Using the definition of the group velocity `u`, derive Rayleigh's fromula `(5.5d)`. Demonstrate that in the vicinity of `lambda = lambda'` the velocity `u` is equal to the segment `upsilon'` cut by the tangent of the curve `upsilon(lambda)` at the point `lambda`' (Fig.)

Explanation

By definition
`u = (d omega)/(dk) = (d)/(dk) (vk)` as `omega = vk = v+ k(dv)/(dk)`
Now `k = (2pi)/(lambda)` so `dk =- (2pi)/(lambda^(2)) d lambda`
Thus `u = v - lambda (dv)/(d lambda)`.
Its interpretation is the following:
`((dv)/(d lambda))_(lambda-lambda')` is the slope of the `v - lambda` curve at `lambda = lambda'`.
Thus as is obvious from the diagram
`v' = v(lambda') - lambda' ((dv)/(d lambda))_(lambda= lambda')` is the group velocity for `lambda = lambda'`