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+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 1: Interference of Light"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.2_1: find_the_wavelength_of_light_in_the_visible_spectrum.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_2_1,pg 1-11\n",
+"\n",
+"i=45                                 //angle of incidence\n",
+"\n",
+"u=1.2                                        //refractive index of soap film\n",
+"\n",
+"t=4*10^-5                                    //thickness of soap film\n",
+"\n",
+"r=asind(sind(i)/u)                             //by  Snell's law\n",
+"\n",
+"//for dark band   '2*u*t*cos(r) = n*lam'\n",
+"\n",
+"wavelength_1=(2*u*t*cosd(r)/1)*10^8                          //for n=1\n",
+"\n",
+"wavelength_2=(2*u*t*cosd(r)/2)*10^8                          //for n=2\n",
+"\n",
+"wavelength_3=(2*u*t*cosd(r)/3)*10^8                          //for n=3\n",
+"\n",
+"//visible range of wavelengths is 4000 A.  to  7000 A.\n",
+"\n",
+"printf('\n for n=1  wavelength = %.1f A.\n',wavelength_1)\n",
+"\n",
+"printf('\n for n=2  wavelength = %.1f A.\n',wavelength_2)\n",
+"\n",
+"printf('\n for n=3  wavelength = %.2f A.\n',wavelength_3)\n",
+"\n",
+"printf('\n hence, none of the wavelengths from the visible region are absent in reflected light ')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.2_2: find_the_wavelength_of_light_in_the_visible_spectrum.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_2_2,pg 1-12\n",
+"\n",
+"u=1.33                                   //refractive index of soap film\n",
+"\n",
+"t=5*10^-5                                //thickness of soap film\n",
+"\n",
+"//for normal incidence \n",
+"\n",
+"r=0                                      //angle of refraction\n",
+"\n",
+"//for constructive interference        '2*u*t*cos(r) = (2*n-1)*wavelength/2'\n",
+"\n",
+"wavelength_1=(2*u*t*cos(r)*2/(2*1-1))*10^8               //for   n=1\n",
+"\n",
+"wavelength_2=(2*u*t*cos(r)*2/(2*2-1))*10^8               //for   n=2\n",
+"\n",
+"wavelength_3=(2*u*t*cos(r)*2/(2*3-1))*10^8               //for   n=3\n",
+"\n",
+"wavelength_4=(2*u*t*cos(r)*2/(2*4-1))*10^8               //for   n=4\n",
+"\n",
+"//visible range of wavelengths is 4000 A.  to  7000 A.\n",
+"\n",
+"printf('\n for n=1  wavelength = %.1f A.\n',wavelength_1)\n",
+"\n",
+"printf('\n for n=2  wavelength = %.1f A.\n',wavelength_2)\n",
+"\n",
+"printf('\n for n=3  wavelength = %.1f A.\n',wavelength_3)\n",
+"\n",
+"printf('\n for n=4  wavelength = %.1f A.\n',wavelength_4)\n",
+"\n",
+"printf('\n The wavelength will be reflected is wavelength = %.1f A.\n',wavelength_3)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.2_3: find_the_order_of_interference_band.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_2_3,pg 1-12\n",
+"\n",
+"u=4/3                                      //refractive index of soap film\n",
+"\n",
+"t=1.5*10^-4                                //thickness of soap film \n",
+"\n",
+"wavelength=5*10^-5                                //wavelength of light \n",
+"\n",
+"i=45                               //angle of incidece\n",
+"\n",
+"r=asind(sind(i)/u)                           //by Snell's law\n",
+"\n",
+"n=2*u*t*cosd(r)/wavelength                         //for nth dark band\n",
+"\n",
+"printf('\n the order of an interference band is n = %.0f',n)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.2_4: find_the_thickness_of_soap_film.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_2_4,pg 1-13\n",
+"\n",
+"//for constructive interference   2 u t cos(r) = (2 n -1) wavelength/2\n",
+"\n",
+"u=1.33\n",
+"\n",
+"i=45\n",
+"\n",
+"r=asind(sind(i)/u)                            //by Snell's law\n",
+"\n",
+"n=1                                         //for minimum thickness \n",
+"\n",
+"wavelength=5896*10^-8\n",
+"\n",
+"t=(2*n-1)*wavelength/(4*u*cosd(r))\n",
+"\n",
+"printf('\n the minimum thickness of soap film is t = %.7f cm',t)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.2_5: find_the_thickness_of_oil_layer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_2_5,pg 1-14\n",
+"\n",
