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
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
|
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 5 Laser"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5_1 pgno:242"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" # PROBLEM 1 # \n",
"\n",
"\n",
" Number of oscillation corresponding to coherent length is \n",
" Coherent time is sec. 50000.0 9.81666666667e-11\n"
]
}
],
"source": [
"# Given that\n",
"l = 2.945e-2 # coherent length of sodium light\n",
"lamda = 5890 # wavelength of light used in angstrom\n",
"c = 3e8 # speed of light\n",
"# Sample Problem 1 on page no. 242\n",
"print(\"\\n # PROBLEM 1 # \\n\")\n",
"n = l/(lamda*1e-10) # number of oscillation corresponding to coherent length\n",
"t = l/c # coherent time\n",
"print\"\\n Number of oscillation corresponding to coherent length is \\n Coherent time is sec.\",n,t\n",
"\n",
"\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5_2 pgno:242"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" # PROBLEM 2 # \n",
"\n",
"\n",
" Angular spread is rad. \n",
" Areal spread is m^2. 0.00016 4096000000.0\n"
]
}
],
"source": [
"\n",
"# Given that\n",
"l = 4e5 # Distance of moon in km\n",
"lamda = 8e-7 # wavelength of light used\n",
"a = 5e-3 # Aperture of laser\n",
"c = 3e8 # speed of light\n",
"# Sample Problem 2 on page no. 242\n",
"print\"\\n # PROBLEM 2 # \\n\"\n",
"theta = lamda/a # Angular of spread \n",
"Areal_spread = (l*1000*theta)**2 # Areal spread\n",
"print\"\\n Angular spread is rad. \\n Areal spread is m^2.\",theta,Areal_spread\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5_3 pgno:242"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" # PROBLEM 3 # \n",
"\n",
"\n",
" Number of oscillation corresponding to coherent length is \n",
" Coherent time is sec. 50000.0 9.81666666667e-11\n"
]
}
],
"source": [
"\n",
"# Given that\n",
"l = 2.945e-2 # coherent length of sodium light\n",
"lamda = 5890 # wavelength of light used\n",
"c = 3e8 # speed of light\n",
"# Sample Problem 3 on page no. 242\n",
"print\"\\n # PROBLEM 3 # \\n\"\n",
"n = l/(lamda *1e-10) # number of oscillation corresponding to coherent length\n",
"t = l/c # coherent time\n",
"print\"\\n Number of oscillation corresponding to coherent length is \\n Coherent time is sec.\",n,t\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5_4 pgno:243"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" # PROBLEM 4 # \n",
"\n",
"\n",
" Energy difference is eV. 0.365641494412\n"
]
}
],
"source": [
"\n",
"# Given that\n",
"k = 12400 # constant\n",
"lamda = 3.3913 # wavelength IR radiation\n",
"\n",
"# Sample Problem 4 on page no. 243\n",
"print\"\\n # PROBLEM 4 # \\n\"\n",
"E = k/(lamda*1e4) # Energy difference\n",
"print\"\\n Energy difference is eV.\",E\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5_5 pgno:"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" # PROBLEM 5 # \n",
"\n",
"\n",
" Energy of one photon is eV. \n",
" Total energy is J 1 4.8\n"
]
}
],
"source": [
"\n",
"k = 12400 # constant\n",
"lamda = 6943 # wavelength of radiation in angstrom\n",
"n = 3e19 # Total number of ions\n",
"# Sample Problem 5 on page no. 243\n",
"print\"\\n # PROBLEM 5 # \\n\"\n",
"E = k/(lamda) # Energy difference\n",
"E_total = E*n*1.6e-19 # Total Energy emitted \n",
"print\"\\n Energy of one photon is eV. \\n Total energy is J\",E,E_total\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5_6 pgno:244"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" # PROBLEM 6 # \n",
"\n",
"\n",
" Required length of cavity is cm. 10.010896\n"
]
}
],
"source": [
"\n",
"# Given that\n",
"h_w = 2e-3 # half width of gain profile of laser in nm\n",
"mu = 1 # refractive index\n",
"lamda = 6328 # wavelength of light used in angstrom\n",
"# Sample Problem 6 on page no. 244\n",
"print\"\\n # PROBLEM 6 # \\n\"\n",
"L = (lamda*1e-10)**2/(2*mu*h_w*1e-9) # Length of cavity \n",
"print\"\\n Required length of cavity is cm.\",L*100\n",
"\n"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.9"
}
},
"nbformat": 4,
"nbformat_minor": 0
}
|
