 Which wavelength enters the derivative formula

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2017/paper-2/section-b/b7/which-wavelength-enters-the-derivative-formula

For the wavelength in the medium,

λ_{m} = 2 π / k

, differentiate

ω = k v_{p} (k)

v_{g} = v_{p} + k \frac{d v _{p}}{d k} = v_{p} + k \frac{d v _{p}}{d λ _{m}} (- \frac{λ _{m}}{k}), v_{g} = v_{p} - λ_{m} \frac{d v _{p}}{d λ _{m}} .

λ_{0} = 2 π c / ω

k = ω / v_{p} (λ_{0})

instead gives

v_{g} = \frac{v _{p}}{1 + ( λ _{0} / v _{p} ) d v _{p} / d λ _{0}} = \frac{c}{n - λ _{0} d n / d λ _{0}} .

Thus the printed derivative identity is valid, but its wavelength coordinate cannot be carried unchanged into data specified before refraction.

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