{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 1: Interference of Light"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.2.1, Page number 1-11"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"i = 45 #angle of incidence(degrees)\n",
"t = 4*10**-5 #thickness of film(cm)\n",
"u = 1.2\n",
"\n",
"#Calculations & Result\n",
"r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n",
"for n in range(1,4):\n",
" lamda = (2*u*t*math.cos(r*math.pi/180))/n\n",
" print \"For n = %d, wavelength = %.2f A\"%(n,lamda/10**-8)\n",
"print \"Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths lie in this range\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"For n = 1, wavelength = 7756.29 A\n",
"For n = 2, wavelength = 3878.14 A\n",
"For n = 3, wavelength = 2585.43 A\n",
"Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths lie in this range\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.2.2, Page number 1-12"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"r = 90 #angle of refraction(degrees)\n",
"t = 5*10**-5 #thickness of film(cm)\n",
"u = 1.33\n",
"\n",
"#Calculations & Result\n",
"for n in range(1,4):\n",
" lamda = (4*u*t*int(math.cos(math.radians(90))))/((2*n)-1)\n",
" print \"For n = %d, wavelength = %.2f A\"%(n,lamda)\n",
"print \"Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths do not lie in this range\"\n",
"\n",
"print \"\\nPlease note: Since r=90, cos(r)=0\\nHence, the answers given in the textbook are incorrect\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"For n = 1, wavelength = 0.00 A\n",
"For n = 2, wavelength = 0.00 A\n",
"For n = 3, wavelength = 0.00 A\n",
"Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths do not lie in this range\n",
"\n",
"Please note: Since r=90, cos(r)=0\n",
"Hence, the answers given in the textbook are incorrect\n"
]
}
],
"prompt_number": 33
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.2.3, Page number 1-12"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"i = 45 #angle of incidence(degrees)\n",
"t = 1.5*10**-4 #thickness of film(cm)\n",
"lamda = 5*10**-5 #wavelength(cm)\n",
"u = 4./3. #refractive index\n",
"\n",
"#Calculations\n",
"r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n",
"n = (2*u*t*math.cos(r*math.pi/180))/lamda\n",
"\n",
"#Result\n",
"print \"The order of interfernce is %.2f, close to 7\"%n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The order of interfernce is 6.78, close to 7\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 12.2.4, Pae number 1-13"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"i = 45 #angle of incidence(degrees)\n",
"lamda = 5896*10**-8 #wavelength(cm)\n",
"u = 1.33 #refractive index\n",
"n = 1\n",
"\n",
"#Calculations\n",
"r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n",
"t = ((2*n-1)*lamda)/(2*u*math.cos(r*math.pi/180)*2)\n",
"\n",
"#Result\n",
"print \"The required thickness is\",round((t/1E-5),2),\"*10^-5 cm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required thickness is 1.31 *10^-5 cm\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.2.5, Page number 1-14"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"lamda1 = 7000 #wavelength(A)\n",
"lamda2 = 5000 #wavelength(A)\n",
"u = 1.3 #R.I. of oil\n",
"\n",
"#Calculations\n",
"'''\n",
"2utcosr = (2n-1)7000/2 ----(1)\n",
"2utcosr = (2n+1)5000/2 ----(2)\n",
"Divinding (1) by (2), we get the following expression\n",
"1 = (2n+1)5000\n",
" -----------\n",
" (2n-1)7000\n",
"Solving the above expression, we get,\n",
"'''\n",
"n = 12000/4000\n",
"t = ((2*n-1)*lamda)/(2*u*math.cos(r*math.pi/180)*2)\n",
"\n",
"#Result\n",
"print \"The required thickness is\",round((t/1E-5),4),\"*10^-5 cm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required thickness is 6.6936 *10^-5 cm\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.2.6, Page number 1-15"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"#Variable declaration\n",
