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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 1: Light"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 1.1, Page Number 10"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The Brewsters Angle of the Material is 56.31 Degrees\n"
]
}
],
"source": [
"import math\n",
"\n",
"n2=1.5 #Given Refractive Index of Glass in Air\n",
"n1=1 #Given Refractive Index of Air\n",
"\n",
"theta=0 #Brewster's Angle\n",
"#From Equation 1.13 (Brewsters angle= Tan Inverse (n2/n1))\n",
"theta=math.degrees(math.atan(1.5))\n",
"print \"The Brewsters Angle of the Material is \"+str(round(theta,2))+\" Degrees\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 1.2, Page Number 13"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"In Coherant Sources The Maximum Irradiance is 16I\n",
"In Incoherant Sources The Maximum Irradiance is 4I\n"
]
}
],
"source": [
"n=4 #Total Number of Sources\n",
"\n",
"#For Coherant Sources\n",
"print \"In Coherant Sources The Maximum Irradiance is \"+str(n*n)+\"I\" #Where I is the Irradiance at any point\n",
"#For Incoherant Sources\n",
"print \"In Incoherant Sources The Maximum Irradiance is \"+str(n)+\"I\" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 1.3, Page Number 23"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(A)The Minimum Seperation Between the Sources is 0.0025 m\n",
"(B)The Minimum Wavelength Difference which may be resolved is 2.08333333333e-11 m\n"
]
}
],
"source": [
"D=0.1 #Diameter of the Objective Lens\n",
"d1=500 #Distance from the source\n",
"l =0.000000500 #Wavelength Provided\n",
"p=1 #First Order\n",
"N=40*600 #The diffraction grating is 40 mm wide and has 600 lines/mm\n",
"\n",
"#From Equation 1.29 we Have\n",
"Smin=(d1*l)/D #Where Smin is the minimum Seperation of the Sources\n",
"print \"(A)The Minimum Seperation Between the Sources is \"+str(Smin)+\" m\"\n",
"\n",
"#We know that Chromatic resolving power is given by l/dl where dl is the Minimum Wavelength Difference\n",
"#From Equation l/dl=p*N\n",
"dl=l/(N*p)\n",
"\n",
"print \"(B)The Minimum Wavelength Difference which may be resolved is \"+str(dl)+\" m\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##Example 1.4, Page Number 29"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The Total Power Radiated from the Source is 6.3504 W\n"
]
}
],
"source": [
"em=0.7 #Emissivity Of the Surface\n",
"T=2000 #Temperature in Kelvin\n",
"A=0.00001 #Area in Meter Square\n",
"S=5.67*(10**-8) #Stefan-Boltzmann Constant\n",
"\n",
"W=S*A*em*(T**4) #Where W is the total power radiated\n",
"\n",
"print \"The Total Power Radiated from the Source is \"+str(W)+\" W\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##Example 1.5, Page Number 31"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The Ionization Energy required to excite the electron from ground to Infinity 13.66 eV\n"
]
}
],
"source": [
"Z=1 #Atomic Number of Hydrogen\n",
"m=9.1*(10**-31) #Mass of a Electron\n",
"e=1.6*(10**-19) #Charge Of a Electron\n",
"p=6.6*(10**-34) #Plancks Constant\n",
"e1=8.85*(10**-12)#Permittivity of Free Space\n",
"#From Equation 1.43\n",
"E=(m*(Z**2)*(e**4))/(8*(p**2)*(e1**2)) #Where E is the Ionization Energy\n",
"E2=E/e #Converting in Electron Volts\n",
"E2=round(E2,2)\n",
"\n",
"print \"The Ionization Energy required to excite the electron from ground to Infinity \"+str(E2)+\" eV\"\n",
" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##Example 1.6, Page Number 32"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The Required Work function is 4.5375 eV\n"
]
}
],
"source": [
"e=1.6*(10**-19) #Charge Of a Electron\n",
"h=6.6*(10**-34) #Plancks Constant\n",
"vo=1.1*(10**15) #Threshold Frequency in Hertz\n",
"\n",
"# We Know h*vo=phi*e where phi is the required Work Function\n",
"# We assume that the ejected electron has zero kinetic energy\n",
"\n",
"phi=h*vo/e\n",
"print \"The Required Work function is \"+str(phi)+\" eV\""
]
}
],
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"language": "python",
"name": "python2"
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|
