Observe that the sum of the angles of the triangle ABC with top angle Aand
base angles π 2 −αand π 2 −βis πradians so that one can sum the angles of the triangle
and write
π
2 −α+
π
2 −β+A=π or A=α+β (12 .12)
Here the deviation angle yis the exterior angle of a triangle with γ and δthe two
opposite interior angles. Consequently one can write
y=γ+δ= (x−α) + (ξ−β) or y=x−A+ξ (12 .13)
Use equation (12.12) to show
sin β= sin(A−α) = sin Acos α−sin αcos A (12 .14)
and then use the equations (12.11) in the form
sin β=
ng
nasin ξ, sin α=
na
ngsin x, cos α=
√
1 −
(
na
ng
) 2
sin^2 x
to express equation (12.14) in the form
na
ngsin ξ= sin A
√
1 −sin^2 α−
na
ngsin xcos A
na
ngsin ξ= sin A
√
1 −
(
na
ng
) 2
sin^2 x−
na
ngsin xcos A
sin ξ=^1
na
sin A
√
n^2 g−n^2 asin^2 x−sin xcos A
(12 .15)
The equations (12.15) together with equation (12.13) can be used to express yas a
function of x. One finds that the angle of deviation in terms of the incident angle x
can be expressed in the form
y=x−A+ sin −^1
[
1
nasin A
√
n^2 g−n^2 asin^2 x−cos Asin x
]
(12 .16)
The figure 12-3 illustrates a graph of the angle of deviation as a function of the
incident value for selected nominal values of A, n a, n g. Observe that there is some
incident angle where the angle of deviation is a minimum. To find out where this
minimum value occurs one can differentiate equation (12.16) to obtain
dy
dx
= 1 −
na√sin(A) sin(x) cos(x)
√ ng^2 −na^2 sin^2 (x) + cos(A) cos(x)
1 −
(
cos(A) sin(x)−sin(A)
√
ng^2 −na^2 sin^2 (x)
na
) 2 (12 .17)