"i = 30 #angle of incidence(degrees)\n",
"lamda = 5890*10**-8 #wavelength(cm)\n",
"u = 1.46 #refractive index\n",
"n = 8\n",
"\n",
"#Calculations\n",
"r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n",
"t = (n*lamda)/(2*u*math.cos(r*math.pi/180))\n",
"\n",
"#Result\n",
"print \"The required thickness is\",round((t/1E-4),3),\"*10^-4 cm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required thickness is 1.718 *10^-4 cm\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.2.7, Page number 1-15"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"#Variable declaration\n",
"r = 60 #angle of refraction(degrees)\n",
"lamda = 5890*10**-8 #wavelength(cm)\n",
"u = 1.5 #refractive index\n",
"n = 1 #for minimumm thickness\n",
"\n",
"#Calculations\n",
"#For r = 60\n",
"t1 = (n*lamda)/(2*u*math.cos(r*math.pi/180))\n",
"\n",
"#For normal incidence \n",
"r = 0\n",
"t2 = (n*lamda)/(2*u*math.cos(r*math.pi/180))\n",
"\n",
"#Result\n",
"print \"For r = 60, the required thickness is\",round((t1/1E-5),2),\"*10^-5 cm\"\n",
"print \"For r = 0, the required thickness is\",round((t2/1E-5),2),\"*10^-5 cm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"For r = 60, the required thickness is 3.93 *10^-5 cm\n",
"For r = 0, the required thickness is 1.96 *10^-5 cm\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.2.8, Page number 1-16"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"#Variable declaration\n",
"r = 0 #for normal incidence(degrees)\n",
"lamda = 5.5*10**-5 #wavelength(cm)\n",
"n = 1 #for minimumm thickness\n",
"A = 10**4 #area(cm^2)\n",
"V = 0.2 #volume(cc)\n",
"\n",
"#Calculations\n",
"t = V/A\n",
"#for nth dark band,\n",
"u = (n*lamda)/(2*t*math.cos(r*math.pi/180))\n",
"\n",
"#Result\n",
"print \"Refractive index =\",u"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Refractive index = 1.375\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.2.9, Page number 1-17"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"#Variable declaration\n",
"r = 60 #angle of incidence(degrees)\n",
"t = 2*10**-7 #thickness of film(cm)\n",
"u = 1.2\n",
"\n",
"#Calculations & Result\n",
"for n in range(1,4):\n",
" lamda = (4*u*t*math.cos(r*math.pi/180))/(2*n-1)\n",
" print \"For n = %d, wavelength = %.2f A\"%(n,lamda/10**-10)\n",
"print \"Since the visible range of wavelengths lie between 4000 to 7000 A, the wavelength in the visible spectrum is 4800 A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"For n = 1, wavelength = 4800.00 A\n",
"For n = 2, wavelength = 1600.00 A\n",
"For n = 3, wavelength = 960.00 A\n",
"Since the visible range of wavelengths lie between 4000 to 7000 A, the wavelength in the visible spectrum is 4800 A\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.3.1, Page number 1-21"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration \n",
"a = 40. #angle(sec)\n",
"lamda = 1.2 #distance between fringes(cm)\n",
"alpha = 10 #no. of fringes\n",
"\n",
"#Calculations\n",
"Bair = lamda/alpha #cm\n",
"alpha = (a*math.pi)/(3600*180) #radians\n",
"lamda = 2*alpha*Bair\n",
"\n",
"#Result\n",
"print \"Wavelength of monochromatic light =\",round((lamda/1E-8),1),\"A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Wavelength of monochromatic light = 4654.2 A\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.3.2, Page number 1-22"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration \n",
"lamda = 5893*10**-8 #wavelength(cm)\n",
"u = 1.52 #refractive index\n",
"B = 0.1 #fringe spacing(cm)\n",
"\n",
"#Calculations\n",
"alpha = (lamda/(2*u*B))*180*3600/math.pi #seconds\n",
"\n",
"#Result\n",
"print \"Angle of wedge =\",round(alpha,2),\"secs\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Angle of wedge = 39.98 secs\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.3.3, Page number 1-22"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration \n",
"u = 1.4 #refractive index\n",
"B = 0.25 #fringe spacing(cm)\n",
