Question

A 1 ns pulse of electromagnetic waves would be 30cmlong.

(a) Consider such a pulse of 633 nmwavelength laser light. Find its central wave number and the range of wave numbers it comprises.

(b) Repeat part (a), but for a 1nspulse of 100 MHzradio waves.

Short answer

(a) The value of the wave number is$9.93x{10}^6{\mathrm{m}}^{-1}$and the range of the wave number is $∆\mathrm{k}\ge 1.67{\mathrm{m}}^{-1}$.

(b) The value of the wave number is $20.9{\mathrm{dm}}^{-1}$and the range of the wave number is $∆\mathrm{k}\ge 1.67{\mathrm{m}}^{-1}$.

Step-by-step solution

Formula for expression for the uncertainty in the wave’s position and the relation between wave numbers.

The expression for the uncertainty in the wave’s position is given by,

$\mathbf{∆}\mathbf{x}\mathbf{∆}\mathbf{k}\mathbf{\ge }\frac{\mathbf{1}}{\mathbf{2}}$

The expression for the relation between the wave number k and $\mathit{\lambda }$is given by,$\mathit{k}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{\lambda }}$

Use the expression ΔxΔk≥12 and k=2πλ for calculation.

(a).

The wavelength$\lambda$of the pulse is$6.33nm$.

The length $∆x$of the electromagnetic wave is$30cm$.

The wavenumber is calculated as,

role="math" localid="1659761406782" $$\begin{gathered} k=\frac{2\mathrm{\pi }}{\lambda } \\ =\frac{2\mathrm{\pi }}{633nm} \\ =\frac{2\mathrm{\pi }}{633nm\left({\displaystyle \frac{{10}^{-9}m}{1nm}}\right)} \\ =9.93\times {10}^6{m}^{-1} \end{gathered}$$

The range of the wavenumber is calculated as,

$$\begin{gathered} ∆x∆k\ge \frac{1}{2} \\ \left(30cm\left(\frac{{10}^{-2}m}{1cm}\right)\right)∆k\ge \frac{1}{2} \\ ∆k\ge 1.67m^{-1} \end{gathered}$$

Therefore, the value of the wave number is $9.93x{10}^6{\mathrm{m}}^{-1}$and the range of the wave number is $\Delta k\ge 1.67m^{-1}$.

Formula for the expression for the relation between the wavenumber and the wave equation of the light.

The expression for the relation between the wave numberkand$\mathit{\lambda }$is given by,

$k=\frac{2\mathrm{\pi }}{\lambda }$

The expression for wave equation of light is given by,

$\mathit{c}\mathbf{=}\mathit{\lambda }\mathit{f}$

Use the expression k=2πλ and c=λf for calculation.

(b)

The frequency f of the pulse is 1Hz.

The expression for the wavenumber is evaluated as,

$$\begin{gathered} k=\frac{2\mathrm{\pi }}{\lambda } \\ =\frac{2\mathrm{\pi }}{c} \\ =\frac{2\mathrm{\pi f}}{c} \end{gathered}$$

The value of wavenumber is calculated as,

$$\begin{gathered} k=\frac{2\mathrm{\pi f}}{c} \\ =\frac{2\mathrm{\pi }\left({\displaystyle \frac{1}{1\mathrm{ms}}}\right)}{3\times {10}^{8}m/s} \\ =\frac{2\mathrm{\pi }\left({\displaystyle \frac{1}{1\mathrm{ns}\left({\displaystyle \frac{1\times {10}^{-9}\mathrm{s}}{1\mathrm{ns}}}\right)}}\right)}{3\times {10}^{8}m/s} \\ =20.9m^{-1} \end{gathered}$$

The range of the wavenumber is calculated as,

$$\begin{gathered} ∆x∆k\ge \frac{1}{2} \\ \left(30cm\left(\frac{{10}^{-2}m}{1cm}\right)\right)∆k\ge \frac{1}{2} \\ ∆k\ge 1.67m^{-1} \end{gathered}$$

Therefore, the value of the wave number is $20.9{\mathrm{m}}^{-1}$and the range of the wave number is $∆k\ge 1.67m^{-1}$