(31)
dy
2
=
Ds
z, y
2a
Now integrate
s across the mirror. The slope of the tangent along the z direction is
(32)
¹Ds
z, y
¹z
=
h
+
(2a

z) tan x
cos
2
x
c
+
O
c
2
The change in angle, averaged over the 2a by 2b plane, is then
(33)
dy
2
=
1
4 a b
°
0
2a
°

b
b
w
z, y
¹Ds
z, y
¹z
dy dz
Evaluating (33) with w(z,y) = 1, we have simply
(34)
dy
2
=
h
+
a tan x
cos
2
x c
+
O
c
2
Solving for
, we find that the temperature gradient constraint is
(35)
b [
cos
2
x
2 a
h
+
a tan x
t
+
O(t
2
)
For the case h = 0,
,
,
, and
, we have
x
=
45 deg t
=
25 lrad a
=
30 cm
a
=
2 $ 10

8
K

1
cos
2
x
2 a a tan x t
=
5.1 mK
m
The tolerance on the transverse temperature gradient due to a constraint on the basic angle devia

tion is similar to that of the longitudinal gradient, differing by a factor and, more impor
tan x
tantly, by a dependence on the distance
h.
3.3. Beam Divergence.
The curvature of the mirrors introduced by a transverse temperature gradient will cause the colli

mated input beam to diverge. Using eq. (16) to calculate the radius of curvature for the surface
function given by eq. (30), we find
(36)
f
a
=

cos
2
x
2c tan x
and
f
b
= 
1
2c tan x
where f
a
and f
b
are again the focal lengths along the long and short dimensions. Note the similari

ties to eqs. (17). A CTE of
and a temperature gradient
= 5 mK/m again pro
a
=
2 $ 10

8
K

1
duce focal lengths of 2.5 and 5.0 gigameters, since I'm using
= 45 deg. The conclusions of
section 2.4 also hold for the transverse case.
Basic Angle Temperature Gradient Sensitivity
FTMUSNO9501
10