1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
|
{
"metadata": {
"name": "Chapter3"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": "3: Interference"
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 3.1, Page number 71"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the velocity of light\n\n#importing modules\nfrom __future__ import division\nimport math\n\n#Variable declaration\nbeta = 0.51; #Fringe width(mm)\nd = 2.2; #Distance between the slits(mm)\nD = 2; #Distance between the slits and the screen(m)\n\n#Calculation\nbeta = beta*10**-1; #Fringe width(cm)\nd = d*10**-1; #Distance between the slits(cm)\nD=D*10**2; #Distance between the slits and the screen(cm)\nlamda = beta*d/D; #Wavelength of light(cm)\nlamda = lamda*10**8; #Wavelength of light(A)\n\n#Result\nprint \"The wavelength of light is\",lamda, \"angstrom\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The wavelength of light is 5610.0 angstrom\n"
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 3.2, Page number 71"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the seperation between the third fringe\n\n#importing modules\nfrom __future__ import division\nimport math\n\n#Variable declaration\nlambda1 = 4250; #First wavelength emitted by source of light(A)\nlambda2 = 5050; #Second wavelength emitted by source of light(A)\nD = 1.5; #Distance between the source and the screen(m)\nd = 0.025; #Distance between the slits(mm)\nn = 3; #Number of fringe from the centre\n\n#Calculation\nlambda1 = lambda1*10**-10; #First wavelength emitted(m)\nlambda2 = lambda2*10**-10; #Second wavelength emitted(m)\nd = d*10**-3; #Distance between the slits(m)\nx3 = n*lambda1*D/d; #Position of third bright fringe due to lambda1(m)\nx3_prime = n*lambda2*D/d; #Position of third bright fringe due to lambda2(m)\nx = x3_prime-x3; #separation between the third bright fringe(m)\nx = x*10**2; #separation between the third bright fringe(cm)\n\n#Result\nprint \"The separation between the third bright fringe due to the two wavelengths is\",x, \"cm\"\n",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The separation between the third bright fringe due to the two wavelengths is 1.44 cm\n"
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 3.3, Page number 71"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the refractive index of the sheet of glass\n\n#importing modules\nimport math\n\n#Variable declaration\nlamda = 5.5*10**-5; #Wavelength emitted by source of light(cm)\nn = 4; #Number of fringes shifted\nt = 3.9*10**-4; #Thickness of the thin glass sheet(cm)\n\n#Calculation\nmew = (n*lamda/t)+1; #Refractive index of the sheet of glass\nmew = math.ceil(mew*10**4)/10**4; #rounding off the value of v to 4 decimals\n\n#Result\nprint \"The refractive index of the sheet of glass is\",mew",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The refractive index of the sheet of glass is 1.5642\n"
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 3.4, Page number 72"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the least thickness of the film that appears bright and dark\n\n#importing modules\nimport math\n\n#Variable declaration\nlamda = 5893; #Wavelength of monochromatic lihgt used(A)\nn = 1; #Number of fringe for the least thickness of the film\ncosr = 1; #for normal incidence\nmew = 1.42; #refractive index of the soap film\n\n#Calculation\n#As for constructive interference, \n#2*mew*t*cos(r) = (2*n-1)*lambda/2, solving for t\nt = (2*n-1)*lamda/(4*mew*cosr); #Thickness of the film that appears bright(A)\n#As for destructive interference, \n#2*mu*t*cos(r) = n*lambda, solving for t\nt1 = n*lamda/(2*mew*cosr); #Thickness of the film that appears bright(A)\n\n#Result\nprint \"The thickness of the film that appears bright is\",t, \"angstrom\"\nprint \"The thickness of the film that appears dark is\",t1, \"angstrom\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The thickness of the film that appears bright is 1037.5 angstrom\nThe thickness of the film that appears dark is 2075.0 angstrom\n"
