Question

A certain helium–neon laser emits red light in a narrow band of wavelengths centered at$\mathbf{632}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{\text{nm}}$ and with a “wavelength width” (such as on the scale of Fig. 33-1) of$\mathbf{0}\mathbf{.}\mathbf{0100}\mathbf{}\mathbf{\text{nm}}$. What is the corresponding “frequency width” for the emission?

Short answer

The frequency width of the emission is$7.49\times {10}^9\mathrm{Hz}$

Step-by-step solution

Step 1: Listing the given quantities

Wavelength, $\lambda =623.8\text{nm}=623.8\times {10}^{-9}\text{m}$.

Wavelength width,$\Delta \lambda =0.0100\mathrm{nm}=0.0100\times {10}^{-9}\mathrm{m}$.

Speed of light, $c=3\times {10}^8\text{m/s}$.

Step 2: Understanding the concepts of frequency and wavelength

In wavelength scale Fig. 33-1, each scale marker represents a change in wavelength and corresponding frequency. A certain scale denotes the electromagnetic spectrum. In this problem, we will calculate the frequency width of the He-Ne laser required for the emission using the formula for the change in frequency in terms of velocity of light and wavelength.

Formula:

$\Delta f=\left|\Delta \left(\frac{c}{\lambda }\right)\right|$

Step 3: Calculations of the frequency width of the emission

Since $\Delta \lambda \ll \lambda ,$ we will find $\Delta f$as,

$\begin{array}{rcl}\Delta f& =& \left|\Delta \left(\frac{c}{\lambda }\right)\right|\\ & =& \frac{c\Delta \lambda }{{\lambda }^2}\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}\Delta f& =& \frac{\left(3\times {10}^8\right)\left(623.8\times {10}^{-9}\right)}{{(0.0100\times {10}^{-9})}^2}\\ & =& 7.49\times {10}^9\mathrm{Hz}\\ & & \end{array}$

The frequency width of the emission is$7.49\times {10}^9\mathrm{Hz}$.