Question

A 1 fs pulse of laser light would be $\mathbf{0}\mathbf{.}\mathbf{3}\mathit{\mu }\mathit{m}$long. What is the range of wavelengths in a 0.3Imlong pulse of (approximately)600nmlaser light?

Short answer

The range of wavelength=95.5nm

Step-by-step solution

Formula of wavelength.

Distance between two corresponding points of two consecutive waves.

$\mathit{E}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{\lambda }}$

Where

c is the speed of light,

$\mathit{\lambda }$is the wavelength.

his Planck’s constant.

Use the wavelength formula for calculation.

Write it as,

$E=\frac{hc}{\lambda }$

On derivation,

$$\begin{gathered} dE=-\frac{hc}{{\lambda }^2}d\lambda \\ d\lambda =-\frac{{\lambda }^2}{hc}dE \end{gathered}$$

After ignoring the negative sign, the equation can be rewritten as:

$$\begin{gathered} d\lambda =\frac{{\lambda }^2}{hc}dE \\ dE=\frac{h}{4\mathrm{\pi }∆\mathrm{t}} \end{gathered}$$

Substitute the given values in the equation

$$\begin{gathered} d\lambda =\frac{{\lambda }^2}{hc}\left(\frac{h}{4\pi \Delta t}\right) \\ d\lambda =\frac{{\lambda }^2}{hc}\left(\frac{1}{4\mathrm{\pi }∆\mathrm{t}}\right) \\ d\lambda =\frac{{\left(600\times {10}^{-9}\right)}^2}{4\mathrm{\pi }\times 3\times {10}^8\times 1\times {10}^{-15}} \\ d\lambda =95.5\mathrm{nm} \end{gathered}$$

Hence, the 1fs pulse covers a range of wavelengths equal to this value, you may note also that we have taken the absolute value throughout the derivation, however, the range in general could be positive or negative.