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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 22:The properties of Light"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex22.1:pg-1067"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The position of the image is i= 6.0  cm\n",
+      "The Size of the image is I= 0.0  cm High\n"
+     ]
+    }
+   ],
+   "source": [
+    "  #Example 22_1\n",
+    " \n",
+    "  \n",
+    "#To find the position and size of the image\n",
+    "d1=5      #units in cm\n",
+    "d2=30       #units in cm\n",
+    "i=(d1*d2)/(d2-d1)     #Units in cm\n",
+    "d3=2             #units in cm\n",
+    "I=(i/d2)*d3        #units in cm\n",
+    "print \"The position of the image is i=\",round(i),\" cm\\nThe Size of the image is I=\",round(I,2),\" cm High\"       \n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex22.2:pg-1068"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The position of the image is i= -10.0  cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "  #Example 22_2\n",
+    " \n",
+    "  \n",
+    "#To find the location of the image\n",
+    "d1=10      #units in cm\n",
+    "d2=5       #units in cm\n",
+    "i=(d1*d2)/(d2-d1)     #Units in cm\n",
+    "print \"The position of the image is i=\",round(i),\" cm\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex22.3:pg-1069"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The location of the image is i= -30.0  cm\n",
+      " The relative size of the image is I_O= 0.4  cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "  #Example 22_3\n",
+    " \n",
+    "  \n",
+    "#To find the location of the image and its relative size\n",
+    "r=100.0              #Unts in cm\n",
+    "d1=-r/2      #units in cm\n",
+    "d2=75.0       #units in cm\n",
+    "i=(d1*d2)/(d2-d1)     #Units in cm\n",
+    "p=75      #units in cm\n",
+    "sizee=-i/p             #units in cm\n",
+    "print \"The location of the image is i=\",round(i),\" cm\\n The relative size of the image is I_O=\",round(sizee,2),\" cm\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex22.4:pg-1069"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The angle at which the light emerges in air is theta= 53.0  degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example 22_4\n",
+    " \n",
+    "import math  \n",
+    "#To find the angle at which the light emerge in to the air\n",
+    "theta=37        #Units in degrees\n",
+    "n1=1.33       #Units in constant\n",
+    "n2=1       #Units in constant\n",
+    "thetaa=math.asin((n1*math.sin(theta*math.pi/180))/n2)*180/math.pi            #units in degrees\n",
+    "print \"The angle at which the light emerges in air is theta=\",round(thetaa),\" degrees\" \n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex22.5:pg-1068"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "We have Theta1=Theta4 \n",
+      "Which shows that A unform layer of transparent material does not change the direction of the beam of light\n"
+     ]
+    }
+   ],
+   "source": [
+    "  #Example 22_5\n",
+    " \n",
+    "  \n",
+    "#At what angle does the light emerges from the bottom of the dish\n",
+    "print \"We have Theta1=Theta4 \\nWhich shows that A unform layer of transparent material does not change the direction of the beam of light\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex22.6:pg-1068"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "From the diagram we notice that eyes will assume that the three rays come from image position indicated and as we see the image is virtual, erect and enlarged\n",
+      "\n",
+      "The image is located at i= -10.0  cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "  #Example 22_6\n",
+    " \n",
+    "  \n",
+    "#To draw a ray diagram to locate the image\n",
+    "print \"From the diagram we notice that eyes will assume that the three rays come from image position indicated and as we see the image is virtual, erect and enlarged\"\n",
+    "d1=10      #units in cm\n",
+    "d2=5       #units in cm\n",
+    "i=(d1*d2)/(d2-d1)     #Units in cm\n",
+    "print \"\\nThe image is located at i=\",round(i,2),\" cm\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex22.7:pg-1068"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "From the ray diagram we have noticed that the image is virtual, erect and dimnished in size\n",
+      "\n",
+      "The image is located at i= 3.0  cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "  #Example 22_7\n",
+    " \n",
+    "  \n",
+    "#To find the image position by means of the ray diagram\n",
+    "print \"From the ray diagram we have noticed that the image is virtual, erect and dimnished in size\"\n",
+    "d1=5      #units in cm\n",
+    "d2=-10       #units in cm\n",
+    "i=(d1*d2)/(d2-d1)     #Units in cm\n",
+    "print \"\\nThe image is located at i=\",round(i,2),\" cm\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex22.8:pg-1069"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "The image is located at i= -13.33  cm\n",
+      "\n",
+      "The Size of the image is I= 1.0  cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "  #Example 22_8\n",
+    " \n",
+    "  \n",
+    "#To find the image positon and size\n",
+    "d1=-20      #units in cm\n",
+    "d2=40.0       #units in cm\n",
+    "i=(d1*d2)/(d2-d1)     #Units in cm\n",
+    "print \"\\nThe image is located at i=\",round(i,2),\" cm\"\n",
+    "d3=3.0            #units in cm\n",
+    "I=(-i*d3)/d2     #units in cm\n",
+    "print \"\\nThe Size of the image is I=\",round(I),\" cm\"\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.11"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}