+"u=1.3                                      //refractive index of liquid\n",
+"  \n",
+"r=0                                        //angle of refraction for normal incidence\n",
+" \n",
+"wavelength_1=7000                          //wavelength of light \n",
+"\n",
+"wavelength_2=5000                          //wavelength of light \n",
+"\n",
+"//for destructive interference    '2*u*t*cos(r) = (2*n-1)*wavelength/2'\n",
+"\n",
+"//'n' order for 'wavelength_1' and 'n+1' order for 'wavelength_2'\n",
+"\n",
+"//as LHS  is same for both the wavelengths,   therefore\n",
+"\n",
+"//(2*n-1)*7000/2 =(2*(n+1)-1)*5000/2\n",
+"\n",
+"n=3                                       //number of orders \n",
+"\n",
+"t=((2*n)-1)*wavelength_1/(4*u*cosd(r))\n",
+"\n",
+"printf('\nThe thickness of oil layer is t = %.2f A.',t)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.2_6: find_the_thickness_of_film.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_2_6,pg 1-15\n",
+"\n",
+"n=8                               \n",
+"\n",
+"wavelength=5890*10^-8                      //wavelength of light \n",
+"\n",
+"u=1.46                               //refractive index of oil\n",
+"\n",
+"i=30                         //angle of incidence\n",
+"\n",
+"r=asind(sind(i)/u)                     //by Snell's law\n",
+"\n",
+"t=n*wavelength/(2*u*cosd(r))\n",
+"\n",
+"printf('\n the thickness of an oil film is t =%.7f cm',t)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.2_7: find_the_minimum_thickness_of_film.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_2_7,pg 1-15\n",
+"\n",
+"u=1.5                                //refractive index of thin film\n",
+"\n",
+"r1=60                         //angle of refraction\n",
+"\n",
+"wavelength=5890*10^-8                      //wavelength of light\n",
+"\n",
+"n=1                                  //for minimum thickness\n",
+"\n",
+"t1=n*wavelength/(2*u*cosd(r1))\n",
+"\n",
+"printf('\n the thickness of an oil film is t =%.7f cm',t1)\n",
+"\n",
+"r2=0                                 //for normal incidence \n",
+"\n",
+"t2=n*wavelength/(2*u*cosd(r2))\n",
+"\n",
+"printf('\n the thickness of an oil film is t =%.7f cm',t2)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.2_8: find_the_refractive_index_of_oil.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_2_8,pg 1-16\n",
+"\n",
+"V=0.2                             //volume of oil\n",
+"\n",
+"A=10^4                             //area\n",
+"\n",
+"t=V/A                              //Thickness of oil film\n",
+"\n",
+"r=0                                //for normal incidence\n",
+"\n",
+"n=1                                //for 1st dark band\n",
+"\n",
+"wavelength=5.5*10^-5                      //wavelength of light \n",
+"\n",
+"u=n*wavelength/(2*t*cosd(r))\n",
+"\n",
+"printf('\nrefractive index of oil is  u = %.3f',u)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.2_9: find_the_wavelength_of_light_in_the_visible_region.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_2_9,pg 1-17\n",
+"\n",
+"u=1.2                                  //refractive index of oil film\n",
+"\n",
+"t=2*10^-7                              //thickness of oil film\n",
+"\n",
+"r=60                           //angle of refraction\n",
+"\n",
+"//for destructive interference    '2*u*t*cos(r) = (2*n-1)*wavelength/2'\n",
+"\n",
+"wavelength_1=(2*u*t*cosd(r)*2/(2*1-1))*10^10               //for   n=1\n",
+"\n",
+"wavelength_2=(2*u*t*cosd(r)*2/(2*2-1))*10^10               //for   n=2\n",
+"\n",
+"wavelength_3=(2*u*t*cosd(r)*2/(2*3-1))*10^10               //for   n=3\n",
+"\n",
+"//visible range of wavelengths is 4000*10^-10 m  to  7000*10^-10 m\n",
+"\n",
+"printf('\n for n=1  wavelength = %.f A.\n',wavelength_1)\n",
+"\n",
+"printf('\n for n=2  wavelength = %.f A.\n',wavelength_2)\n",
+"\n",
+"printf('\n for n=3  wavelength = %.f A.\n',wavelength_3)\n",
+"\n",
+"printf('\n The wavelength will be reflected is wavelength = %.f A.\n',wavelength_1)\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.3_1: find_wavelength_of_monochromatic_light.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_3_1,pg 1-21\n",
+"\n",
