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diff --git a/Engineering_Physics_-_I/README.txt b/Engineering_Physics_-_I/README.txt
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+Contributed By: saikomal chanagam
+Course: btech
+College/Institute/Organization: SASTRA University
+Department/Designation: ECE
+Book Title: Engineering Physics - I
+Author: G. SenthilKumar
+Publisher: VRB Publishers, Chennai
+Year of publication: 2011
+Isbn: 9789380241852
+Edition: 1
\ No newline at end of file
diff --git a/Engineering_Physics_-_I/chapter1.ipynb b/Engineering_Physics_-_I/chapter1.ipynb
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+{
+ "metadata": {
+  "name": "Chapter1(s)"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 1: Ultrasonics"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1, Page No:1.29"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math\n\n# variable declaration\nP           = 1;            # for fundamental mode\nt           = 0.1*10**-2;    # thickness of piezo electric crystal\nE           = 80*10**9       # young's modulus\np           = 2654          # density in kg/m^3\n\n# Calculations\n\nf           = (P/(2*t))*math.sqrt(E/p);      # frequency of the oscillator circuit\n\n# Result\nprint 'The Frequency of the oscillator circuit %3.4g' %f,'Hz';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Frequency of the oscillator circuit 2.745e+06 Hz\n"
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2, Page No:1.29"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math\n# variable declaration\nP           = 1;             # for fundamental mode\nt           = 0.1*10**-2;    # thickness of piezo electric crystal\nE           = 7.9*10**10     # young's modulus\np           = 2650           # density in kg/m^3\n\n# Calculations\n\nf           = (P/(2*t))*math.sqrt(E/p);      # frequency of the oscillator circuit\n\n#Result\nprint 'The Frequency of the vibrating crystal %0.2f'%(f/(10**6)),'MHz';\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Frequency of the vibrating crystal 2.73 MHz\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3, Page No:1.30"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math\n\n# variable Declaration\nf       = 1.5*10**6;         # frequency of ultrasonics in Hz\nd6      = 2.75*10**-3;       # distance between 6 consecutive nodes\n\n# Calculations\nd       = d6/5;             # distance b/w two nodes\nlamda   = 2*d;              #  wavelength in m\nv       = f*lamda;          # velocity of ultrasonics\n\n# Result\nprint 'Velocity of ultrasonics ' ,v,'m/sec';\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Velocity of ultrasonics  1650.0 m/sec\n"
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Add_example 1, Page No: 1.31"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math\n\n# Variable Declaration\nP           = 1;            # for fundamental mode\nt           = 1.5*10**-3;    # thickness of quartz crystal\nE           = 7.9*10**10     # young's modulus in N/m^2\np           = 2650          # density in kg/m^3\n\n# Calculations\n\nf           = (P/(2*t))*math.sqrt(E/p);      # frequency of the oscillator circuit\n\n# Result\nprint 'The Fundamental Frequency of the Quartz crystal %3.2f'%(f/10**6), 'MHz';\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Fundamental Frequency of the Quartz crystal 1.82 MHz\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Add_example 2, Page No: 1.31"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math\n\n# Variable Declaration\nv       = 5000;         # velocity of ultrasonics in m/s\ndf      = 60*10**3;      # difference b/w two adjacent harmonic freq. in Hz\n\n# Calculations\n\nd       = (float(v)/(2*df))  ;       # thickness of steel plate\n\n# Result\nprint 'The thickness of steel plate %3.4f'%(d),'m';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The thickness of steel plate 0.0417 m\n"
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Add_example 3, Page No: 1.32"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math\n\n# Variable Declaration\nv       = 1440;         # velocity of ultrasonics in  sea water in m/s\nt       = 0.33          # time taken b/w tx and rx in sec\n\n# Calculations\n\nd       = v*t;          # distance travelled by ultrasonics\nD       = d/2;          # depth of submerged submarine in m\n\n# Result\nprint 'Depth of submerged submarine',D,'m';\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Depth of submerged submarine 237.6 m\n"
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Add_example 4, Page No: 1.33"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math\n\n# Variable Declaration\nd           = 0.55*10**-3;        #  distance b/w two antinodes\nf           = 1.5*10**6;         # freq of the crystal\n\n# Calculations\n\nlamda       = 2*d;              # wavelength\nv           = f*lamda;          # velocity of ultronics\n\n# Result\nprint 'Velocity of waves in sea water',v,'m/s';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Velocity of waves in sea water 1650.0 m/s\n"
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Add_example 5, Page No: 1.33\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math\n\n# Variable Declaration\nP           = 1;            # for fundamental mode\np           = 2660          # density of quartz in kg/m^3\nf           = 1300*10**3     # freq of quartz plate for sub division ii\nk           = 2.87*10**3\n\n#f1        = (k)/t  # freq for sub division i\n\n# Calculations\n\n#f           = (P/(2*t))*sqrt(E/p);  \nE             = p*4*(k)**2;      # Youngs modulus in N/m^2\nt           = (float(P)/(2*f))*math.sqrt(E/p);       \n\n\n# Result\nprint 'Youngs modulus of quartz plate %3.5g'%E,'Nm^-2'\nprint 'Thickness of the crystal %.4e'%t,'m';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Youngs modulus of quartz plate 8.7641e+10 Nm^-2\nThickness of the crystal 2.2077e-03 m\n"
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Physics_-_I/chapter2.ipynb b/Engineering_Physics_-_I/chapter2.ipynb
new file mode 100755
index 00000000..84198c1f
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+++ b/Engineering_Physics_-_I/chapter2.ipynb
@@ -0,0 +1,293 @@
+{
+ "metadata": {
+  "name": "Chapter 2(s)"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 2: Lasers"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1, Page No: 2.40"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding Number of electron hole pairs\n\nimport math\n\n# Variable Declaration\nA           = 4*10**-6;              #  Receiving area of photo detector\nI           = 200;                   # Intensity in W/m^2\nh           = 6.625*10**-34;          # planck's constant\nc           = 3*10**8;                # vel. of light in m/s\nlamda       = 0.4*10**-6;             # wavelength of light in m\n\n#Calculations\nv           = c/lamda;              # frequency\nNOP         = I*A/(h*v)             # number of photons\n\n#since each photon generates an electron hole pair, the number of photons is equal to number of electron hole pairs\n\n# Result\n\nprint 'Number of electron hole pairs = %3.2e '%NOP;\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Number of electron hole pairs = 1.61e+15 \n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2, Page No:2.40"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding Wavelength\n\nimport math\n\n# Variable Declaration\nEg          = 2.8;                   # bandgap energy in eV\nh           = 6.625*10**-34;          # plank's constant\nc           = 3*10**8;                # vel. of light in m/s\nq           = 1.602*10**-19;          # charge of electron\n\n# Calculations\nE           = Eg*q                 # eV to joules conversion\nlamda       = h*c/E;               # wavelength\n\n#Result\nprint 'wavelength = %3.1f' %(lamda*10**10), '\u00c5(Blue Colour)';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "wavelength = 4430.8 \u00c5(Blue Colour)\n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3, Page No:2.41"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding Energy bandgap\n\nimport math;\n\n# Variable Declaration\nh           = 6.625*10**-34;          # plank's constant\nc           = 3*10**8;                # vel. of light in m/s\nlamda       = 1.55*10**-6;            # wavelength of light in m\nq           = 1.6*10**-19;            # charge of electron\n\n#Calculations\nEg          = (h*c)/lamda;           # band gap energy in joules\nE           = Eg/q                   # bang gap energy in eV\n\n# Result\nprint 'Energy bandgap Eg   = %3.4f'%E, 'eV';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Energy bandgap Eg   = 0.8014 eV\n"
