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{
"metadata": {
"name": "",
"signature": "sha256:614bd66037ed01713a261d0e06bb9f5175d6e2a9e3ef900d57af3a2e6efdf43f"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"2: Interference and Diffraction"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.1, Page number 75"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"t = 12*10**-5; #thickness of mica sheet(cm)\n",
"lamda = 6000; #wavelength(Angstrom)\n",
"n = 1;\n",
"\n",
"#Calculation\n",
"lamda = lamda*10**-10; #wavelength(m)\n",
"mew_1 = n*lamda/t;\n",
"mew = mew_1+1; #refractive index of mica\n",
"\n",
"#Result\n",
"print \"refractive index of mica is\",mew\n",
"print \"answer given in the book is wrong\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"refractive index of mica is 1.005\n",
"answer given in the book is wrong\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.2, Page number 75"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"D = 0.53; #distance of fringes from slit(m)\n",
"lamda = 5890; #wavelength of light(angstrom)\n",
"two_d = 0.6*10**-3; #separation of slits(m)\n",
"\n",
"#Calculation\n",
"lamda = lamda*10**-10; #wavelength(m)\n",
"beta = D*lamda/two_d; #width of fringes(m)\n",
"beta = beta*10**3;\n",
"beta = math.ceil(beta*10**3)/10**3; #rounding off to 3 decimals\n",
"\n",
"#Result\n",
"print \"width of fringes is\",beta,\"*10**-3 m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"width of fringes is 0.521 *10**-3 m\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.3, Page number 75"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"beta = 9*10**-4; #width of fringes(m)\n",
"d1 = 75; #distance of fringes from biprism(cm)\n",
"d2 = 5; #distance of biprism from slit(cm)\n",
"lamda = 5890; #wavelength of light(angstrom)\n",
"two_d = 0.6*10**-3; #separation of slits(m)\n",
"\n",
"#Calculation\n",
"lamda = lamda*10**-10; #wavelength(m)\n",
"d1 = d1*10**-2; #distance of fringes from biprism(m)\n",
"d2 = d2*10**-2; #distance of biprism from slit(m)\n",
"D = d1+d2; #distance of fringes from slit(m)\n",
"two_d = D*lamda/beta; #separation of slits(m)\n",
"two_d = two_d*10**4;\n",
"two_d = math.ceil(two_d*10**2)/10**2; #rounding off to 2 decimals\n",
"\n",
"#Result\n",
"print \"distance between slits is\",two_d,\"*10**-4 m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"distance between slits is 5.24 *10**-4 m\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.4, Page number 75"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"lamda = 6*10**-7; #wavelength(m)\n",
"t = 7.2*10**-6; #thickness(m)\n",
"n = 6;\n",
"\n",
"#Calculation\n",
"mew_1 = n*lamda/t;\n",
"mew = mew_1+1; #refractive index of sheet\n",
"\n",
"#Result\n",
"print \"refractive index of sheet is\",mew"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"refractive index of sheet is 1.5\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.5, Page number 76"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"beta = 3; #fringe separation(mm)\n",
"mew = 1; #refractive index\n",
"lamda = 6000; #wavelength(angstrom)\n",
"\n",
"#Calculation\n",
"lamda = lamda*10**-10; #wavelength(m)\n",
"beta = beta*10**-3; #fringe separation(m)\n",
"theta = lamda/(2*mew*beta); #angle between plates(sec)\n",
"theeta = theta*180*3600/math.pi; #angle between plates(sec \")\n",
"theta = theta*10**4;\n",
"theeta = math.ceil(theeta*10**3)/10**3; #rounding off to 3 decimals\n",
"\n",
"#Result\n",
"print \"angle between plates is\",theta,\"*10**-4 sec or\",theeta,\"'\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"angle between plates is 1.0 *10**-4 sec or 20.627 '\n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.6, Page number 76"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"lamda = 5900*10**-7; #wavelength of light(m)\n",
"mew = 1; #refractive index\n",
"n = 7.4; #number of fringes\n",
"\n",
