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diff --git a/sample_notebooks/Ashish KumarSingh/Chapter_First.ipynb b/sample_notebooks/Ashish KumarSingh/Chapter_First.ipynb new file mode 100755 index 00000000..26e2a823 --- /dev/null +++ b/sample_notebooks/Ashish KumarSingh/Chapter_First.ipynb @@ -0,0 +1,243 @@ +{
+ "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\""
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.10"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
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