Chapter01 - Home

Exercise. This is the solution to exercise 1.1 in the book.

Solution. Recall that:

$$E=h\nu =\frac{hc}{\lambda}$$where $h$ is Planck’s constant, $c$ is the speed of light and $\lambda $ is the wavelength of the photons with energy $E$. Hence:

$$E=\frac{6.626\cdot {10}^{-34}J\cdot s\cdot 3\cdot {10}^{8}m\u2215s}{1064\cdot {10}^{-9}m}=1.868\cdot {10}^{-19}J=1.166eV$$The energy of the pulse is:

$${E}_{p}=P\cdot \tau =5\cdot {10}^{6}W\cdot 2\cdot {10}^{-8}s=0.1J$$Hence the number of photons of wavelength 1064-nm contained in the pulse is:

$$N=\frac{{E}_{p}}{E}=\frac{0.1J}{1.868\cdot {10}^{-19}J}=5.35\cdot {10}^{17}$$