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4, Page No:2.41"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding Number of photons required to do one Joule of work\n\nimport math\n\n# Variable Declaration\nh           = 6.625*10**-34;         # plank's constant\nc           = 3*10**8;               # vel. of light in m/s\nlamda       = 4961*10**-10;          # wavelength of light in m\n\n# Calculations\nE           = (h*c)/lamda;           # energy in joules\nN           = 1/E\n\n# Result\nprint 'Number of photons required to do one Joule of work  = %3.4e'%N,'/m^3';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Number of photons required to do one Joule of work  = 2.4961e+18 /m^3\n"
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 5, Page No:2.41"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding Wavelength Limit\n# import math\n\n# Variable Declaration\nE           = 0.02;                 # ionisation energy in eV\nh           = 6.625*10**-34;         # plank's constant\nc           = 3*10**8;               # vel. of light in m/s\nq           = 1.6*10**-19;           # charge of electron\n\n# Calculations\n\nlamda       = h*c/(E*q)              # long wavelength limit in m\n\n# Result\n\nprint 'long wavelength limit = %3.3e' %lamda,' m';\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "long wavelength limit = 6.211e-05  m\n"
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6, Page No:2.42"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding Wavelength\n\nimport math;\n\n# Variable Declaration\nE           = 1.44;                 # Bandgap energy in eV\nh           = 6.625*10**-34;        # plank's constant\nc           = 3*10**8;              # vel. of light in m/s\nq           = 1.6*10**-19;          # charge of electron\n\n# Calculations\n\nlamda       = h*c/(E*q)              # Wavelength of GaAs laser\n\n# Result\nprint 'Wavelength of GaAs laser = %3.1f'%(lamda*10**10),' \u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Wavelength of GaAs laser = 8626.3  \u00c5\n"
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 1, Page No:2.42"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding Energy of the first excited state\n\nimport math;\n\n# Variable Declaration\nh           = 6.625*10**-34;         # planck's constant\nc           = 3*10**8;                # vel. of light in m/s\nlamda       = 5890*10**-10;           # wavelength of light in m\nq           = 1.6*10**-19;            # charge of electron\n\n\n# Calculations\nEg          = (h*c)/lamda;          # energy in joules\nE           = Eg/q                  # energy in eV\n\n# Result\nprint 'Energy of the first excited state  = %3.3f' %E,'eV';\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Energy of the first excited state  = 2.109 eV\n"
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 2, Page No:2.43"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding The ratio between the stimulated emission and apontaneous emission\n\nimport math;\n\n# Variable Declaration\nh           = 6.625*10**-34;          # planck's constant\nc           = 3*10**8;                # vel. of light in m/s\nlamda       = 5890*10**-10;           # wavelength of light in m\nk           = 1.38*10**-23;           # Boltzmann constant\nTc          = 280                     # Temperature in centigrades\n\n# Calculations\nT           = Tc+273;               # temperature in kelvin\nR           = 1/((math.exp((h*c)/(k*T*lamda))) - 1);     # ratio of stimulated emission to spontaneous emission\n\n# Result\nprint 'The ratio between the stimulated emission and apontaneous emission = %3.3e' %R;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The ratio between the stimulated emission and apontaneous emission = 6.264e-20\n"
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 3, Page No:2.43"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding The No. of Photons emitted per minute\n\nimport math;\n\n# Variable Declaration\nh           = 6.625*10**-34;          # planck's constant\nc           = 3*10**8;                # vel. of light in m/s\nlamda       = 6328*10**-10;           # wavelength of He-Ne laser source in m\nq           = 1.6*10**-19;            # charge of electron\nP           = 3*10**-3                # output power of the He-Ne source in watts or J/sec\n\n\n# Calculations\nv           = c/lamda               # frequency of the photon emitted by the laser beam\nE           = h*v;                  # energy of a photon in joules\nPo          = P*60;                 # conversion from J/sec to J/min\nN           = Po/E;                 # No of photons emitted per minute \n\n# Result\nprint 'The No. of Photons emitted per minute = %3.3e' %N, 'photons/minute';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The No. of Photons emitted per minute = 5.731e+17 photons/minute\n"
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 4, Page No:2.44"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding The No. of Photons emitted per hour\n\nimport math;\n\n# Variable Declaration\nh           = 6.625*10**-34;          # planck's constant\nc           = 3*10**8;                # vel. of light in m/s\nlamda       = 9.6*10**-6;             # wavelength of CO2 laser source in m\nq           = 1.6*10**-19;            # charge of electron\nP           = 10*10**3                # output power of the CO2 laser source in watts or J/sec\n\n\n# Calculations\nv           = c/lamda               # frequency of the photon emitted by the laser beam\nE           = h*v;                  # energy of a photon in joules\nPo          = P*60*60;              # conversion fro J/sec to J/hour\nN           = Po/E;                 # No of photons emitted per hour \n\n# Result\nprint 'The No. of Photons emitted per hour = %3.3e'%N,' photons/hour';\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The No. of Photons emitted per hour = 1.739e+27  photons/hour\n"
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 5, Page No:2.45"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nh           = 6.625*10**-34;          # planck's constant\nc           = 3*10**8;                # vel. of light in m/s\nlamda       = 10*10**-2;              # wavelength for microwave region in m\nT           = 300                     # Temperature in Kelvin\nKb          = 1.38*10**-23            # Boltzmann constant\n\n# Calculations\n#  let R    = Rsp/Rst \nR           = math.exp((h*c)/(lamda*Kb*T)) - 1 ;     # ratio of spontaneous to stimulated emission\nif R<1:\n    print 'Since the spontaneous emission is lesser than stimulated emission \\n hence MASER action is possible at thermal equilibrium'",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Since the spontaneous emission is lesser than stimulated emission \n hence MASER action is possible at thermal equilibrium\n"
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 6, Page No:2.45"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nh           = 6.625*10**-34;          # planck's constant\nc           = 3*10**8;                # vel. of light in m/s\nlamda       = 5000*10*8-10;           # wavelength for optical region in m\nT           = 300                     # Temperature in Kelvin\nKb          = 1.38*10**-23            # Boltzmann constant\n\n# Calculations\n# let R    = Rsp/Rst \nR           = math.exp((h*c)/(lamda*Kb*T)) - 1;     # ratio of spontaneous to stimulated emission\nif R<1:\n    print 'Since the spontaneous emission is lesser than stimulated emission \\n hence LASER action is possible at thermal equilibrium'\nelse:\n    print 'Since the spontaneous emission is more predominant than stimulated emission\\nhence LASER action is not possible at optical frequencies under thermal equilibrium';\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Since the spontaneous emission is lesser than stimulated emission \n hence LASER action is possible at thermal equilibrium\n"
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 7, Page No:2.46"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math\n\n# Variable Declaration\nh           = 6.625*10**-34;         # plank's constant\nc           = 3*10**8;               # vel. of light in m/s\nlamda       = 5511.11*10**-10;        # wavelength of green LED light in m\nq           = 1.6*10**-19;            # charge of electron\n\n# Calculations\nEg          = (h*c)/lamda;          # band gap energy in joules\nE           = Eg/q                  # bang gap energy in eV\n\n# Result\nprint ' Eg   = %3.2f' %E,'eV';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " Eg   = 2.25 eV\n"