"a = 20 #angle(secs)\n",
"\n",
"#Calculations\n",
"alpha = (a*math.pi)/(3600*180) #radians\n",
"lamda = 2*u*alpha*B\n",
"\n",
"#Result\n",
"print \"Wavelength of light =\",round((lamda/1E-8),2),\"A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Wavelength of light = 6787.39 A\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.3.4, Page number 1-23"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration \n",
"u = 1.5 #refractive index\n",
"lamda = 5.82*10**-5 #wavelength(cm)\n",
"a = 20 #angle(secs)\n",
"\n",
"#Calculations\n",
"alpha = (a*math.pi)/(3600*180) #radians\n",
"B = lamda/(2*u*alpha)\n",
"N = 1/B\n",
"\n",
"#Result\n",
"print \"Number of interfernce fronges pr cm is\",round(N)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Number of interfernce fronges pr cm is 5.0\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.3.5, Page number 1-24"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration \n",
"u = 1 #refractive index for air film\n",
"lamda = 6*10**-5 #wavelength(cm)\n",
"B = 1./10 #distance between fringes(cm)\n",
"\n",
"#Calculations\n",
"alpha = lamda/(2*u*B) #radians\n",
"d = alpha*10\n",
"\n",
"#Result\n",
"print \"Daimeter of wire =\",d,\"cm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Daimeter of wire = 0.003 cm\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.3.6, Page number 1-24"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"alpha = 0.01*10**-1/10 #angle(radians)\n",
"u = 1 #refractive index for air film\n",
"lamda = 5900*10**-10 #wavlength(m)\n",
"\n",
"#Calculation\n",
"B = lamda/(2*u*alpha)\n",
"\n",
"#Result\n",
"print \"Seperation between fringes is\",B/10**-3,\"mm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Seperation between fringes is 2.95 mm\n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4.1, Page number 1-32"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration \n",
"n = 40\n",
"\n",
"#Calculation\n",
"#Equating the equation 4*R*n*lamda=4*4R*n*lamda, we get\n",
"\n",
"N = (4*4*n)/4\n",
"\n",
"#result\n",
"print \"Ring number =\",N"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Ring number = 160\n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4.2, Page number 1-32"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration \n",
"n = 10\n",
"Dn = 0.5 #diameter of dark ring(cm)\n",
"lamda = 5*10**-5 #waelength(cm)\n",
"\n",
"#Calculations\n",
"R = Dn**2/(4*n*lamda)\n",
"\n",
"#Result\n",
"print \"Radius of curvature =\",R,\"cm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Radius of curvature = 125.0 cm\n"
]
}
],
"prompt_number": 28
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4.3, Page number 1-33"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration \n",
"n = 5\n",
"p = 10\n",
"D5 = 0.336 #diameter of 5th ring(cm)\n",
"lamda = 5890*10**-8 #waelength(cm)\n",
"D15 = 0.59 #diameter of 15th ringcm\n",
"\n",
"#Calculations\n",
"R = (D15**2-D5**2)/(4*p*lamda)\n",
"\n",
"#Result\n",
"print \"Radius of curvature =\",round(R,2),\"cm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Radius of curvature = 99.83 cm\n"
]
}
],
"prompt_number": 34
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4.4, Page number 1-33"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Varaible declaration\n",
"Dn = 0.42 #diameter of dark ring(cm)\n",
"p = 8 \n",
"R = 200 #radius of curvature(cm)\n",
"Dn8 = 0.7 #diameter of (n+8)th ring(cm)\n",
"\n",
"#Calculations\n",
"lamda = (Dn8**2-Dn**2)/(4*R*p)\n",
"\n",
"#Result\n",
"print \"Wavelength =\",lamda/1E-8,\"A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Wavelength = 4900.0 A\n"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4.5, Page number 1-34"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Varaible declaration\n",
"Dn = 0.218 #cm\n",
"Dn10 = 0.451 #cm\n",
"lamda = 5893*10**-8 #wavelength(cm)\n",
"R = 90 #radius of curvature(cm)\n",
"p = 10\n",
"\n",
"#Calculation\n",
