6.4. Solutions to Problems 221
6.10. In the optical configuration shown in Fig. 6.7a the imaging lens has a focal
length of 100 mm and creates an interference pattern at magnification V =−1/3.
Assuming the test object is of 50 mm in length and 15 mm in height and the
maximum temperature gradient in it is 25
C/mm, find the minimum required
diameter of the lens.
[Note: The refractive index of the object material is n = 1. 3; and dn/dT = 10
4
].
6.11. The temperature profile in an object of 160 mm in length and 38 mm in
height made of transparent material with n = 1. 5 and dn/dt = 10
6
is measured
using the interferometric system shown in Fig. 6.7a. The interference pattern cre-
ated at optical magnification V =−0. 5 in the plane P includes 18 fringes with the
following spacing: d
1
=0. 12 mm; d
2
=0. 20 mm; d
3
=0. 27 mm; d
4
=0. 41 mm;
d
5
=0. 63 mm; d
6
=0. 76 mm; d
7
=0. 94 mm; d
8
=1. 39 mm; d
9
=1. 56 mm;
d
10
=1. 80 mm; d
11
=1. 82 mm; d
12
=1. 88 mm; d
13
=1. 80 mm; d
14
=
1. 74 mm; d
15
=1. 74 mm; d
16
=1. 04 mm; d
17
=0. 48 mm; d
18
=0. 48 mm. Find
the temperature profile in the sample if the minimum temperature (at the bottom)
is 600
C.
6.4. Solutions to Problems
6.1. (a) The optical system is operated at magnification V = S
/S f
/S =
50/5, 000 =−0. 01. Hence the area of the wall from which radiation reaches
the detector is determined as
A =
π(d
det
)
2
4V
2
=
9π
4 × 10
4
= 0. 0707 m
2
.
Due to the final size of the lens only a fraction of radiation emitted by A enters the
detector and this fraction is
ω
2π
=
πD
2
2π × 4 × S
2
=
30
2
× 10
6
8 × 5
2
= 4. 5 × 10
6
.
The total energy emitted by each 1 m
2
of the wall of T = 1, 750 +273 = 2, 023 K
in the spectral interval 0.4–1.1 μm can be calculated using the black body radiation
table presented in Appendix 3:
E = ε
1.1
0.4
e
B
(λ, T )dλ = ε
1.1
0
e
B
(λ, T )dλ
0.4
0
e
B
(λ, T )dλ
= ε(I
2
I
1
)
where I
1
corresponds to λ
1
T =0. 4 × 2, 023 =809 μm K and I
2
correspo-
nds to λ
2
T =1. 1 × 2, 023 =2, 225 μm K, so that interpolation between the

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