+      }
+     ],
+     "prompt_number": 38
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Physics_-_I/chapter3.ipynb b/Engineering_Physics_-_I/chapter3.ipynb
new file mode 100755
index 00000000..068dcc95
--- /dev/null
+++ b/Engineering_Physics_-_I/chapter3.ipynb
@@ -0,0 +1,209 @@
+{
+ "metadata": {
+  "name": "chapter 3(s)"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 3: Fibre Optics And Applications"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1, Page No: 3.40"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding Numerical aperture\n\nimport math;\n\n# Variable Declaration\nn1      = 1.6;      # Refractive index of core\nn2      = 1.5;      # Refractive index of cladding\n\n# Calculations\nNA      = math.sqrt(n1**2 - n2**2);        # Numerical Aperture of optical fiber\n\n# Result\nprint 'Numerical Aperture of the optical fiber = %3.4f' %NA;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Numerical Aperture of the optical fiber = 0.5568\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2, Page No: 3.40"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding Numerical aperture and acceptance Angle\n\nimport math;\n\n# Variable declaration\nn1      = 1.55;      # Refractive index of core\nn2      = 1.5;       # Refractive index of cladding\n\n# Calculations\nNA      = math.sqrt(n1**2 - n2**2);        # Numerical Aperture of optical fiber\nim      = math.asin(NA);                   # Acceptance angle\nim_d    = im*180/math.pi                   # radian to degree conversion\n\n# Result\nprint 'Numerical Aperture of the optical fiber = %3.4f'%NA,' Acceptance angle = %3.2f' %im_d,'degrees ';\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Numerical Aperture of the optical fiber = 0.3905  Acceptance angle = 22.99 degrees \n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": " Example 3, Page No: 3.41"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding Refractive index of cladding\n\nimport math;\n\n# Variable declaration\nNA      = 0.26;      # Numerical aperture \nn1      = 1.5 ;      # Refractive index of core\nd       = 100*10^-6; # diameter of the core in m\n\n# Calculations\nn2      = math.sqrt( n1**2 - NA**2);       # Refractive index of cladding\n\n# Result\nprint 'Refractive index of cladding = %3.4f' %n2;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Refractive index of cladding = 1.4773\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4, Page No: 3.41"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding Numerical aperture\n\nimport math;\n\n# variable Declaration\nn1      = 1.54;      # Refractive index of core\nn2      = 1.5;      # Refractive index of cladding\n\n# Calculations\nNA      = math.sqrt(n1**2 - n2**2);        # Numerical Aperture of optical fiber\n\n# Result\nprint 'Numerical Aperture of the optical fiber = %3.4f' %NA",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Numerical Aperture of the optical fiber = 0.3487\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 1, Page No: 3.42"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding Refractive index, Acceptance angle, Maximum number of modes that fibre allows\n\nimport math;\n\n# Variable Declaration\nn1      = 1.5;       # Refractive index of core\nNA      = float(0.26);      # Numerical aperture \nd       = 100*10**-6  # core diameter\nlamda   = float(10**-6);     # wavelength in m\n\n# Calculations\nn2      = math.sqrt( n1**2 - NA**2);       # Refractive index of cladding\nim      = math.asin(NA);                   # Acceptance angle\nim_d    = im*180/math.pi                   # radian to degree conversion\nN       = 4.9*(d*NA/lamda)**2;              # maximum no of modes\n\n# Result\nprint ' Refractive index of cladding n2 = %3.4f' %n2,'\\n Acceptance angle = %3.2f' %im_d, 'degrees','\\n Maximum number of modes that fibre allows = %d '%N;\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " Refractive index of cladding n2 = 1.4773 \n Acceptance angle = 15.07 degrees \n Maximum number of modes that fibre allows = 3312 \n"
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 2, Page No: 3.43"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": " # Finding Numerical aperture and Critical angle\n\nimport math;\n\n# Varible Declaration\ndelta       = 0.02;         # relative refractive index\nn1          = 1.48;         # refractive index of core\n\n# Calculations\nNA          = n1*(2*delta)**0.5;           # Numerical aperture\nn2          = math.sqrt( n1**2 - NA**2);   # Refractive index of cladding\ncri_ang     = math.asin(n2/n1);            # critical angle\ncri_ang_d   = cri_ang*180/math.pi;         # critical angle in degrees\n\n# Result\nprint ' Numerical Aperture = %3.3f'%NA,'\\n The Critical angle = %3.2f' %cri_ang_d,' degrees';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " Numerical Aperture = 0.296 \n The Critical angle = 78.46  degrees\n"
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 3, Page No: 3.43"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding Refractive Index\n\nimport math\n\n# Variable declaration\ndelta       = 0.015;        # relative refractive index\nNA          = 0.27;         # Numerical aperture\n\n# Calculations\n# we know that NA = n1*sqrt(2*\u0394)\nn1          = NA/math.sqrt(2*delta)          # refractive index of core\nn2          = math.sqrt( n1**2 - NA**2);     #  Refractive index of cladding\n\n# Result\nprint ' Refractive index of the core = %3.3f' %n1,'\\n Refractive index of the cladding = %3.3f\\n' %n2;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " Refractive index of the core = 1.559 \n Refractive index of the cladding = 1.535\n\n"
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 4, Page No: 3.44"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# No. of total modes propagating in a multimode step index fibre\n\nimport math;\n\n# variable Declaration\nNA      = 0.25;        # Numerical aperture \nd       = 60*10**-6       # core diameter\nlamda   = 2.7*10**-6;     # wavelength in m\n\n# calculations\nN       = 4.9*(d*NA/lamda)**2;       # no of modes for step index fibre\n\n# Result\nprint 'No. of total modes propagating in a multimode step index fibre = %d' %N;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "No. of total modes propagating in a multimode step index fibre = 151\n"
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 5, Page No: 3.44"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Finding No. of total modes propagating in the fibre\nimport math;\n\n# Variable Declaration\nNA      = 0.25;          # Numerical aperture \nd       = 6*10**-6        # core diameter\nlamda   = 1.5*10**-6;     # wavelength of laser source\nn1      = 1.47;          # refractive index of core\nn2      = 1.43           # refractive index of cladding\n\n# calculations\nNA      = math.sqrt( n1**2 - n2**2);       # Numerical Aperture\nN       = 4.9*(d*NA/lamda)**2;      # no of modes for step index fibre\n\n# Result\nprint 'No. of total modes propagating in the fibre = %d' %N;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "No. of total modes propagating in the fibre = 9\n"
+      }
+     ],
+     "prompt_number": 30
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Physics_-_I/chapter4.ipynb b/Engineering_Physics_-_I/chapter4.ipynb
new file mode 100755
index 00000000..995bbd50
--- /dev/null
+++ b/Engineering_Physics_-_I/chapter4.ipynb
@@ -0,0 +1,532 @@
+{
+ "metadata": {
