Question

If a laser pulse is of short enough duration, it becomes rather superfluous to refer to its specific color. How short a duration must a light pulse be for its range of frequencies to cover the entire visible spectrum? (The visible spectrum covers frequencies of -4.5to$\mathbf{7}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{14}}\mathbf{Hz}$.)

Short answer

Duration must be$2.65\times {10}^{-16}\mathrm{s}$for the light pulse of a given range of frequencies to cover the entire visible spectrum.

Step-by-step solution

Heisenberg’s Uncertainty Principle.

Using uncertainty principle $\mathbf{∆}\mathit{E}\mathbf{∆}\mathit{t}\mathbf{\ge }\frac{\mathbf{h}}{\mathbf{2}}$

Where,$\mathbf{∆}\mathit{E}$is the uncertainty in energy, $\mathbf{∆}\mathit{t}$is the uncertainty in time,his Planck’s constant.

Use Heisenberg’s Uncertainty Principle for calculation.

Write is as,

$$\begin{gathered} ∆E∆t\ge \frac{h}{2} \\ ∆\left(hf\right)∆t\ge \frac{\left({\displaystyle \frac{h}{2\mathrm{\pi }}}\right)}{2} \\ h∆f∆t\ge \frac{h}{4\mathrm{\pi }} \\ ∆f∆t\ge \frac{1}{4\mathrm{\pi }} \\ ∆t\ge \frac{1}{4\mathrm{\pi }∆\mathrm{f}} \end{gathered}$$

Plug in the values,

$$\begin{gathered} ∆t\ge \frac{1}{4\mathrm{\pi }\left(3\times {10}^{14}\right)} \\ ∆t\ge 2.65\times {10}^{-16}s \end{gathered}$$

Therefore, the duration must be $2.65x{10}^{-16}s$for the light pulse of a given range of frequencies to cover the entire visible spectrum.