Question

The mirrors in the laser of Fig. 40-20, which are separated by 8.0 cm, form an optical cavity in which standing waves of laser light can be set up. Each standing wave has an integral number n of half wavelengths in the 8.0 cm length, where n is large and the waves differ slightly in wavelength. Near $\mathit{\lambda }\mathbf{=}\mathbf{533}\mathbf{}\mathit{n}\mathit{m}$, how far apart in wavelength are the standing waves?

Short answer

The standing waves are 1.8 pm far apart in wavelength.

Step-by-step solution

The given data

a) The separation between the mirrors of the laser, $\mathrm{L}=8\mathrm{cm}=8\times {10}^7\mathrm{nm}$

b) Each wavelength has an integral number n of half wavelengths in the length L=8 cm

c) The wavelength of the laser, $\mathrm{\lambda }=533\mathrm{nm}$

Understanding the concept of standing waves:

A standing wave is a combination of two waves moving in opposite directions, each with the same amplitude and frequency.

Using the concept of standing waves, get the differential equation of the wave equation. This determines the differential of lambda which is the separation between the lambda by substituting the standing wave equation and the given data to it.

Formula:

The standing waves equation of a body is,

$\mathrm{n\lambda }=2\mathrm{L}$ ….. (1)

Calculation of the separation between the wavelengths

Now to obtain the separation between the wavelengths, first differentiate the equation (1) that is the nth harmonic of the standing wave of wavelength $\lambda$in the cavity of width L , such as:

role="math" localid="1661499710483" $$\begin{gathered} \mathrm{n}∆\mathrm{\lambda }+\mathrm{\lambda }∆\mathrm{n}=0 \\ ∆\mathrm{\lambda }=\frac{\mathrm{\lambda }∆\mathrm{n}}{\mathrm{n}} \\ =\frac{\mathrm{\lambda }}{\mathrm{n}}\left(\mathrm{separation}\mathrm{between}\mathrm{any}\mathrm{two}\mathrm{harmonics},∆\mathrm{n}=1\right) \\ =\frac{{\mathrm{\lambda }}^2}{2\mathrm{L}}\left(\mathrm{from}\mathrm{equation}\left(1\right)\right) \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} ∆\mathrm{\lambda }=\frac{{\left(533\mathrm{nm}\right)}^2}{2\left(8\times {10}^7\mathrm{nm}\right)} \\ =1.8\times {10}^{-12}\mathrm{m} \\ =1.8\mathrm{pm} \end{gathered}$$

Hence, the value of the separation is 1.8 pm .