+  "name": "chapter4s"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 4: Quantum Physics"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1, Page No: 4.52"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "\nimport math;\n\n# Variable Declaration\nlamda   = 3*10**-10;         # wavelength of incident photons in m\ntheta   = 60;                # viewing angle in degrees\nh       = 6.625*10**-34       # plancks constant\nmo      = 9.11*10**-31        # mass in Kg\nc       = 3*10**8;            # vel. of light \n\n# Calculatioms\n# from Compton theory ,Compton shift is given by\n# lamda' - lamda = (h/(mo*c))*(1-cos\u03b8)\n\ntheta_r = theta*math.pi/180;     # degree to radian conversion\nlamda1  = lamda+( (h/(mo*c))*(1-math.cos(theta_r)))  # wavelength of scattered photons\n\n# Result\nprint 'Wavelength of Scattered photons = %3.4f'%(lamda1*10**10),'\u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Wavelength of Scattered photons = 3.0121 \u00c5\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2, Page No:4.52"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "\nimport math;\n\n# Variable declaration\ntheta   = 135;               #  angle in degrees\nh       = 6.625*10**-34       # plancks constant\nmo      = 9.1*10**-31         # mass in Kg\nc       = 3*10**8;            # vel. of light in m/s\n\n# Calculatioms\n# from Compton theory ,Compton shift is given by\n# lamda' - lamda = (h/(mo*c))*(1-cos\u03b8)\ntheta_r = theta*math.pi/180;     # degree to radian conversion\nc_lamda = ( (h/(mo*c))*(1-math.cos(theta_r))) # Change in wavelength in m\n\n# Result\nprint 'Change in Wavelength  = %3.5f' %(c_lamda*10**10),' \u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Change in Wavelength  = 0.04143  \u00c5\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3, Page No:4.53"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "\nimport math;\n\n# Variable Declaration\nlamda   = 0.1*10**-9;         # wavelength of X-rays in m\ntheta   = 90;                # angle with incident beam in degrees\nh       = 6.625*10**-34       # plancks constant\nmo      = 9.11*10**-31        # mass in Kg\nc       = 3*10**8;            # vel. of light \n\n# Calculatioms\n# from Compton theory ,Compton shift is given by\n# lamda' - lamda = (h/(mo*c))*(1-cos\u03b8)\ntheta_r = theta*math.pi/180;     # degree to radian conversion\nlamda1  = lamda+( (h/(mo*c))*(1-math.cos(theta_r)))  #wavelength of scattered beam\n\n# Result\nprint 'Wavelength of Scattered beam = %3.4f' %(lamda1*10**10),' \u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Wavelength of Scattered beam = 1.0242  \u00c5\n"
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4, Page No:4.53"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "\nimport math;\n\n# Variable Declaration\nh       = 6.625*10**-34         # plancks constant\nm       = 9.11*10**-31          #  mass of electron in Kg\ne       = 1.6*10**-19           # charge of electron\nV       = 150;                  # potential difference in volts\n\n# Calculations\n\nlamda   = h/(math.sqrt(2*m*e*V))  # de Broglie wavelength\n\n#Result\nprint 'The de-Broglie wavelength = %d' %(lamda*10**10), '\u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The de-Broglie wavelength = 1 \u00c5\n"
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 5, Page No:4.54"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "\nimport math;\n\n# Variable Declaration\nh       = 6.625*10**-34        # plancks constant\nm       = 9.11*10**-31         # mass of electron in Kg\ne       = 1.6*10**-19          # charge of electron\nV       = 5000;                # potential in volts\n\n# Calculations\n\nlamda   = h/(math.sqrt(2*m*e*V))     #de Broglie wavelength\n\n# Result\nprint 'The de-Broglie wavelength of electron = %3.5f' %(lamda*10**10),' \u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The de-Broglie wavelength of electron = 0.17353  \u00c5\n"
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6, Page No:4.55"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "\nimport math;\n\n# Variable Declaration\nE       = 100                 # Energy of electron in eV\nh       = 6.625*10**-34       # plancks constant\nm       = 9.11*10**-31        # mass of electron in Kg\ne       = 1.6*10**-19         # Charge of electron in Columbs\n\n# Calculations\n\nE1      = E*e                    # Energy conversion from eV to Joule\nlamda   = h/(math.sqrt(2*m*E1))  # de Broglie wavelength\n\n# Result\nprint 'The de-Broglie wavelength  = %3.3f' %(lamda*10**10),' \u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The de-Broglie wavelength  = 1.227  \u00c5\n"
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7, Page No:4.55"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "\nimport math;\n\n# Variable Declaration\nm       = 1.675*10**-27;     # Mass of proton in kg\nc       = 3*10**8;           # velocity of light in m/s\nh       = 6.625*10**-34      # plancks constant\n\n# Calculations\n\nvp      = c/20;             # velocity of proton in m/s\nlamda   = h/(m*vp)          # de-Broglie wavelength in m\n\n# Result\nprint 'de-Broglie wavelength = %e'%(lamda),'lamda';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "de-Broglie wavelength = 2.636816e-14 lamda\n"
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8, Page No:4.56"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "\nimport math;\n\n# Variable declaration\nE       = 10000               # Energy of neutron in eV\nh       = 6.625*10**-34       # plancks constant\nm       = 1.675*10**-27       # mass of neutron in Kg\ne       = 1.6*10**-19   \n\n# Calculations\n\nE1      = E*e               # Energy conversion from eV to Joule\nlamda   = h/(math.sqrt(2*m*E1))  # de Broglie wavelength\n\n# Result\nprint 'The de-Broglie wavelength of neutron = %3.3e' %lamda,' m';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The de-Broglie wavelength of neutron = 2.862e-13  m\n"
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 10, Page No:4.58"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable decalaration\nl       = 0.1*10**-9;        # side of cubical box\nh       = 6.625*10**-34      # plancks constant in Jsec\nm       = 9.11*10**-31       # mass of electron in Kg\nKb      = 1.38*10**-23       # Boltzmann constant \n\n# Calculations\n# for cubical box the energy eigen value is Enx ny nz = (h^2/(8*m*l^2))*(nx^2 + ny^2 +nz^2)\n# For the next energy level to the lowest energy level nx = 1 , ny = 1 and nz = 2\nnx      = 1\nny      = 1\nnz      = 2\nE112    = (h**2/(8*m*l**2))*( nx**2 + ny**2 + nz**2);\n\n# We know the average energy of molecules of aperfect gas = (3/2)*(Kb*T)\nT       = (2*E112)/(3*Kb);      # Temperature in kelvin\n\n# Result\nprint 'E112 = %3.4e' %E112,'Joules'',\\n','Temperature of the molecules T = %3.4e' %T, 'K';\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "E112 = 3.6134e-17 Joules,\nTemperature of the molecules T = 1.7456e+06 K\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 11, Page No:4.59"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable declaration\nl       = 4*10**-9;           # width of infinitely deep potential\nh       = 6.625*10**-34       # plancks constant in Jsec\nm       = 9.11*10**-31        # mass of electron in Kg\nn       = 1;                  # minimum energy\ne       = 1.6*10**-19         # charge of electron in columbs\n\n# Calculations\nE       = (h**2 * n**2)/(8*m*l**2)     # Energy of electron in an infinitely deep potential well \nE1      = E/e                          #energy conversion from joules to eV\n\n# Result\nprint 'Minimum energy of an electron = %3.4f' %E1,' eV';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Minimum energy of an electron = 0.0235  eV\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12, Page No:4.61"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nl       = 0.1*10**-9;         # length of one dimensional box \nh       = 6.625*10**-34       # plancks constant in Jsec\nm       = 9.11*10**-31        # mass of electron in Kg\nn       = 1;                  # for ground state\nn5      = 6;                  # n value for fifth excited state\ne       = 1.6*10**-19         # charge of electron in columbs\n\n# Calculations\nEg      = (h**2 * n**2)/(8*m*l**2 * e )   # Energy in ground state in eV \nEe      = (h**2 * n5**2)/(8*m*l**2 * e)   # Energy in excited state  in eV\nE       = Ee - Eg;                       # energy req to excite electrons from ground state to fifth excited state\n\n# Result\nprint 'Energy required to excite an electron from ground state to fifth excited state = %3.2f' %E, 'eV';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Energy required to excite an electron from ground state to fifth excited state = 1317.38 eV\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 13, Page No:4.62"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable decalration\nl       = 0.1*10**-9;          # length of one dimensional box \nh       = 6.625*10**-34        # plancks constant in Jsec\nm       = 9.11*10**-31         # mass of electron in Kg\nn       = 1;                   # for ground state\ne       = 1.6*10**-19          # charge of electron in columbs\n\n# Calculations\nE       = (h**2 * n**2)/(8*m*l**2 *e )   # Energy of electron in eV \n\n# Result\nprint 'Energy of an electron = %3.3f' %E,' eV';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Energy of an electron = 37.639  eV\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 14, Page No:4.63"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable declaration\nl       = 0.5*10**-9;          # width of one dimensional box in m \nh       = 6.625*10**-34        # plancks constant in Jsec\nm       = 9.11*10**-31         # mass of electron in Kg\nn       = 1;                   # for ground state\ne       = 1.6*10**-19          # charge of electron in columbs\n\n# Calculations\nE       = (h**2 * n**2)/(8*m*l**2 *e )   # Energy of electron in eV \n\n# Result\nprint 'Least Energy of an electron = %3.4f' %E,' eV';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Least Energy of an electron = 1.5056  eV\n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 1, Page No:4.64"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# variable declaration\nh       = 6.625*10**-34      # plancks constant\nc       = 3*10**8;           # vel. of light\nlamda   = 5893*10**-10;      # wavelength in m\nP       = 100                # power of sodium vapour lamp\n\n# Calculations\nE       = (h*c)/lamda;       # Energy in joules\nN       =  P/E               # Number of photons emitted\n\n# Result\nprint 'Number of Photons emitted = %3.4e' %N,' per second';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Number of Photons emitted = 2.9650e+20  per second\n"