"#Calculation\n",
"t2_t1 = n*lamda/(2*mew); #difference of film thickness(m)\n",
"t2_t1 = t2_t1*10**2;\n",
"\n",
"#Result\n",
"print \"difference of film thickness is\",t2_t1,\"*10**-2 m\"\n",
"print \"answer given in the book is wrong\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"difference of film thickness is 0.2183 *10**-2 m\n",
"answer given in the book is wrong\n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.7, Page number 77"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"lamda = 5.9*10**-7; #wavelength of light(m)\n",
"n = 10; #10th ring\n",
"D10 = 0.5; #diameter of 10th ring(cm)\n",
"\n",
"#Calculation\n",
"D10 = D10*10**-2; #diameter of 10th ring(m)\n",
"R = D10**2/(4*n*lamda); #radius of curvature of lens(m)\n",
"R = math.ceil(R*10**4)/10**4; #rounding off to 4 decimals\n",
"t = D10**2/(8*R); #thickness of the air film(m)\n",
"\n",
"#Result\n",
"print \"radius of curvature of lens is\",R,\"m\"\n",
"print \"thickness of the air film is\",round(t/1e-6,2),\"*10**-6 m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"radius of curvature of lens is 1.0594 m\n",
"thickness of the air film is 2.95 *10**-6 m\n"
]
}
],
"prompt_number": 39
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.8, Page number 77"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"n = 20; #number of fringes\n",
"lamda = 5890; #wavelength(angstrom)\n",
"\n",
"#Calculation\n",
"lamda = lamda*10**-8; #wavelength(cm)\n",
"t = n*lamda/2; #thickness of wire(cm)\n",
"t = t*10**4;\n",
"\n",
"#Result\n",
"print \"thickness of wire is\",t,\"*10**-4 cm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"thickness of wire is 5.89 *10**-4 cm\n"
]
}
],
"prompt_number": 41
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.9, Page number 77"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"lamda = 5880; #wavelength(angstrom)\n",
"n = 1; #number of fringes\n",
"mew = 1.5; #refractive index\n",
"r = 60; #angle of refraction(degree)\n",
"\n",
"#Calculation\n",
"r = r*math.pi/180; #angle of refraction(radian)\n",
"lamda = lamda*10**-10; #wavelength(m)\n",
"t = n*lamda/(2*mew*math.cos(r)); #smallest thickness of the plate(m)\n",
"t = t*10**10; #smallest thickness of the plate(angstrom)\n",
"\n",
"#Result\n",
"print \"smallest thickness of the plate is\",t,\"angstrom\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"smallest thickness of the plate is 3920.0 angstrom\n"
]
}
],
"prompt_number": 46
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.10, Page number 78"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"n1 = 4; #fourth ring\n",
"n2 = 12; #12th ring\n",
"n3 = 20; #20th ring\n",
"D4 = 0.4; #diameter of 4th ring(cm)\n",
"D12 = 0.7; #diameter of 12th ring(cm)\n",
"\n",
"#Calculation\n",
"p1 = n2-n1;\n",
"p2 = n3-n2;\n",
"#D12**2-D4**2 = 4*p1*lamda*R and D20**2-D12**2 = 4*p2*lamda*R\n",
"#therefore D12**2-D4**2 = D20**2-D12**2\n",
"D20 = math.sqrt((2*D12**2)-(D4**2)); #diameter of 20th ring(cm)\n",
"D20 = math.ceil(D20*100)/100; #rounding off to 2 decimals\n",
"\n",
"#Result\n",
"print \"diameter of 20th ring is\",D20,\"cm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"diameter of 20th ring is 0.91 cm\n"
]
}
],
"prompt_number": 50
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 2.11, Page number 78"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"lamda1 = 6*10**-5; #wavelength of light 1(cm)\n",
"lamda2 = 4.5*10**-5; #wavelength of light 2(cm)\n",
"R = 90; #radius of curvature(cm)\n",
"\n",
"#Calculation\n",
"n = lamda2/(lamda1-lamda2); #number of fringes\n",
"Dn = math.sqrt(4*n*lamda1*R); #diameter of nth ring(cm)\n",
"Dn = math.ceil(Dn*10**4)/10**4; #rounding off to 4 decimals\n",
"\n",
"\n",
"#Result\n",
"print \"diameter of nth ring is\",Dn,\"cm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"diameter of nth ring is 0.2546 cm\n"
]
}
],
"prompt_number": 53
}
],
"metadata": {}
}
]
}
|