At a minimum value the derivative must equal zero and so equation (12.17) must be
set equal to zero and xmust be solved for. This is not an easy task and so numerical
methods must be resorted to.
Figure 12-3.
Deviation angle versus incident angle for prism using nominal values above.
An alternative approach of finding the value of xwhich produces a minimum
value is as follows. Observe that when y=ymin , then dydx = 0. Assume that yhas the
value y=ymin and differentiate the equations (12.12) and (12.13) with respect to x
and show
dA
dx =
dα
dx +
dβ
dx = 0 or
dβ
dx =−
dα
dx (12 .18)
because the angle Ais a constant. One also finds that when y=ymin, then
dy
dx = 1 +
dξ
dx = 0 or
dξ
dx =−^1 (12 .19)
because of our assumption that y=ymin and hence dydx = 0. Next differentiate the
equations (12.11) with respect to xand show
nacos x=ngcos αdα
dx
(12 .20)
and
ngcos β
dβ
dx =nacos ξ
dξ
dx (12 .21)