+"N=10                                                //no of dark fringes \n",
+"\n",
+"d=1.2                                               //distance between consecutive fringes \n",
+"\n",
+"B_air=d/N                                            //fringe width in air\n",
+"\n",
+"a=(40/3600)*(%pi/180)                               //angle made by film in radians \n",
+"\n",
+"wavelength=2*a*B_air                                        //as fringe width in air is     'B_air = wavelength/(2*a)'\n",
+"\n",
+"printf('\nThe wavelength of monochromatic light is = %.8f cm\n',wavelength)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.3_2: calculate_the_angle_of_wedge.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_3_2,pg 1-22\n",
+"\n",
+"wavelength=5893*10^-8                                                  //wavelength of light \n",
+"\n",
+"B=0.1                                                        //fringe width\n",
+"\n",
+"u=1.52                                                           //refractive index of glass wedge\n",
+"\n",
+"a=(wavelength/(2*u*B))*3600*(180/%pi)                                   //as fringe spacing is    'B = wavelength/(2*a*u)'\n",
+"\n",
+"printf('\nThe angle of wedge is  a =%.2f seconds \n',a)\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.3_3: calculate_the_wavelength_of_light.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_3_3,pg 1-22\n",
+"\n",
+"B=0.25                                                        //fringe width\n",
+"\n",
+"u=1.4                                                              //refractive index of film\n",
+"\n",
+"a=(20/3600)*(%pi/180)                                              //angle made by film in radians\n",
+"\n",
+"wavelength=2*a*B*u                                                        //as fringe width is     'B = wavelength/(2*a*u)'\n",
+"\n",
+"printf('\nThe wavelength of monochromatic light is = %.8f cm\n',wavelength)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.3_4: find_the_number_of_fringes.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_3_4,pg 1-23\n",
+"\n",
+"wavelength=5.82*10^-5                                                      //wavelength of a monochromatic light\n",
+"\n",
+"u=1.5                                                               //refractive index of glass\n",
+"\n",
+"a=(20/3600)*(%pi/180)                                               //angle made by glass film in radians\n",
+"\n",
+"B=wavelength/(2*u*a)                                                       //The fringe width \n",
+"\n",
+"N=1/B                                                               //the number of fringes per cm\n",
+"\n",
+"printf('\nThe number of fringes per cm = %.f \n',N)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.3_5: find_the_diameter_of_wire.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_3_5,pg 1-24\n",
+"\n",
+"wavelength=6*10^-5                                                      //wavelength of light \n",
+"\n",
+"B=0.1                                                            //fringe width(as there are 10 fringes)\n",
+"\n",
+"u=1                                                              //refractive index of air wedge\n",
+"\n",
+"a=wavelength/(2*u*B)                                                    //as fringe spacing is    'B = wavelength/(2*a*u)'\n",
+"\n",
+"dist=10                                                          //distance of plane of rectangular pieces from wire \n",
+"\n",
+"d=a*dist                                                         //as for small angle 'tan(a) = a = d/dist'\n",
+"\n",
+"printf('\nThe diameter of wire is  d = %.3f cm\n',d)\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.3_6: find_the_separation_between_consecutive_bright_fringes.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_3_6,pg 1-24\n",
+"\n",
+"a=10^-4                                                  //as for small angle 'tan(a) = a'\n",
+"\n",
+"wavelength=5900*10^-10                                           //wavelength of light in air\n",
+"\n",
+"u=1                                                      //refractive index of air \n",