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 3.5, Page number 72"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the angle of the wedge\n\n#importing modules\nimport math\n\n#Variable declaration\nlamda = 5893; #Wavelength of monochromatic lihgt used(A)\nn = 10; #Number of fringe that are found \nd = 1; #Distance of 10 fringes(cm)\n\n#Calculation\nbeta = d/n; #Fringe width(cm)\nlamda = lamda*10**-8; #Wavelength of monochromatic lihgt used(cm)\ntheta = lamda/(2*beta); #Angle of the wedge(rad)\ntheta = theta*10**4;\ntheta = math.ceil(theta*10**4)/10**4; #rounding off the value of theta to 4 decimals\n\n#Result\nprint \"The angle of the wedge is\",theta,\"*10**-4 rad\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The angle of the wedge is 2.9465 *10**-4 rad\n"
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 3.6, Page number 72"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the separation between consecutive bright fringes\n\n#importing modules\nimport math\nfrom __future__ import division\n\n#Variable declaration\nlamda = 5900; #Wavelength of monochromatic lihgt used(A)\nt = 0.010; #Spacer thickness(mm)\nl = 10; #Wedge length(cm)\n\n#Calculation\nt = t*10**-1; #Spacer thickness(cm)\ntheta = t/l; #Angle of the wedge(rad)\nlamda = lamda*10**-8; #Wavelength of monochromatic lihgt used(cm)\nbeta = lamda/(2*theta); #Fringe width(cm)\n\n#Result\nprint \"The separation between consecutive bright fringes is\",beta, \"cm\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The separation between consecutive bright fringes is 0.295 cm\n"
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 3.7, Page number 72"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the diameter of 20th dark ring\n\n#importing modules\nimport math\n\n#Variable declaration\nD4 = 0.4; #Diameter of 4th dark ring(cm)\nD12 = 0.7; #Diameter of 12th dark ring(cm)\n\n#Calculation\n#We have (dn_plus_k**2)-Dn**2 = 4*k*R*lamda\n#D12**2-D4**2 = 32*R*lamda and D20**2-D12**2 = 32*R*lamda for k = 8\n#since RHS are equal, by equating the LHS we get D12**2-D4**2 = D20**2-D12**2\nD20 = math.sqrt((2*D12**2)-D4**2); #Diameter of 20th dark ring(cm)\nD20 = math.ceil(D20*10**4)/10**4; #rounding off the value of D20 to 4 decimals\n\n#Result\nprint \"The diameter of 20th dark ring is\",D20, \"cm\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The diameter of 20th dark ring is 0.9056 cm\n"
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 3.8, Page number 73"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the refractive index of the liquid\n\n#importing modules\nimport math\nfrom __future__ import division\n\n#Variable declaration\nDn = 0.30; #Diameter of nth dark ring with air film(cm)\ndn = 0.25; #Diameter of nth dark ring with liquid film(cm)\n\n#Calculation\nmew = (Dn/dn)**2; #Refractive index of the liquid\n\n#Result\nprint \"The refractive index of the liquid is\", mew\n",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The refractive index of the liquid is 1.44\n"
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 3.9, Page number 73"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the wavelength of light\n\n#importing modules\nimport math\n\n#Variable declaration\nx = 0.002945; #Distance through which movable mirror is shifted(cm)\nN = 100; #Number of fringes shifted\n\n#Calculation\nx = x*10**-2; #Distance through which movable mirror is shifted(m)\nlamda = 2*x/N; #Wavelength of light(m)\nlamda = lamda*10**10; #Wavelength of light(A)\n\n#Result\nprint \"The wavelength of light is\",lamda, \"angstrom\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The wavelength of light is 5890.0 angstrom\n"
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 3.10, Page number 73"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the shift in movable mirror\n\n#importing modules\nimport math\n\n#Variable declaration\nlambda1 = 5896; #Wavelength of D1 line of sodium(A)\nlambda2 = 5890; #Wavelength of D2 line of sodium(A)\n\n#Calculation\nlamda = (lambda1+lambda2)/2;\nx = (lamda**2)/(2*(lambda1-lambda2)); #Shift in movable mirror of Michelson Interferometer(A)\nx = x*10**-7; #Shift in movable mirror of Michelson Interferometer(mm)\nx = math.ceil(x*10**4)/10**4; #rounding off the value of D20 to 4 decimals\n\n#Result\nprint \"The shift in movable mirror is\",x, \"mm\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The shift in movable mirror is 0.2894 mm\n"
}
],
"prompt_number": 17
},
{
"cell_type": "code",
"collapsed": false,
"input": "",
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
