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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.) |
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Answer» Solution :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'` 
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