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{
"metadata": {
"name": "",
"signature": "sha256:7c369c2593aa95d4aac8153f9063553bb327aa743223d1fff41c98aa45d77ebe"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter4:POLARISATION"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4.1:pg-147"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# compare the intensities of ordinary and extraordinary rays\n",
"#intensity of ordinary rays is given by Io=a**2 *(sin theta)**2\n",
"#where theta=30 degree\n",
"#we get Io=a**2/4\n",
"Io=1.0/4\n",
"#intensity of extraordinary ray is given by IE=(a*cos theta)**2\n",
"#we get IE=3*a**2/4\n",
"IE=3.0/4\n",
"I=IE/Io\n",
"print \"the intensities of ordinary and extraordinary rays is I=\",I,\"unitless\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the intensities of ordinary and extraordinary rays is I= 3.0 unitless\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4.2:pg-147"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate angle of refraction\n",
"#according to brewster's law mu=tan ip\n",
"mu=1.732 #refractive index\n",
"ip=math.degrees(math.atan(mu)) #polarising angle in degree\n",
"r=90-ip\n",
"print \"angle of refraction of ray is r=\",round(r),\"degree\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"angle of refraction of ray is r= 30.0 degree\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4.3:pg-147"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate polarising angle and angle of refraction\n",
"mu=1.345 #refractive index, mu=1/sinc=1/sin48degree=1/0.7431 \n",
"ip=math.degrees(math.atan(mu))\n",
"r=90-ip\n",
"print \"polarising angle is ip=\",round(ip,3),\"degree\"\n",
"print \"angle of refraction is r=\",round(r,3),\"degree\" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"polarising angle is ip= 53.369 degree\n",
"angle of refraction is r= 36.631 degree\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4.4:pg-147"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate thickness of a half wave plate of quartz\n",
"lamda=5*10**-5 #wavelength in cm\n",
"mue=1.553 \n",
" #refractive index (unitless)\n",
"muo=1.544\n",
"#for a half plate of positive crystal\n",
"t=lamda/(2*(mue-muo))\n",
"print \"thickness of a half wave plate of quartz is t=\",\"{:.2e}\".format(t),\"cm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"thickness of a half wave plate of quartz is t= 2.78e-03 cm\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4.5:pg-148"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate thickness of quarter wave plate\n",
"lamda=5.890*10**-5 #wavelength of light in cm\n",
"mue=1.553\n",
" #refractive index\n",
"muo=1.544\n",
"t=lamda/(4*(mue-muo)) \n",
"print \"thickness of quarter wave plate is t=\",\"{:.3e}\".format(t),\"cm\" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"thickness of quarter wave plate is t= 1.636e-03 cm\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4.6:pg-148"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate thickness of a doubly refracting plate\n",
"lamda=5.890*10**-5 #wavelength in cm\n",
"muo=1.53 \n",
" #refractive index\n",
"mue=1.54\n",
"t=lamda/(4*(mue-muo))\n",
"print \"thickness of a plate is t=\",\"{:.2e}\".format(t),\"cm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"thickness of a plate is t= 1.47e-03 cm\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4.7:pg-152"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate angle of rotation\n",
"alpha=66 #specific rotation of cane sugar in degree\n",
"c=15.0/100 #concentration of the solution in gm/cc\n",
"l=20 #length of tube in cm\n",
"theta=alpha*l*c/10\n",
"print \"the angle of rotation of the plane of polarisation is theta=\",theta,\"degree\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the angle of rotation of the plane of polarisation is theta= 19.8 degree\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4.8:pg-153"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate specific rotation \n",
"theta=26.4 #in degree\n",
"l=20 #length in cm\n",
"c=0.2 #gm/cm**3\n",
"alpha=10*theta/(l*c)\n",
"print \"the specific rotation is alpha=\",alpha,\"degree\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the specific rotation is alpha= 66.0 degree\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4.9:pg-153"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate strength of solution\n",
"theta=11 #degree\n",
"l=20.0 #length in cm\n",
"alpha=66 #specific rotation of sugar in degree\n",
"c=10*theta/(l*alpha)\n",
"print \"strength of solution is c=\",round(c,4),\"gm/cm**3\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"strength of solution is c= 0.0833 gm/cm**3\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4.10:pg-153"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate difference in the refractive indices\n",
"#specific rotation is theta/d=29.73 degree/mm\n",
"theta=29.73 #where theta=theta/d\n",
"lamda=5.086*10**-4 #wavelength in mm\n",
"#optical rotation is given by theta=math.pi*d*(mul-mur)/lamda\n",
"#where mul and mur are refractive indices for anti-clockwise and clockwise polarised lights\n",
"mu=theta*lamda/180 #where mu=mul-mur\n",
"print \"difference in refractive indices is mu=\",\"{:.1e}\".format(mu),\"unitless\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"difference in refractive indices is mu= 8.4e-05 unitless\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4.11:pg-153"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate optical rotation\n",
"#let theta' be the optical rotation by a solution of strength c' in a tube of length l' then\n",
"#we get 10*theta'/l'*c'=10*theta/l*c\n",
"c=1.0/3 #it is given that solution is 1/3 of its previous concentration i.e. c'/c=1/3,where c=c'/c\n",
"l1=30 #where l1=l'\n",
" #length in cm \n",
"l=20.0\n",
"theta=13 #degree\n",
"#formula is theta'=l'*c'*theta/(l*c)\n",
"theta1=l1*c*theta/l\n",
"print \"optical rotation is theta1=\",theta1,\"degree\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"optical rotation is theta1= 6.5 degree\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4.12:pg-154"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate specific rotation\n",
"theta=52.8 #optical rotation in degree\n",
"l=20.0 #length of the solution in cm\n",
"c=20/50.0 #concentration of the solution in gm/cc\n",
"alpha=10*theta/(l*c)\n",
"print \"the specific rotation is alpha=\",alpha,\"degree\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the specific rotation is alpha= 66.0 degree\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4.13:pg-154"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate length \n",
"l=40 #length in cm\n",
"c=5.0/100 #concentration in percentage\n",
"theta1=35 #optical rotation in degree ,where theta1=theta'\n",
"c1=10.0/100 #concentration in % ,where c1=c'\n",
"theta=20\n",
"#formula of specific rotation is alpha=10*theta/l*c\n",
"l1=l*c*theta1/(c1*theta)\n",
"print \"length is l1=\",l1,\"cm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"length is l1= 35.0 cm\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4.14:pg-155"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate rotation of plane of polarisation of light\n",
"mur=1.53914\n",
" #refractive index\n",
"mul=1.53920\n",
"lamda=6.5*10**-5 #wavelength in cm\n",
"d=0.02 #distance in cm\n",
"thetaR=180*(mul-mur)*d/lamda\n",
"print \"rotation of plane of polarisation of light is thetaR=\",round(thetaR,3),\"degree\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"rotation of plane of polarisation of light is thetaR= 3.323 degree\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4.15:pg-155"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#to calculate % purity of the sugar sample\n",
"theta=9.9 #optical rotation in degree\n",
"alpha=66 #specific roation of pure sugar solution in dm**-1(gm/cc)**-1\n",
"l=20 #length of tube in cm\n",
"c=10*theta/(l*alpha) #concentration of solution in gm/c.c\n",
"#it is given that 80 gm of impure sugar is dissolved in a litre of water\n",
"per=(c*100*10**3)/80 #here c is in gm/litre\n",
"print \"percentage of the sugar sample is per=\",per,\"%\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"percentage of the sugar sample is per= 93.75 %\n"
]
}
],
"prompt_number": 17
}
],
"metadata": {}
}
]
}
|
