A pulse of light is created by the superposition of many waves that span the frequency range $f_0-\frac{1}{2}\Delta f\le f\le f_0+\frac{1}{2}\Delta f$, wherc $f_0=c/\lambda$is called thc center frequency of thc pulsc. Lascr technology can generate a pulse of light that has a wavelength of $600\mathrm{nm}$and lasts a mere $6.0$fs $1\mathrm{fs}=1$femtosecond localid="1650804865678" $\left.={10}^{-15}\mathrm{s}\right)$.

a. What is the center frequency of this pulse of light?

b. How many cycles, or oscillations, of the light wave are completed during the $6.0$fs pulse?

c. What range of frequencies must be superimposed to create this pulse?

d. What is the spatial length of the laser pulse as it travels through space?

e. Draw a snapshot graph of this wave packet.

Frequency range, $f_o-\frac{1}{2}\Delta f\le f\le f_o+\frac{1}{2}\Delta f$

Wavelength, $\lambda =600\mathrm{nm}$

Time,$t=1\times {10}^{-15}\sec$

Formula used:

Central frequency is calculated as

$f_o=\frac{c}{\lambda }$

Here,

$c$is the speed of light

$\lambda$is the wavelength

Calculation:

Central frequency is calculated as

$f_o=\frac{c}{\lambda }$

Plugging the values in the above equation

$$\begin{gathered} f_o=\frac{3\times {10}^8}{600\times {10}^{-4}} \\ =5\times {10}^{14}\mathrm{Hz} \end{gathered}$$

Conclusion:Center frequency of this pulse light is$5\times {10}^{14}\mathrm{Hz}$

Frequency range, $f_{\theta }-\frac{1}{2}\Delta f\le f\le f_o+\frac{1}{2}\Delta f$

Wavelength, $\lambda =600\mathrm{nm}$

Time, $t=1\times {10}^{-15}\sec$

Number of oscillation completed during the period of $6\mathrm{fs}$is calculated as

$$\begin{gathered} n=f\Delta t \\ =5\times {10}^{14}\times 6\times {10}^{-15} \\ =3 \end{gathered}$$

Conclusion:Number of cycles completed during $6\times {10}^{-15}\sec$is 3

Time period,$\Delta t=6\times {10}^{-15}\sec$

The frequency range of this time period is calculated as

$$\begin{gathered} \Delta f=\frac{1}{\Delta t}=\frac{1}{6\times {10}^{-13}} \\ =1.6\times {10}^{14}\mathrm{Hz} \end{gathered}$$

Frequency range is calculated as

$$\begin{gathered} f_o-\frac{1}{2}\Delta f\le f\le f_o+\frac{1}{2}\Delta f \\ 5\times {10}^{14}-\frac{1}{2}\times 1.6\times {10}^{14}\le f\le 5\times {10}^{14}+\frac{1}{2}\times 1.6\times {10}^{14} \\ 4.2\times {10}^{14}\le f\le 5.8\times {10}^{14}\mathrm{Hz} \end{gathered}$$

Conclusion:Frequency range is$4.2\times {10}^{14}\le f\le 5.8\times {10}^{14}\mathrm{Hz}$

Time period,$\Delta t=6\times {10}^{-15}\sec$

Spatial length of the wave is calculated as

$$\begin{gathered} L=c\Delta t \\ =3\times {10}^8\times 6\times {10}^{-15} \\ =1.8\times {10}^{-6}\mathrm{m} \\ =1.8\mu \mathrm{m} \end{gathered}$$

Conclusion:Spatial length of the laser pulse is$1.8\mu \mathrm{m}$

Conclusion:Snapshot is drawn above