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diff --git a/Engineering_Physics_by_Shyam_Singh_and_Rajeev_Singh/Chapter2.ipynb b/Engineering_Physics_by_Shyam_Singh_and_Rajeev_Singh/Chapter2.ipynb new file mode 100755 index 00000000..ef49544b --- /dev/null +++ b/Engineering_Physics_by_Shyam_Singh_and_Rajeev_Singh/Chapter2.ipynb @@ -0,0 +1,507 @@ +{
+ "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": {}
+ }
+ ]
+}
\ No newline at end of file |