Past exam of the physics A course of the University of Cambridge / 2017 / Paper 2 / Section B / B7

Superpose two nearby harmonic waves. The identityseparates a rapid carrier from a slow beat envelope. A constant point of the envelope moves at $\Delta \omega /\Delta k$, which tends toas the components approach one another. Equivalently, expand a narrow wave packet spectrum around $k_0$: $\omega (k)={\omega }_0+v_g(k-k_0)+\cdots$. Factoring the carrier out of its Fourier transform representation leaves an envelope depending on $x-v_{g}t$. This interpretation requires a narrow spectrum and little attenuation or distortion; higher derivatives of the dispersion relation can change the envelope shape. The carrier phase velocity is $v_p=\omega /k$ and need not equal $v_g$.

For the wavelength in the medium, ${\lambda }_m=2\pi /k$, differentiate $\omega =k{v}_p(k)$:The distinction in group velocity in material and vacuum wavelength coordinates is essential for the numerical data. For incident vacuum wavelength ${\lambda }_0=2\pi c/\omega$, differentiation of $k=\omega /v_p({\lambda }_0)$ instead givesThus the printed derivative identity is valid, but its wavelength coordinate cannot be carried unchanged into data specified before refraction.

The wavelengths specified for incidence from vacuum are ${\lambda }_0$. With fixed incident angle ${\theta }_i$, Snell's law gives $n=\sin {\theta }_i/\sin {\theta }_t$ and $v_p=c\sin {\theta }_t/\sin {\theta }_i$. At the central wavelength,Use a central difference, converting the angle change to radians:The group index from refracted-angle dispersion isDirectly inserting the incident vacuum wavelengths into $v_p-\lambda d{v}_p/d\lambda$ would instead give $1.120\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$. That is a plausible intended rough answer, but it mixes wavelength coordinates; the boxed result consistently uses the incident data. The two agree only to first order in a sufficiently small dimensionless dispersion correction.

Taking $L=1.00\mathrm{cm}$ as the distance travelled in the medium, the narrowband group delay isIf $1\mathrm{cm}$ denotes a slab thickness measured normal to the interface instead, the central ray travels $L/\cos 17.80^{\circ }$, giving about $92.1\mathrm{ps}$. The coordinate-mixed velocity above would give $89.3\mathrm{ps}$ for a $1\mathrm{cm}$ path; it should not replace the consistent calculation silently.

The pulse envelope travels at the local group velocity, while individual carrier crests move at the phase velocity. In general, variation of group velocity across the pulse spectrum produces different delays, temporal broadening and chirp. Absorption can reshape the spectrum and envelope as well. The refraction data also show wavelength-dependent angles, so a finite-width incident beam acquires spatial separation of its spectral components. The supplied first derivative determines the central delay but does not specify a pulse duration, absorption spectrum or reliable second derivative of the dispersion relation; it therefore does not determine a numerical broadening, or its sign. If propagation phase is effectively linear across a sufficiently narrow spectrum, the leading approximation is a delayed pulse with unchanged shape.