I'm learning with Kenneth Krane Modern Physics 3ed.

When white light travels through glass, the phase velocity of each wavelength depends on the wavelength. (This is the origin of dispersion and the breaking up of white light into its component colors—different wavelengths travel at different speeds and have different indices of refraction.) How does vphase depend on λ? Is dvphase/dλ positive or negative? Therefore, is vgroup >vphase or < vphase?

The index of refraction n for light in glass decreases as λ increases (shorter wavelengths are refracted more than longer wavelengths); that is dn / dλ < 0 . Because n = c / vphase , dn / dλ and dvphase / dλ have opposite signs and so dvphase / dλ > 0. Thus vgroup > vphase.

I don't understand the last formula [vgroup > vphase]. I think it should be [vgroup < vphase] because of the equation in problem 28-(a)[vgroup=vphase - λ(dvphase/dλ)]. Is the solution wrong?


1 Answer 1


The solution does indeed look wrong if everything you have supplied is correct. If the equation $$v_\text{group}=v_\text{phase} - \lambda\frac{dv_\text{phase}}{dλ}$$ holds true, and $$\frac{dv_\text{phase}}{d\lambda}\gt 0$$ then it is certainly true that $$\boxed{v_\text{phase}\gt v_\text{group}}$$

where of course $\lambda\gt 0$. It is true (in general) that for dispersive mediums like glass, the group velocity is usually smaller than the phase velocity.

  • $\begingroup$ I'm really sorry, but I still don't understand. Why it is certainly true that vg>vp? Subtracting λ(dvp/dλ) from vp is vg. Clearly dvp/dλ>0 and λ>0 , then vg<vp. Why am I wrong? $\endgroup$
    – Pulsar
    Commented Apr 3, 2022 at 4:35
  • $\begingroup$ @Pulsar Please excuse my sloppy algebra. I should have had $v_p\gt v_g$ or $v_g\lt v_p$ You are correct and it's probably a typo. $\endgroup$
    – joseph h
    Commented Apr 3, 2022 at 4:51
  • $\begingroup$ Thank you. I thought I was wrong because there is a principle I don't know, but the solution was wrong. $\endgroup$
    – Pulsar
    Commented Apr 3, 2022 at 7:59
  • $\begingroup$ No worries. Good luck with your studies. $\endgroup$
    – joseph h
    Commented Apr 3, 2022 at 8:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.