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   "source": [
    "# Chapter 2 DIFFRACTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 2_1 pg.no:29"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "No of lines per centimeter is 5000\n"
     ]
    }
   ],
   "source": [
    "#To calculate the no of lines in one cm of grating surface\n",
    "from math import pi,sin\n",
    "k=2.\n",
    "lamda=5*10**-5 #units in cm\n",
    "theta=30 # units in degrees\n",
    "#We have nooflines=1/e=(k∗lamda)/sin(theta)\n",
    "nooflines=sin(theta*pi/180)/(k*lamda) #units in cm\n",
    "print \"No of lines per centimeter is %.f\"%nooflines\n",
    "#In text book the answer is printed wrong as 10ˆ3\n",
    "#The correct answer is 5∗10ˆ3"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 2_2 pg.no:30"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "For First order spectra theta1=17.5 degrees\n",
      "For Third order spectra theta3=64.2 degrees\n",
      "Difference in Angles of deviation in first and third order spectra is theta3−theta1=46.70 degrees\n"
     ]
    }
   ],
   "source": [
    "#To Find the difference in angles of deviation in first and third order spectra\n",
    "from math import pi,asin\n",
    "lamda=5000. # units in armstrongs\n",
    "lamda=lamda*10**-8 # units in cm\n",
    "e=1./6000.\n",
    "#For first order e∗sin(theta1)=1∗lamda\n",
    "theta1=asin(lamda/e) # units in radians\n",
    "theta1=theta1*180./pi # units in degrees\n",
    "print \"For First order spectra theta1=%.1f degrees\"%theta1\n",
    "#For third order e∗sin(theta3)=3∗lamda\n",
    "theta3=asin(3.*lamda/e) # units in radians\n",
    "theta3=theta3*180/pi # units in degrees\n",
    "print \"For Third order spectra theta3=%.1f degrees\"%theta3\n",
    "diffe=theta3-theta1 #units in degrees\n",
    "print \"Difference in Angles of deviation in first and third order spectra is theta3−theta1=%.2f degrees\"%diffe"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 2_3 pg.no:30"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false,
    "scrolled": true
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "No of lines per cm=196.0 \n"
     ]
    }
   ],
   "source": [
    "#To calculate minimum no of lines per centimeter\n",
    "lamda1=5890 # units in armstrongs\n",
    "lamda2=5896 # units in armstrongs\n",
    "dlamda=lamda2-lamda1 #units in armstrongs\n",
    "k=2\n",
    "n=lamda1/(k*dlamda)\n",
    "width=2.5 #units in cm\n",
    "nooflines=n/width\n",
    "print \"No of lines per cm=%.1f \"%nooflines"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 2_4 pg.no:31"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "As total no of lines required for resolution in first order is 981 and total no of lines in grating is 850 the lines will not be resolved in first order\n",
      "As total no of lines required for resolution in first order is 490 and total no of lines in grating is 850 the lines will be resolved in second order\n"
     ]
    }
   ],
   "source": [
    "#To examine two spectral lines are clearly resolved in first order and second order\n",
    "n=425.\n",
    "tno=2.*n\n",
    "lamda1=5890 # units in armstrongs\n",
    "lamda2=5896 # units in armstrongs\n",
    "dlamda=lamda2 -lamda1\n",
    "#For first order\n",
    "n=lamda1/dlamda\n",
    "print\"As total no of lines required for resolution in first order is %.f and total no of lines in grating is %d the lines will not be resolved in first order\"%(n,tno)\n",
    "#For second order\n",
    "n=lamda1/(2*dlamda)\n",
    "print\"As total no of lines required for resolution in first order is %.f and total no of lines in grating is %d the lines will be resolved in second order\"%(n,tno)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 2_5 pg.no:32"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Angle of separation is 16 minutes\n"
     ]
    }
   ],
   "source": [
    "#To find the angle of separation\n",
    "from math import asin,pi\n",
    "\n",
    "lamda1=5016.  # units in armstrongs\n",
    "lamda2=5048.  # units in armstrongs\n",
    "lamda1=lamda1*10**-8   # units in cm\n",
    "lamda2=lamda2*10**-8   # units in cm\n",
    "k=2.\n",
    "n=15000\n",
    "e=2.54/n     # units in cm \n",
    "theta1=asin((2*lamda1)/e)*(180/pi)    # units in in degrees\n",
    "theta2=asin((2*lamda2)/e)*(180/pi)  # units in in degrees\n",
    "diffe=theta2-theta1 # units in in degrees\n",
    "diffe=diffe*60 # units in minutes\n",
    "print \"Angle of separation is %.f minutes\"%diffe"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 2_6 pg.no:32"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The dispersive power of the grating is 15000\n"
     ]
    }
   ],
   "source": [
    "#To Calculate the dispersive power of the grating\n",
    "from math import pi,asin,cos\n",
    "n=4000.\n",
    "e=1/n #units in cm\n",
    "k=3.\n",
    "lamda=5000 # units in armstrongs\n",
    "lamda=lamda*10**-8 # units in cm\n",
    "theta=asin((k*lamda)/e)*(180/pi) # units in degrees\n",
    "costheta=cos(theta*pi/180)\n",
    "disppower=(k*n)/costheta\n",
    "print \"The dispersive power of the grating is %.f\"%disppower"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 2_7 pg.no:33"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The highest order spectrum Seen with monochromatic light is 3.33\n"
     ]
    }
   ],
   "source": [
    "#To Calculate highest power of spectrum seen with mono chromaic light\n",
    "lamda=6000.    # units in armstrongs\n",
    "lamda=lamda*10**-8   #units in cm\n",
    "n=5000.\n",
    "e=1/n #units in cm\n",
    "k=e/lamda\n",
    "print \"The highest order spectrum Seen with monochromatic light is %.2f\"%k"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 2_8 pg.no:35"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Wavelength of the lines is 0.0000606 cms\n",
      "Minimum grating width required is 4.2 cm \n"
     ]
    }
   ],
   "source": [
    "#To calculate the wavelength\n",
    "from math import pi,sin\n",
    "k=2.\n",
    "theta1=10.\n",
    "dtheta=3.\n",
    "dlamda=5*10**-9\n",
    "lamda=(sin((theta1*pi)/180)*dlamda*60*60)/(cos((theta1*pi)/180)*dtheta*(pi/180)) # units in cm\n",
    "print \"Wavelength of the lines is %.7f cms\"%lamda\n",
    "lamda_dlamda=lamda+dlamda # units in cm\n",
    "N=6063\n",
    "Ne=(N*k*lamda)/sin((theta1*pi)/180) # units in cm\n",
    "print \"Minimum grating width required is %.1f cm \"%Ne"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 2_9 pg.no:35"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Resolving power is 10000 \n"
     ]
    }
   ],
   "source": [
    "#To calculate resolving power in second order\n",
    "#We have e∗sin(theta)=k∗lamda\n",
    "#We have e∗0.2=k∗lamda −>1\n",
    "#And e∗0.3=(k+1)∗lamda −>2\n",
    "#Subtracting one and two 3∗0.1=lamda\n",
    "lamda=5000. # units in armstrongs\n",
    "lamda=lamda*10**-8 # units in cm\n",
    "e=lamda/0.1 # units in cm\n",
    "width=2.5 #units in cm\n",
    "N=width/e\n",
    "respower=2*N\n",
    "print \"Resolving power is %.f \"%respower"
   ]
  }
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