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 2, Page No:4.64"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable declaration\nlamda1  = 0.022*10**-10;      # wavelength of scatterd X-rays in m\ntheta   = 45;                 # scatterring angle in degrees\nh       = 6.625*10**-34       # plancks constant\nmo      = 9.11*10**-31        # mass in Kg\nc       = 3*10**8;            # vel. of light \n\n# Calculatioms\n# from Compton theory ,Compton shift is given by\n# lamda' - lamda = (h/(mo*c))*(1-cos\u03b8)\n\ntheta_r = theta*math.pi/180;     # degree to radian conversion\nlamda  = lamda1-( (h/(mo*c))*(1-math.cos(theta_r))) # incident Wavelength\n\n# Result\nprint 'Wavelength of incident beam = %3.4f' %(lamda*10**10),' \u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Wavelength of incident beam = 0.0149  \u00c5\n"
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 3 , Page No:4.65"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nEi      = 1.02*10**6          # photon energy in eV\ntheta   = 90;                 # scattered angle in degrees\nh       = 6.625*10**-34       # plancks constant\nmo      = 9.1*10**-31         # mass of electron in Kg\ne       = 1.6*10**-19         # charge of electron\nc       = 3*10**8;            # vel. of light in m/s\n\n# Calculations\n# from Compton theory ,Compton shift is given by\n# lamda' - lamda = (h/(mo*c))*(1-cos\u03b8)\ntheta_r = theta*math.pi/180;     # degree to radian conversion\nc_lamda = ( (h/(mo*c))*(1-math.cos(theta_r)))  #Change in wavelength in m\ndv      = c/c_lamda;         # change in frequency of the scattered photon\ndE      = (h*dv)/e           # change in energy of scattered photon in eV\n# This change in energy is transferred as the KE of the recoil electron\nEr      = dE;                # Energy of recoil electron\nEs      = Ei - Er            # Energy of scattered photon\n\n\n# Result\nprint 'Energy of the recoil electron = %3.4f' %(Er*10**-6),' MeV','\\n','Energy of the Scattered photon = %3.4f' %(Es*10**-6),'MeV';\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Energy of the recoil electron = 0.5119  MeV \nEnergy of the Scattered photon = 0.5081 MeV\n"
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 4, Page No:4.65"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nlamda   = 0.124*10**-10;     # wavelength of X-rays in  m\ntheta   = 180;               # Scattering angle in degrees\nh       = 6.625*10**-34      # plancks constant\nmo      = 9.11*10**-31       # mass in Kg\nc       = 3*10**8;           # vel. of light \n\n# Calculatioms\n# from Compton theory ,Compton shift is given by\n# lamda' - lamda = (h/(mo*c))*(1-cos\u03b8)\n\ntheta_r = theta*math.pi/180;     # degree to radian conversion\nlamda1  = lamda+( (h/(mo*c))*(1-math.cos(theta_r))) # wavelength of scattered X-rays\n\n# Result\nprint 'Wavelength of Scattered X-rays = %3.4f' %(lamda1*10**10),' \u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Wavelength of Scattered X-rays = 0.1725  \u00c5\n"
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 5, Page No:4.65"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nh       = 6.625*10**-34        # plancks constant\nm       = 9.11*10**-31         # mass of electron in Kg\ne       = 1.6*10**-19          # charge of electron\nV       = 2000;                # potential in volts\n\n# Calculations\n\nlamda   = h/(math.sqrt(2*m*e*V))  # de Broglie wavelength\n\n# Result\nprint 'The de-Broglie wavelength of electron = %3.4f' %(lamda*10**10),' \u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The de-Broglie wavelength of electron = 0.2744  \u00c5\n"
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 6, Page No:4.66"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# variable Declaration \nh       = 6.625*10**-34       # plancks constant\nm       = 1.678*10**-27       # mass of proton in Kg\ne       = 1.6*10**-19         # charge of electron\nKb      = 1.38*10**-23;       # boltzmann constant\nT       = 300                 # Temperature in kelvin\n\n#Calculations\nlamda   = h/(math.sqrt(3*m*Kb*T))  #de Broglie wavelength\n\n#Result\nprint 'The de-Broglie wavelength = %3.4f' %(lamda*10**10),' \u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The de-Broglie wavelength = 1.4512  \u00c5\n"
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 7, Page No:4.66"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nh       = 6.625*10**-34         # plancks constant\nm       = 9.11*10**-31          # mass of electron in Kg\nlamda   = 3*10**-2;             # wavelength of electron wave\ne       = 1.6*10**-19;          # charge of electron\n\n# Calculations\nE       = (h**2)/(2*m*lamda**2);   # Energy in Joules\nE1      = E/e;\n\n# Result\nprint 'Energy of the electron E = %3.4e' %E1,'eV';\nprint 'Note: Calculation mistake in textbook'",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Energy of the electron E = 1.6729e-15 eV\nNote: Calculation mistake in textbook\n"
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 8, Page No:4.67"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable declaration\nh       = 6.625*10**-34       # plancks constant\nm       = 9.11*10**-31        # mass of electron in Kg\nc       = 3*10**8;            # velocity of light in m/s\n\n# Calculations\nve      = 0.7071*c           # velocity of electron\nlamda   = h/(m*ve*math.sqrt(1-(ve/c)**2))    #de Broglie wavelength\n\n# we know Compton wavelength  ,lamda' - lamda = (h/(mo*c))*(1-cos\u03b8)\n# maximum shift \u03b8 = 180\ntheta    = 180\ntheta1   = theta*math.pi/180;\nd_lamda  = (h/(m*c))*(1-math.cos(theta1))\nprint 'de Broglie wavelength = %e' %lamda,' m';\nprint 'compton wavelength    = %e' %d_lamda,'m';\nprint 'The de-Broglie wacelength is equal to the compton wavelength';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "de Broglie wavelength = 4.848152e-12  m\ncompton wavelength    = 4.848152e-12 m\nThe de-Broglie wacelength is equal to the compton wavelength\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 9, Page no:4.68"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nl       = 10**-10;            # side of one dimensional box \nh       = 6.625*10**-34       # plancks constant in Jsec\nm       = 9.11*10**-31        # mass of electron in Kg\nn1      = 1;                  # for 1st eigen value\nn2      = 2;                  # for 2nd eigen value\nn3      = 3;                  # for 3rd eigen value\nn4      = 4;                  # for 4th eigen value\ne       = 1.6*10**-19         # charge of electron in columbs\n\n# Calculations\nE1      = (h**2 * n1**2)/(8*m*l**2 *e )   #first Eigen value\nE2      = (h**2 * n2**2)/(8*m*l**2 *e )   # second Eigen value\nE3      = (h**2 * n3**2)/(8*m*l**2 *e )     # third Eigen value\nE4      = (h**2 * n4**2)/(8*m*l**2 *e )    # fourth Eigen value\n \n# Result\nprint '1st Eigen value  = %3.1f' %E1,'eV';\nprint '2nd Eigen value  = %3.1f' %E2,'eV';\nprint '3rd Eigen value  = %3.1f' %E3,'eV';\nprint '4th Eigen value  = %3.1f' %E4,'eV';\n\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "1st Eigen value  = 37.6 eV\n2nd Eigen value  = 150.6 eV\n3rd Eigen value  = 338.8 eV\n4th Eigen value  = 602.2 eV\n"