"u = (4*p*lamda*R)/(Dn10**2-Dn**2)\n",
"\n",
"#Result\n",
"print \"Refractive index =\",round(u,3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Refractive index = 1.361\n"
]
}
],
"prompt_number": 37
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4.6, Page number 1-34"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Varaible declaration\n",
"D5 = 0.42 #diameter of dark ring(cm)\n",
"\n",
"#Calculations\n",
"'''\n",
"For 5th dark ring,\n",
"D5^2 = 20*R*lamda -----1\n",
"\n",
"For 10th dark ring,\n",
"D10^2 = 40*R*lamda -----2\n",
"\n",
"Substituting 1 in 2,\n",
"'''\n",
"\n",
"D10 = math.sqrt((40*D5**2)/20)\n",
"\n",
"#Result\n",
"print \"Diameter of the 10th dark ring =\",round(D10,3),\"cm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Diameter of the 10th dark ring = 0.594 cm\n"
]
}
],
"prompt_number": 35
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4.7, Page number 1-35"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Varaible declaration\n",
"lamda_n = 6000 #wavelength of nth ring(A)\n",
"lamda_n1 = 5000 #wavelength for (n+1)th ring(cm)\n",
"\n",
"#Calculations\n",
"'''\n",
"Dn^2 = 4*R*n*lamda_n ---1\n",
"\n",
"Dn+1^2 = 4*R(n+1)*lamda_n1 ---2\n",
"\n",
"Equating 1 and 2, we get,\n",
"'''\n",
"\n",
"n = 5\n",
"R = 2\n",
"Dn = math.sqrt(4*R*n*lamda_n*10**-8)\n",
"\n",
"#Result\n",
"print \"Diameter =\",round(Dn,3),\"cm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Diameter = 0.049 cm\n"
]
}
],
"prompt_number": 42
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4.8, Page number 1-35"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Varaible declaration\n",
"Dair = 2.3 #diameter of ring in air(cm)\n",
"Dliq = 2 #diameter of ring in liquid(cm)\n",
"\n",
"#Calculations\n",
"u = Dair**2/Dliq**2\n",
"\n",
"#Result\n",
"print \"Refractive index =\",u"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Refractive index = 1.3225\n"
]
}
],
"prompt_number": 43
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4.11, Page number 1-37"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"D4 = 0.4 #diameter of 4th ring(cm)\n",
"D12 = 0.7 #diameter of 12th ring(cm)\n",
"\n",
"#Calculations\n",
"'''\n",
"For D4, \n",
"D4 = math.sqrt(4R*4*lamda)\n",
"'''\n",
"rt_Rl = 0.1\n",
"R = 80 \n",
"\n",
"#For D20,\n",
"D20 = math.sqrt(R)*rt_Rl\n",
"\n",
"#Result\n",
"print \"Diameter of 20th ring =\",round(D20,3),\"cm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Diameter of 20th ring = 0.894 cm\n"
]
}
],
"prompt_number": 46
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4.12, Page number 1-36"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration \n",
"n = 5\n",
"p = 10\n",
"D5 = 0.336 #diameter of 5th ring(cm)\n",
"D15 = 0.590 #diameter of 15th ring(cm)\n",
"R = 100 #radius of curvature(cm)\n",
"\n",
"#Calculations\n",
"lamda = (D15**2-D5**2)/(4*R*p)\n",
"\n",
"#Result\n",
"print \"Wavlength =\",lamda/1E-8,\"A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Wavlength = 5880.1 A\n"
]
}
],
"prompt_number": 50
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.7.1, Page number 1-42"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration\n",
"lamda = 560 #wavelength(nm)\n",
"u = 2 #refractive index\n",
"\n",
"#Calculations\n",
"lamda_dash = lamda/u\n",
"t = lamda_dash/4\n",
"\n",
"#Result\n",
"print \"Thickness of film =\",t,\"nm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thickness of film = 70 nm\n"
]
}
],
"prompt_number": 55
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.7.2, Page number 1-42"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration\n",
"lamda = 6000 #wavelength(E)\n",
"u = 1.2 #refractive index\n",
"\n",
"#Calculations\n",
"lamda_dash = lamda/u\n",
"t = lamda_dash/4\n",
"\n",
"#Result\n",
"print \"Thickness of film =\",t,\"A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thickness of film = 1250.0 A\n"
]
}
],
"prompt_number": 56
}
],
"metadata": {}
}
]
}