface resistance is given by Rs=1/(t*of) in Ohms per square\n",
"\n",
"\n",
"print\"The gold film surface resistance in Ohms per square is=\",round(Rs,2),\"Ohms/square\"\n",
"\n",
"\n",
"#(b) Program to find the microwave attenuation \n",
"\n",
"Attenuation=40-20*log10(Rs) #Microwave attenuation \n",
"\n",
"print\"Microwave Attenuation in db is=\",int(Attenuation),\"db\"\n",
"\n",
"\n",
"#(c)Light transmittance T\n",
"\n",
"print\"From figure No.2-6-5 of Light transmittance T and light attenuation loss L versus wavelength with film thickness t as parameter for gold film, we find that for given gold film of thickness 80 angstrom ,the LIGHT TRANSMITTANCE T is estimated to be 75%\"\n",
"\n",
"\n",
"#(d)light reflection loss R\n",
"\n",
"print\"From the same figure the LIGHT REFLECTION LOSS R is about 25%\"\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The gold film surface resistance in Ohms per square is= 12.14 Ohms/square\n",
"Microwave Attenuation in db is= 18 db\n",
"From figure No.2-6-5 of Light transmittance T and light attenuation loss L versus wavelength with film thickness t as parameter for gold film, we find that for given gold film of thickness 80 angstrom ,the LIGHT TRANSMITTANCE T is estimated to be 75%\n",
"From the same figure the LIGHT REFLECTION LOSS R is about 25%\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2.6.6:pg-74"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"import math \n",
"#(a) Program to find copper-film surface resistance \n",
" \n",
" \n",
"t=60*(10**(-10)) #Film Thickness\n",
"o=5.8*(10**7) #Bulk conductivity \n",
"p=420*(10**(-10)) #Electron mean free path \n",
"of=((3*t*o)/(4*p))*(0.4228 + math.log(p/t)) #the copper-film conductivity is of=(3*t*o/4*p)*(0.4228 + ln(p/t))\n",
"Rs=1/(t*of) #the copper-film surface resistance is given by Rs=1/(t*of) in Ohms per square\n",
"\n",
"print\"The copper-film surface resistance in Ohms per square is=\",round(Rs,2),\"Ohms/square\"\n",
"\n",
"\n",
"#(b) Program to find the microwave attenuation \n",
"\n",
"Attenuation=40-20*log10(Rs) #Microwave attenuation \n",
"\n",
"print\"Microwave Attenuation in db is=\",int(round(Attenuation)),\"db\"\n",
"\n",
"#(c)Light transmittance T\n",
"\n",
"print\"From figure No.2-6-11 of Light transmittance T and light attenuation loss L versus wavelength with film thickness t as parameter for copper film, we find that for given copper film of thickness 60 angstrom ,the LIGHT TRANSMITTANCE T is estimated to be 82%\"\n",
"\n",
"#(d)light reflection loss R\n",
"\n",
"print\"From the same figure the LIGHT REFLECTION LOSS R is about 18%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The copper-film surface resistance in Ohms per square is= 11.32 Ohms/square\n",
"Microwave Attenuation in db is= 19 db\n",
"From figure No.2-6-11 of Light transmittance T and light attenuation loss L versus wavelength with film thickness t as parameter for copper film, we find that for given copper film of thickness 60 angstrom ,the LIGHT TRANSMITTANCE T is estimated to be 82%\n",
"From the same figure the LIGHT REFLECTION LOSS R is about 18%\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
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|