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 10 , Page No:4.68"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nl       = 10**-10   ;         # length of one dimensional box in m \nh       = 6.625*10**-34       # plancks constant in Jsec\nm       = 9.11*10**-31        # mass of electron in Kg\nn       = 1;                  # for ground state\ne       = 1.6*10**-19         # charge of electron in columbs\n\n# Calculations\nE       = 2*(h**2 * n**2)/(8*m*l**2 *e )   #Energy of system having two electrons\n\n# Result\nprint 'Energy of the system having two electrons = %3.4f' %E,' eV';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Energy of the system having two electrons = 75.2789  eV\n"
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 11 , Page No:4.69"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nb       = 40;       # angle subtended by final images at eye in degrees\na       = 10        # angle subtended by the object at the eye kept at near point in degrees\n\n# Calculations\nb_r     = b*math.pi/180;    # degree to radian conversion\na_r     = a*math.pi/180;    # degree to radian conversion\nM       = math.tan(b_r)/math.tan(a_r);    # magnifying power\n\n#Result\nprint 'Magnifying power = %3.3f' %M;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Magnifying power = 4.759\n"
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Physics_-_I/chapter5.ipynb b/Engineering_Physics_-_I/chapter5.ipynb
new file mode 100755
index 00000000..78a1015e
--- /dev/null
+++ b/Engineering_Physics_-_I/chapter5.ipynb
@@ -0,0 +1,700 @@
+{
+ "metadata": {
+  "name": "Chapter5(s)"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 5: Crystal Physics"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1, Page no:5.72"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\n\n#Copper has FCC structure\nr       = 1.273;            # Atomic radius in angstrom\nN       = 6.023*10**26;     # Avagadros number in atoms/kilomole\nA       = 63.5;             # Atomic weight of copper in grams\nn       = 4;                # No. of atoms per unit cell for FCC\n\n# Calculations\nr1      = r*10**-10;           # Radius conversion from angstrom to m\na       = (4*r1)/math.sqrt(2); # lattice parameter for FCC\np       = (n*A)/(N*a**3);      # Density of copper\n\n# Result\nprint 'Lattice Constant a = %3.1e' %a,' m','\\n', 'Density of copper = %3.1f' %p,' kg/m^3';\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Lattice Constant a = 3.6e-10  m \nDensity of copper = 9034.4  kg/m^3\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2, Page no:5.73"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\n# given intercepts 3,4 and \u221e, the recipocals of intercepts is\n# (1/3):(1/4):(1/\u221e)\n# LCM = 12\n# multiplying by LCM we get miller indices\n# miller indices of a plane are the smallest integers of the reciprocals of its intercerpts\n# therefore miller indices(h k l) is (4 3 0);\n\nh   = 4;       # miller indice\nk   = 3;       # miller indice\nl   = 0;       # miller indice\na   = 2;       # primitive vector of lattice in angstrom\n\n#Calculations\n\ndhkl    = a/math.sqrt((h**2)+(k**2)+(l**2));   #interplanar distance\n\n# Result\nprint 'Miller indices = (4 3 0)';\nprint 'The interplanar distance d = %3.1f' %dhkl,' \u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Miller indices = (4 3 0)\nThe interplanar distance d = 0.4  \u00c5\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3, Page no:5.74"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\n# \u03b1-Iron solidifies to BCC structure\n\nr       = 1.273;             # Atomic radius in angstrom\nN       = 6.023*10**26;      # Avagadros number in atoms/kilomole\nA       = 55.85;             # Atomic weight of \u03b1-Iron in kilograms\nn       = 2;                 # No. of atoms per unit cell for BCC\np       = 7860;              # density in kg/m^-3\n\n#Calculations\n\n# p    = (n*A)/(N*a^3);    density\n\na       = ((n*A)/(N*p))**(0.333);  # lattice constant\na1      = a*10**10;              # m to angstrom conversion\nr       = (a1*math.sqrt(3))/4    #  atomic radius for BCC\n\n#Output\nprint 'The Radius of the atom = %3.5f' %r,' \u00c5'\nprint 'Note : atomic wt taken as 55.58*10^-3 instead of 55.85 in calculation'",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Radius of the atom = 1.26955  \u00c5\nNote : atomic wt taken as 55.58*10^-3 instead of 55.85 in calculation\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4, Page no:5.75"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nlamda   = 1.5418;       # wavelength in \u00c5\nh       = 1;            # miller indice\nk       = 1;            # miller indice\nl       = 1;            # miller indice\nn       = 1;            # given first order\ntheta   = 30;           # diffraction angle in degrees\n\n# Calculations\ntheta1   = theta*math.pi/180;   # degree to radian conversion\n# d     = (n*lamda)/(2*sin\u03b8);    by Braggs law               ------------- 1\n# d     = a/sqrt((h^2)+(k^2)+(l^2));  interplanar distance    ------------ 2\n# equating 1 and 2\n\na       = (n*lamda*math.sqrt((h**2)+(k**2)+(l**2))/(2*math.sin(theta1)))\n\n# Result\nprint 'Interatomic spacing a = %f \u00c5' %a;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Interatomic spacing a = 2.670476 \u00c5\n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 5, page no:5.76"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# variable Declaration\nh1       = 1;            # miller indice\nk1       = 1;            # miller indice\nl1       = 1;            # miller indice\nh0       = 0;            # miller indice\nk0       = 0;            # miller indice\nl0       = 0;            # miller indice\n\n# Calculations\n# dhkl    = a/sqrt((h^2)+(k^2)+(l^2)); // interplanar distance\n# assume a = 1(constant) for easier calculation in scilab\n\na        = 1;\nd100     = a/math.sqrt((h1**2)+(k0**2)+(l0**2)); # interplanar distance\nd110     = a/math.sqrt((h1**2)+(k1**2)+(l0**2)); # interplanar distance\nd111     = a/math.sqrt((h1**2)+(k1**2)+(l1**2)); # interplanar distance\n\n# Result\nprint 'd100 : d110 : d111 = ','%d ' %d100,':','%3.2f' %d110,':', '%3.2f' %d111;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "d100 : d110 : d111 =  1  : 0.71 : 0.58\n"
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6, page no:5.76"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\n# Aluminium is FCC\n\na       = 0.405*10**-9;          # lattice constant of aluminium\nt       = 0.005*10**-2;          # thickness of aluminium foil in m\ns       = 25*10**-2;             # side of square in m\n\n# Calculations\nVUC     = a**3;                  # volume of unit cell\nVal     = (s**2)*t               # volume of aluminium foil (area*thickness)\nN       = Val/VUC                # Number if unit cells\n\n# Result\nprint 'Number of unit cells = %3.3e' %N",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Number of unit cells = 4.704e+22\n"
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7, page no:5.77"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# variable declaration\n# metallic iron changes from  BCC to FCC form at 910 degress\nrb      = 0.1258*10**-9;     # atomic radius of BCC iron atom\nrf      = 0.1292*10**-9;     # atomic radius of FCC iron atom\n\n# Calculations\n\nab      = (4*rb)/(math.sqrt(3));      # lattice constant for BCC\nVbcc    = (ab**3)/2;                  # volume occupied by one BCC atom\naf      = (4*rf)/(math.sqrt(2))       # lattice constant for FCC\nVfcc    = (af**3)/4;                  # volume occupied by one FCC atom\ndv      = ((Vbcc-Vfcc)/Vbcc)*100      # percentage change in volume\n\n# Result\nprint 'During the structural change the percentage change in volume = %3.4f' %dv;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "During the structural change the percentage change in volume = 0.4933\n"
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8, page no:5.78"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math\n#variable declaration\n#Copper Crystallines in  FCC structure\n\np       = 8960;             # Density of copper in kg/m^3\nN       = 6.023*10**26;      # Avagadros number in atoms/kilomole\nA       = 63.5;             # Atomic weight of copper in kg/mol\nn       = 4;                # No. of atoms per unit cell for FCC\n\n# Calculations\n\na       = ((n*A)/(N*p))**(0.333);\n\n# Result\n\nprint 'Lattice Constant a = %3.4f' %(a*10**10),' \u00c5';\nprint 'atomic wt of copper is taken as 63.5*10^-3 instead of 63.5 in textbook';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Lattice Constant a = 3.6899  \u00c5\natomic wt of copper is taken as 63.5*10^-3 instead of 63.5 in textbook\n"
