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+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 36: Interference"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 36.1: Sample_Problem_1.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Given that\n",
+"l = 550*10^-9  //in meter\n",
+"n2 = 1.60\n",
+"n1 = 1.00\n",
+"t = 2.6*10^-6  //in meter\n",
+"\n",
+"//Sample Problem 36-1a\n",
+"printf('**Sample Problem 36-1a**\n')\n",
+"deltaPHI = t/l*(n2 - n1)*360\n",
+"printf('The phase difference is equal to %1.2fdegrees\n', deltaPHI)\n",
+"\n",
+"//Sample Problem 36-1b\n",
+"printf('\n**Sample Problem 36-1b**\n')\n",
+"printf('The interference produced would be constructive')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 36.2: Sample_Problem_2.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Given that\n",
+"l = 546*10^-9  //in meter\n",
+"d = 12*10^-5  //in meter\n",
+"D = 55*10^-2  //in meter\n",
+"\n",
+"//Sample Problem 36-2\n",
+"printf('**Sample Problem 36-2**\n')\n",
+"beeta = l*D/d\n",
+"printf('The difference between two adjacent maxima is %1.2em', beeta)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 36.3: Sample_Problem_3.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"exec('degree_rad.sci', -1)\n",
+"\n",
+"//Given that\n",
+"Eo = 1  //(say)\n",
+"E1 = Eo\n",
+"E2 = Eo\n",
+"E3 = Eo\n",
+"phi1 = dtor(0)\n",
+"phi2 = dtor(60)\n",
+"phi3 = dtor(-30)\n",
+"\n",
+"//Sample Problem 36-3\n",
+"printf('**Sample Problem 36-3**\n')\n",
+"Eh = E1*cos(phi1) + E2*cos(phi2) + E3*cos(phi3)\n",
+"Ev = E1*sin(phi1) + E2*sin(phi2) + E3*sin(phi3)\n",
+"Er = sqrt(Ev^2 + Eh^2)\n",
+"theta = rtod(atan(Ev/Eh))\n",
+"printf('The resultant electric field is E=%1.2f*Eo*sin(w*t + %1.2f)', Er, theta)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 36.4: Sample_Problem_4.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Given that\n",
+"lmin = 400  //in nm\n",
+"lmax = 690  //in nm\n",
+"n2 = 1.33\n",
+"L = 320  //in nm\n",
+"\n",
+"//Sample Problem 36-4\n",
+"printf('**Sample Problem 36-4**\n')\n",
+"flag = 1\n",
+"odd_number = 1\n",
+"while flag == 1\n",
+"    lambda = 4*L*n2/odd_number\n",
+"    if lambda > lmin & lambda < lmax then\n",
+"        flag = 0\n",
+"    end\n",
+"    odd_number = odd_number + 2\n",
+"end\n",
+"printf('The wavelength of the light is %1.2enm', lambda)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 36.5: Sample_Problem_5.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Given that\n",
+"n1 = 1.38\n",
+"n2 = 1.50\n",
+"lambda = 550  //in nm\n",
+"\n",
+"//Sample Problem 36-5\n",
+"printf('**Sample Problem 36-5**\n')\n",
+"Lmin = lambda/4/n1\n",
+"printf('The minimum value of wavelength possible is %1.2fnm', Lmin)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 36.6: Sample_Problem_6.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Given that\n",
+"lambda = 632.8*10^-9  //in meter\n",
+"i = 0  //in rad\n",
+"dFringes = 6\n",
+"bFringes = 5\n",
+"\n",
+"//Sample Problem 36-6\n",
+"printf('**Sample Problem 36-6**\n')\n",
+"//Assume the difference in thickness is 't'\n",
+"//then the path difference will be 2*L\n",
+"//hence\n",
+"deltaL = bFringes/2*lambda\n",
+"printf('The difference in thickness is equal to %fnm', deltaL*10^9)"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}