+"\n",
+"B=wavelength/(2*u*a)                                            //The fringe width \n",
+"\n",
+"printf('\nThe fringe width is  B = %.5f m\n',B)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.4_11: find_the_diameter_of_ring.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_4_11,pg 1-37\n",
+"\n",
+"D_4=0.4                                  //diameter of 4th dark ring\n",
+"\n",
+"D_12=0.7                                 //diameter of 12th dark ring \n",
+"\n",
+"const=D_4^2/(4*4)                                 //assume (R*wavelength = const)  for 4th dark ring\n",
+"\n",
+"D_20=sqrt(4*20*const)                             //For 20th dark ring \n",
+"\n",
+"printf('\nDiameter of 20th dark ring is   D20 = %.3f cm\n',D_20)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.4_12: calculate_the_wavelength_of_light.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_4_12,pg 1-38\n",
+"\n",
+"n_1=5                                       //5th ring \n",
+"\n",
+"n_2=15                                      //15th ring \n",
+"\n",
+"p=n_2-n_1                                    //difference between rings\n",
+"\n",
+"Dn_1=0.336                            //diameter of 5th ring \n",
+"\n",
+"Dn_2=0.59                             //diameter of 15th ring\n",
+"\n",
+"R=100                                        //Radius of curvature\n",
+"\n",
+"wavelength=(Dn_2^2-Dn_1^2)/(4*p*R)*10^8                  //wavelength of light \n",
+"\n",
+"printf('\nwavelength of light is = %.f A.',wavelength)\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.4_1: find_the_ring_number.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_4_1,pg 1-32\n",
+"\n",
+"//let the diameter of nth dark ring be double the diameter of that of 40th ring \n",
+"\n",
+"//as   Dn^2 = 4*R*n*wavelength\n",
+"\n",
+"n_1=40                                                     //40 th dark ring\n",
+"\n",
+"n=4*n_1                                                    //as diameter is double\n",
+"\n",
+"printf('\nThe ring number is   n= %.f',n)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.4_2: find_radius_of_curvature_and_thickness_of_film.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_4_2,pg 1-32\n",
+"\n",
+"//For dark rings Dn=sqrt(4*R*n*wavelength)\n",
+"\n",
+"n=10                    //10th ring\n",
+"\n",
+"Dn=0.5                  //diameter of 10th ring \n",
+"\n",
+"wavelength=5*10^-5             //wavelength of light\n",
+"\n",
+"R=Dn^2/(4*n*wavelength)        //radius of curvature \n",
+"\n",
+"t=Dn^2/(8*R)            //thickness of film\n",
+"\n",
+"printf('\nThe radius of curvature is  R = %.2f cm\n',R)\n",
+"\n",
+"printf('\nThe thickness of film is  t = %.5f cm\n',t)\n",
+"\n",
+"//mistake in textbook"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.4_3: find_the_radius_of_curvature.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_4_3,pg 1-33\n",
+"\n",
+"n_1=5                                       //5th ring \n",
+"\n",
+"n_2=15                                      //15th ring \n",
+"\n",
+"p=n_2-n_1                                    //difference between rings\n",
+"\n",
+"Dn_1=0.336                            //diameter of 5th ring \n",
+"\n",
+"Dn_2=0.59                             //diameter of 15th ring\n",
+"\n",
+"wavelength=5890*10^-8                            //wavelength of light\n",
+"\n",
+"R=(Dn_2^2-Dn_1^2)/(4*p*wavelength)                  //radius of curvature\n",
+"\n",
+"printf('\nThe radius of curvature is  R = %.2f cm\n',R)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.4_4: find_the_wavelength_of_light.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_4_4,pg 1-33\n",
+"\n",
+"//as   n1 = nth ring      n2 = (n+8)th ring\n",
+"\n",
+"p=8                                        //difference between rings\n",
+"\n",
+"Dn_1=0.42                            //diameter of 5th ring \n",
+"\n",
+"Dn_2=0.7                             //diameter of 15th ring\n",
+"\n",
+"R=200                                        //radius of curvature\n",
+"\n",
+"wavelength=(Dn_2^2-Dn_1^2)/(4*p*R)                  //wavelength of light\n",
+"\n",