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9, page no:5.79"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable declaration\n# (100) planes in rock crystal\nh       = 1;            # miller indice\nk       = 0;            # miller indice\nl       = 0;            # miller indice\na       = 2.814         # lattice constant in \u00c5\n\n# Calculations\ndhkl    = a/math.sqrt((h**2)+(k**2)+(l**2)); # interplanar distance\n\n#Result\nprint 'd-spacing for (100) plane in rock salt = %3.3f' %dhkl,' \u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "d-spacing for (100) plane in rock salt = 2.814  \u00c5\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 10, page no:5.79"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\n# FCC structured crystal\n\np       = 6250;             # Density of crystal in kg/m^3\nN       = 6.023*10**26;      # Avagadros number in atoms/kilomole\nA       = 60.2;             # molecular weight\nn       = 4;                # No. of atoms per unit cell for FCC\n\n# Calculations\n\na       = ((n*A)/(N*p))**(0.333);\n\n# Result\n\nprint 'Lattice Constant a = %3.3e' %a, 'm';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Lattice Constant a = 4.087e-10 m\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 11, page no:5.80"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\n# (321) plane in simple cubic lattice\nh       = 3;            # miller indice\nk       = 2;            # miller indice\nl       = 1;            # miller indice\na       = 4.12          # inter atomic space \u00c5\n\n# Calculations\ndhkl    = a/math.sqrt((h**2)+(k**2)+(l**2)); #interplanar distance\n\n# Result\nprint 'd = %3.2f' %dhkl,' \u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "d = 1.10  \u00c5\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12, page no:5.81"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\n# BCC structured crystal\n\np       = 7860;             # Density of iron in kg/m^3\nN       = 6.023*10**26;     # Avagadros number in atoms/kilomole\nA       = 55.85;            # Atomic weight\nn       = 2;                # No. of atoms per unit cell for BCC\n\n# Calculations\n\na       = ((n*A)/(N*p))**(0.333); #lattice constant\n\n# Result\n\nprint 'Lattice Constant of Fe = %3.3f' %(a*10**10),' \u00c5 ';\nprint 'Note: density of iron is taken as 7.86 instead of 7860 in calculation'",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Lattice Constant of Fe = 2.932  \u00c5 \nNote: density of iron is taken as 7.86 instead of 7860 in calculation\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 14, page no:5.82"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# variable Declaration\nr       = 0.123*10**-10;         # Radius of the atom\n\n# Calculations\na       = (4*r)/math.sqrt(3);        # Lattice constant in m For a BCC structure\nV       = a*a*a;                     # Volume of BCC\n\n# Result\nprint 'Volume of the unit cell = %3.4e' %V,' m^3';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Volume of the unit cell = 2.2920e-32  m^3\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 15, page no:5.82"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable declaration\na   = 0.05;     # unit cell edge of an orthorhombic crystal in nm\nb   = 0.05;     # unit cell edge of an orthorhombic crystal in nm\nc   = 0.03;     # unit cell edge of an orthorhombic crystal in nm\nIa  = 0.025     # intercept on 'a' in nm\nIb  = 0.02      # intercept on 'b' in nm\nIc  = 0.01      # intercept on 'c' in nm\n\n# Calculations\n\nh   = a/Ia;     # miller indice h\nk   = b/Ib;     # miller indice k\nl   = c/Ic      # miller indice l\n\n# Result\nprint 'Miller indices (h k l) =', '%d' %h,'%d' %k, '%d' %l; ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Miller indices (h k l) = 2 2 3\n"
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 16, page no:5.83"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Magnesium has HCP structure\n# for HCF(Hexagonal closed packed structure) consider the relation between 'c' and 'a';\n# c/a = sqrt(8/3) = 1.6329 \n\n# Variable Declaration\nr   = 0.1605*10**-9;     # radius of magnesium atom in m\n\n# Calculations\n\na   = 2*r               # lattice constant of HCP\nc   = a*math.sqrt(float(8)/3);      # relation b/w c and a in HCP\nV   = (3*(3**0.5))*(a*a*c)/2;         #Volume of unit cell in m^3\n\n# Result\nprint 'Volume of the unit cell of magnesium = %g' %V,' m^3';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Volume of the unit cell of magnesium = 1.4033e-28  m^3\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 17, page no:5.84"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math\n\n# Variable Declaration\n#  (101),(221) planes in simple cubic lattice\nh1       = 1;            # miller indice\nk0       = 0;            # miller indice\nl1       = 1;            # miller indice\nh2       = 2;            # miller indice\nk2       = 2;            # miller indice\nl1       = 1;            # miller indice\na        = 4.2           # inter atomic space \u00c5\n\n# Calculations\n\nd101    = a/math.sqrt((h1**2)+(k0**2)+(l1**2)); # interplanar distance\nd221    = a/math.sqrt((h2**2)+(k2**2)+(l1**2)); # interplanar distance\n\n\n# Result\nprint 'd(101) = %3.4f' %d101,' \u00c5','\\n','d(221) = %3.1f' %d221,' \u00c5 ';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "d(101) = 2.9698  \u00c5 \nd(221) = 1.4  \u00c5 \n"
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 1, page no:5.85"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable declaration\n\n# Copper has FCC structure\na   = 3.6;      #lattice parameter of copper in \u00c5\n\n# Calculations\n\nr   = a*math.sqrt(2)/4;      # atomic radius of copper\n\n# Result\nprint 'Atomic Radius of copper = %3.3f' %r,'\u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Atomic Radius of copper = 1.273 \u00c5\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 2, page no:5.85"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# variable Declaration\n\n# Copper has FCC structure\n\nr       = 1.278;            # Atomic radius in angstrom\nN       = 6.023*10**26;     # Avagadros number in atoms/kilomole\nA       = 63.54;            # Atomic weight of copper \nn       = 4;                # No. of atoms per unit cell for FCC\n\n# Calculations\nr1      = r*10**-10;           # Radius conversion from angstrom to m\na       = (4*r1)/math.sqrt(2); # lattice parameter for FCC\np       = (n*A)/(N*a**3);      # Density of copper\n\n# Result\n\nprint ' Density of copper = %3.2f' %p,' kg/m^3';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " Density of copper = 8934.43  kg/m^3\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 3, page no:5.86"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nANa     = 23;               # atomic wt of sodiim\nACl     = 35.45             # atomic wt of chlorine\nN       = 6.023*10**26;     # Avagadros number in atoms/kilomole\nn       = 4                 # No. of atoms per unit cell for FCC\np       = 2180;             # density in kg/m^-3\n\n# Calculations\n\n# p    = (n*A)/(N*a^3);    density\nA       = ANa+ACl;              # atomic wt of NaCl\na       = ((n*A)/(N*p))**(0.33333); # lattice constant\nr       = a/2                   # Distance b/w two adjacent atoms\n\n# Result\nprint 'Distance between two adjacent atoms is r = %3.2e' %r,' m';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Distance between two adjacent atoms is r = 2.81e-10  m\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 4, page no:5.87"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nr       = 1.273;            # Atomic radius in angstrom\nN       = 6.023*10**26;      # Avagadros number in atoms/kilomole\nA       = 55.85      ;      # Atomic weight of Fe \nn       = 2;                # No. of atoms per unit cell for BCC\np       = 7860;             # density in kg/m^-3\n\n# Calculations\n\n#  p    = (n*A)/(N*a^3);    density\n\na       = ((n*A)/(N*p))**(0.33333);      # lattice constant\na1      = a*10**10;                      #  m to angstrom conversion\nr       = (a1*math.sqrt(3))/4                 # atomic radius for BCC\n\n# Result\nprint 'The Radius of the Fe = %3.3f' %r,' \u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Radius of the Fe = 1.242  \u00c5\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 5, page no:5.88"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nN       = 6.023*10**26;      # Avagadros number in atoms/kilomole\nA       = 119;               # Atomic weight of pottasium bromide\nn       = 4;                 # No. of atoms per unit cell for FCC\np       = 2700;              # density in kg/m^-3\n\n# Calculations\n\n# p    = (n*A)/(N*a^3);    density\n\na       = ((n*A)/(N*p))**(0.33333);  # lattice constant\na1      = a*10**10;                   # m to angstrom conversion\n\n# Output\nprint 'Lattice constant = %3.1f' %a1,' \u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Lattice constant = 6.6  \u00c5\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 6, page no:5.88"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\na       = 4.3*10**-10;      # Lattice constant in \u00c5\np       = 960;              # Density of crystal in kg/m^3\nA       = 23;               # Atomic wt\nN       = 6.023*10**26;     # avogadros no in atoms/kilomole\n\n# Calculations\n\nn       = (p*N*(a**3))/A;      # No. of atoms per unit cell\n\n# result\nprint 'No. of atoms per unit cell = %3.0f' %n,' (BCC)';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "No. of atoms per unit cell =   2  (BCC)\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 7, page no:5.89"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration \n\n# given crystal has BCC structure\nr   = 1.2*10**-10;       # atomic radius in m\n\n# Calculations\n\na   = (4*r)/math.sqrt(3);    # lattice constant\nV   = a**3;                  #  volume of cell\n\n# Result\nprint 'Volume of the cell = %3.3e' %V,' m^3';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Volume of the cell = 2.128e-29  m^3\n"
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 8, page no:5.89"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\na   = 4*10**-10;     # lattice constant of the crystal\nh   = 1              # miller indice\nk   = 0              # miller indice\nl   = 0              # miller indice\n\n# Calculations\n\n# in fig consider (100) plane. the no of atoms in plane ABCD\nN   = 4*(float(1)/4);      # Number of atoms\np   = N/(a*a);      # planar atomic density in atoms/m^2\np1  = p*10**-6      # planar atomic density in atoms/mm^2\n\n# Result\nprint 'planar atomic density = %3.2e' %p1,' atoms/mm^2';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "planar atomic density = 6.25e+12  atoms/mm^2\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 9, page no:5.90"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\n# in fig 5(b) the given plane is parallel to X and Z axes.Thus,its numerical intercepts on these axes is infinity\n#The numerical intercept on y axis is 1/2. Thus the numerical intercepts of plane is (\u221e 1/2 \u221e)\nprint ' Miller indices of plane shown in fig 5.(b) = (0 2 0)';\n# in fig 5(c) the given plane is parallel to Z axis.Thus its numerical intercept on z axis is infinity\n# The numerical intercept on x axis is 1 and y axis is 1/2. this numerical intercepts on plane is (1 1/2 \u221e )\nprint ' Miller indices of plane shown in fig 5.(c) = (1 2 0)'\n# in fig 5(d) the given plane is parallel to Z axis.Thus its numerical intercept on z axis is infinity\n# The numerical intercept on x axis is 1/2 and y axis is 1/2. this numerical intercepts on plane is (1/2 1/2 \u221e )\nprint ' Miller indices of plane shown in fig 5.(d) = (2 2 0)'",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " Miller indices of plane shown in fig 5.(b) = (0 2 0)\n Miller indices of plane shown in fig 5.(c) = (1 2 0)\n Miller indices of plane shown in fig 5.(d) = (2 2 0)\n"
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 11, page no:5.91"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\n\n# (311) plane in simple cubic lattice\nh       = 3;            # miller indice\nk       = 1;            # miller indice\nl       = 1;            # miller indice\na       = 2.109*10**-10 # lattice constant in m\n\n# Calculations\ndhkl    = a/math.sqrt((h**2)+(k**2)+(l**2)); # interplanar distance\n\n# Result\nprint 'd = %3.3e' %dhkl,' m';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "d = 6.359e-11  m\n"
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 12, page no:5.92"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nh       = 1;            # miller indice\nk       = 1;            # miller indice\nl       = 0;            # miller indice\nd       = 2.86*10**-10  # interplanar distance in m\n\n# Calculations\na       = d*math.sqrt((h**2)+(k**2)+(l**2));   # interplanar distance\n\n# Result\nprint 'Lattice constant a = %3.3e' %a,' m';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Lattice constant a = 4.045e-10  m\n"
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 13, page no:5.93"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nh1      = 1;\nh0      = 0;\nk0      = 0;\nl0      = 0;\nl1      = 1;\n\n# Calculations\n\n# we know that dhkl = a/sqrt( h^2 + k^2 + l^2)\n# let sqrt( h^2 + k^2 + l^2) = p\np101    = math.sqrt( h1**2 + k0**2 + l1**2);\np100    = math.sqrt( h1**2 + k0**2 + l0**2);\np001    = math.sqrt( h0**2 + k0**2 + l1**2);\n\n# Result\nprint 'd101 : d100 : d001 :: a/%3.4f' %p101,' : ','a/%d' %p100,':',' a/%d ' %p001;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " d101 : d100 : d001 :: a/1.4142  :  a/1 :  a/1 \n"
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 14, page no:5.93"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# if a plane cut intercepts of lengths l1,l2,l3 the on three crystal axes ,then\n# l1 : l2 : l3 = pa : pq :rc\n# where a,b and c are primitive vectors of the unit cell and p,q and r are numbers related to miller indices (hkl) of plane by relation\n# 1/p : 1/q : 1/r = h : k : l\n# since, the crystal is simple cubic a = b = c and given that h = 1, k = 1 and l = 1\n# p : q : r = 1/h : 1/k : 1/l = 1/1 : 1/1 : 1/1 \n# p : q : r = 1 : 1 : 1\n# similarly l1 : l2 : l3 = 1a : 1a : 1a\nprint 'ratio of intercepts on the three axes by (111) plane is l1 : l2 : l3 = 1 : 1 : 1';\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "ratio of intercepts on the three axes by (111) plane is l1 : l2 : l3 = 1 : 1 : 1\n"
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 15, page no:5.94"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\n\nr   = 1.246*10**-10;     # atomic radius in m\nh1  = 1                  # miller indice\nh2  = 2                  # miller indice\nk0  = 0                  # miller indice\nk1  = 1                  # miller indice\nk2  = 2                  # miller indice\nl0  = 0                  # miller indice\nl1  = 1                  # miller indice\n\n# Calculations\na    = (4*r)/math.sqrt(2);    # lattice constant\nd111 = a/math.sqrt((h1**2)+(k1**2)+(l1**2)); # interplanar distance\nd200 = a/math.sqrt((h2**2)+(k0**2)+(l0**2)); # interplanar distance\nd220 = a/math.sqrt((h2**2)+(k2**2)+(l0**2)); # interplanar distance\n\n# Result\nprint 'd111 = %3.3e' %d111,' m','\\n' 'd200 = %3.4e' %d200,' m','\\n''d220 = %3.3e' %d220,' m';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "d111 = 2.035e-10  m \nd200 = 1.7621e-10  m \nd220 = 1.246e-10  m\n"
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 16, page no:5.95"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# variable Declaration\n# the intercept along X-axis be c1 = a\n# the intercept along Y-axis be c2 = b/2 and\n# the intercept along Z-axis be c3 = 3c\n# Therefore, p = c1/a = a/a = 1\n# q = c2/b = (b/2)/b = 1/2\n# r = c3/c = (3c)/c  = 3\n# therefore h = 1/p = 1\n# k = 1/q = 2\n# l = 1/r = 1/3\n# lcm of 1 1 and 3 = 3\nh = 1\nk = 2\nl = float(1)/3\ns = 3 ;   # lcm\nh1= s*h\nk1= s*k\nl1= s*l;\n\n# Result\nprint '(h k l) =', '%d' %h1,' %d' %k1,'%3.0f' %l1;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "(h k l) = 3  6   1\n"
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Addl_Example 17, page no:5.96"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "import math;\n\n# Variable Declaration\nd       = 1.3*10**-10    # interplanar distance\nn       = 1;             # given first order\ntheta   = 23;            # Bragg reflection angle in degrees\n\n# Calculations\ntheta1   = theta*math.pi/180;   # degree to radian conversion\n# d     = (n*lamda)/(2*sin\u03b8);    by Braggs law               ------------- 1\nlamda    = (2*d*math.sin(theta1)/n)\n\n# Result\nprint 'Wavelength of X-ray = %3.4f' %(lamda*10**10),' \u00c5';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Wavelength of X-ray = 1.0159  \u00c5\n"
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
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diff --git a/Engineering_Physics_-_I/screenshots/Young's_modulus.PNG b/Engineering_Physics_-_I/screenshots/Young's_modulus.PNG
new file mode 100755
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diff --git a/Engineering_Physics_-_I/screenshots/numerical_aperture.PNG b/Engineering_Physics_-_I/screenshots/numerical_aperture.PNG
new file mode 100755
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