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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