+"printf('\nThe wavelength of light  is  wavelength = %.6f cm\n',wavelength)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.4_5: find_the_refractive_index_of_liquid.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//chapter-1,Example1_4_5,pg 1-34\n",
+"\n",
+"Dn_1=0.218                           //Diameter of nth ring\n",
+"\n",
+"Dn_2=0.451                           //Diameter of (n+10)th ring\n",
+"\n",
+"wavelength=5893*10^-8                          //wavelength of light \n",
+"\n",
+"R=90                               //Radius of curvature \n",
+"\n",
+"p=10\n",
+"\n",
+"u=(4*p*wavelength*R)/(Dn_2^2-Dn_1^2)              //Refractive index of liquid \n",
+"\n",
+"printf('\nRefractive index of liquid is  u = %.3f',u)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.4_6: find_the_diameter_of_dark_ring.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//chapter-1,Example1_4_6,pg 1-34\n",
+"\n",
+"//For nth dark ring       Dn^2 =4*R*n*wavelength\n",
+"\n",
+"D_5=0.42                  //Diameter of 5th dark ring\n",
+"\n",
+"D_10=sqrt(2*D_5^2)               //as number of ring double, the diameter is sqrt(2) times the diameter of original ring\n",
+"\n",
+"printf('\nThe diameter of 10th dark ring is  D10 = %.3f cm',D_10)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.4_7: find_the_diameter_of_dark_ring.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_4_7,pg 1-35\n",
+"\n",
+"R=200                                                  //radius of curvature\n",
+"\n",
+"wavelength_1=6000*10^-8                                     //wavelength of light for nth dark ring\n",
+"\n",
+"wavelength_2=5000*10^-8                                     //wavelength of light for (n+1)th dark ring\n",
+"\n",
+"//as nth ring due to wavelength_1= 6000*10^-8 cm   is coincide with (n+1)th ring  due to wavelength_2=5000*10^-8 cm\n",
+"\n",
+"//therefore  6*n = 5*(n+1)\n",
+"\n",
+"n=5\n",
+"\n",
+"Dn=sqrt(4*R*n*wavelength_1)\n",
+"\n",
+"printf('\nDiameter of nth dark ring due to wavelength 6000 A. is   Dn = %.4f cm\n',Dn)\n",
+"\n",
+"//wrong ans in textbook"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.4_8: find_the_refractive_index_of_liquid.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_4_8,pg 1-35\n",
+"\n",
+"D_air=2.3                                  //Diameter of bright ring in air\n",
+"\n",
+"D_liquid=2                                    //Diameter of bright ring in liquid\n",
+"\n",
+"u=D_air^2/D_liquid^2                                   //Refractive index of liquid\n",
+"\n",
+"printf('\n The refractive index of liquid is  u = %.4f \n',u)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.7_1: find_the_thickness_of_coating.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_7_1,pg 1-42\n",
+"\n",
+"wavelength=560                                                  //wavelength of light in air\n",
+"\n",
+"u=2.0                                                          //refractive index of silicon monoxide material\n",
+"\n",
+"//The wavelength of 'wavelength_1' in a medium of refractive index 'u' is \n",
+"\n",
+"wavelength_1=wavelength/u\n",
+"\n",
+"t=wavelength_1/4                                                       //thickness of the film \n",
+"\n",
+"printf('\nThe thickness of the film is = %.f nm\n',t)\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.7_2: find_the_thickness_of_coating.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"//Chapter-1,Example1_7_2,pg 1-42\n",
+"\n",
+"wavelength=6000                                                    //wavelength of light in air\n",
+"\n",
+"u=1.2                                                              //refractive index of transparant material\n",
+"\n",
+"wavelength_1=wavelength/u                                                        //The wavelength of wavelength_1 in a medium of refractive index 'u'\n",
+" \n",
+"t=wavelength_1/4                                                          //thickness of coating\n",
+"\n",
+"printf('\nThe thickness of coating to eliminate reflection is  t = %.f A.\n',t)"
+   ]
+   }
+],
+"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
+}