114 3 Sources of Light and Illumination Systems
Figure 3.20 Problem 3.14 Configuration with on-axis dark field illumination.
What should be the illumination angle of each LED and what is the maximum
number of LEDs in a single line ring array?
3.5. Solutions to Problems
3.1. (a) The radiation flux emitted by the filament can be calculated as follows:
P
λ
= ε
λ
e
B
(λ, T ) × S
ω × λ
2π
where S = 4. 2×2. 3 ×10
6
= 9. 96 ×10
6
m
2
is the filament irradiated surface,
ω = 1/0. 5
2
= 4 sr is the solid angle of irradiance measurement, λ = 1 nm, and
e
B
(λ, T ) is the hemispherical black body radiation at a wavelength of 0.5 μm. To
calculate the last value we use tables of black body radiation (Appendix 3). Since
λT = 0. 5 × 3, 234 = 1, 617 we should use two lines of the table, of λT = 1, 600
and λT = 1, 700, and also take into account the factor σ T
5
. Then finally we get
e
B
(λ, T ) = 81. 7 × 5. 668 × 10
8
× 3. 234
5
× 10
15
= 163, 814 × 10
7
W/m
3
and
e
B
(λ, T ) × λ = 1, 638. 14 W/m
2
, which results in P
λ
= 0. 8 × 1, 638. 14 ×
9. 96 × 10
6
× 4/2π = 8. 06 mW (at 0.5 m for 1 nm spectral bandwidth).
(b) For U/U
0
= 0. 05 we have T/T = 0. 4(U/U
0
) = 0. 4 × 0. 05 = 0. 02
and therefore the new temperature is T =3, 169 K. Then we proceed using the
table of Appendix 3 exactly in the same way as in (a) above and finally obtain
P
λ
= 6. 73 mW.
3.2. The optical configuration addressed in this problem is presented in Fig. 3.21.
Starting with the cylindrical lens we find first its optical power in the vertical plane:
1
f
= (n 1)/R
1
= 0. 5163/20. 65; f
= 40 mm
3.5. Solutions to Problems 115
Figure 3.21 Problem 3.2 A line generator with a QTH lamp.
and then the location of the line of light along the OZ axis and its height y
:
S
1
=
40 × 60
(60 40)
= 120 mm; V =−120/60 =−2; y
= 4. 2 × 2 = 8. 4 mm.
Obviously in the horizontal plane the lens has no optical power and the ray bundle
of 1 mm width on the line has a width x = 60/(60+120) = 0. 33 mm in the plane
of the lens. Hence, all rays concentrated by the lens in the segment of area B of the
line are those which are transferred by the lens strip A = 20 ×0. 33 = 6. 67 mm
2
.
They constitute the solid angle = A cos ϕ/ρ
2
, where ρ = 60/ cos ϕ. For the
light intensity along the created line we get
I(x) = εe
B
(λ, T ) × λ × s ×
(x)
2πB
cos ϕ(x)
= εe
B
(λ, T ) × λ × s
20 × x
2πB ×60
2
cos
4
ϕ(x) = I
0
cos
4
ϕ(x).
To calculate I
0
we use the data from Problem 3.1:
e
B
(λ, T ) × λ = 1, 628. 14 W/m
2
/nm; s = 9. 96 × 10
6
m
2
and also keep in mind that B = 1 × y
= 8. 4 × 10
6
m
2
; tan ϕ
max
= 150/180;
cos
4
ϕ
max
=0. 348. Therefore, the intensity along the line varies from the max-
imum value I
0
= 0. 439 W/m
2
/nm (in the center) to the minimum value I =
0. 348 × I
0
(at the side).
3.3. (a) As explained in Section 3.2, the two-lens configuration enables one to get
a uniform illumination of the object if the first lens, L
1
, creates the image of the
source S on the second lens, L
2
, and the image of lens L
1
coincides with the object
plane. Such a configuration is shown in Fig. 3.22. To optimize the system with

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