4.222. (P)I (S) = (e21m)2416a.
4.223. (P)/(S) =
(e ((:)/ m)2 0; 4
)2.
4.224. R = 3P/163-tcypMc x 0.6 p.m.
5.1. (a) 3 and 9 mW; (b) cp = 1/2 (V1 + V2 ) clue
= 1.6 lm,
where A = 1.6 mW/lm, V1 and V2 are the values of relative spectral
response of an eye for the given wavelengths.
5.2. a = V l-LoicoAcD/2 3-cr2 Vx, hence Em = 1.1 V/m, Hm
= 3.0 mA/m. Here A = 1.6 mW/lm, Vx is the relative spectral
response of an eye for the given wavelength.
0.—(R/02 I
5.3. (a) (E)= 1/2E0; (b) (E):=
1—R11
R2 —
50 lx.
5.4. M = 2/3o-cLo.
5.5. (a) cD = nLAS sin2 0; (b) M =
5.6. h R, E = LS/4R2 = 40 lx.
5.7. I = /0/cos3 0,
= nI 0R21h2 = 3.102 lm.
5.8. Erna, = (9/161C) pES/R,2 = 0.21 lx, at the distance
R/V-3 from the ceiling.
5.9. E =
5.10. E
5.11. M = E, (1 + h2/R2, ) = 7.102 lm/m2.
5.12. E0 = aLR2 /h,2 = 25 lx.
5.13. e' = e — 2 (en) n.
5.14. Suppose n1, n2, n3 are the unit vectors of the normals to
the planes of the given mirrors, and e0, e1, e2, e3 are the unit vectors
of the incident ray and the rays reflected from the first, second,
and the third mirror. Then (see the answer to the foregoing problem):
el = e, — 2 (eons) n1, e2 =et — 2 (e1n2) n2, e3 = e2 — 2 (e2n3) ns.
Summing termwise the left-hand and right-hand sides of these
expressions, it can be readily shown that e3 =
5.15. 01 = arctan n = 53°.
5.16. n1/n2 = 1/1/-112 — 1 =1.25.
5.17. x= [1 --V (1— sin2 0)/(n2 — sin2 0)] d sin 0 = 3.1 cm.
5.18. h' = (hn2 cos3 0)/(n2 — sin2 0)3/2.
5.21. 0 = 83°.
5.22. From 37 to 58°.
5.23. a = 8.7°.
2 sin (8/2)
5.24. Aa —
An = 0.44°.
- 171— n2 sin2(0/2)
5.27. (a) f = 11341 — p2) = 10 cm; (b) f = 111132132 -
=
= 2.5 cm.
5.28. I' = P 1012 (f — sr = 2.0.103 cd.
5.29. Suppose S is a point source of light and S' its image
(Fig. 38). According to Fermat's principle the optical paths of all
rays originating at S and converging at S' are equal. Let us draw
335
4.223. (P)/(S) =
(e ((:)/ m)2 0; 4
)2.
4.224. R = 3P/163-tcypMc x 0.6 p.m.
5.1. (a) 3 and 9 mW; (b) cp = 1/2 (V1 + V2 ) clue
= 1.6 lm,
where A = 1.6 mW/lm, V1 and V2 are the values of relative spectral
response of an eye for the given wavelengths.
5.2. a = V l-LoicoAcD/2 3-cr2 Vx, hence Em = 1.1 V/m, Hm
= 3.0 mA/m. Here A = 1.6 mW/lm, Vx is the relative spectral
response of an eye for the given wavelength.
0.—(R/02 I
5.3. (a) (E)= 1/2E0; (b) (E):=
1—R11
R2 —
50 lx.
5.4. M = 2/3o-cLo.
5.5. (a) cD = nLAS sin2 0; (b) M =
5.6. h R, E = LS/4R2 = 40 lx.
5.7. I = /0/cos3 0,
= nI 0R21h2 = 3.102 lm.
5.8. Erna, = (9/161C) pES/R,2 = 0.21 lx, at the distance
R/V-3 from the ceiling.
5.9. E =
5.10. E
5.11. M = E, (1 + h2/R2, ) = 7.102 lm/m2.
5.12. E0 = aLR2 /h,2 = 25 lx.
5.13. e' = e — 2 (en) n.
5.14. Suppose n1, n2, n3 are the unit vectors of the normals to
the planes of the given mirrors, and e0, e1, e2, e3 are the unit vectors
of the incident ray and the rays reflected from the first, second,
and the third mirror. Then (see the answer to the foregoing problem):
el = e, — 2 (eons) n1, e2 =et — 2 (e1n2) n2, e3 = e2 — 2 (e2n3) ns.
Summing termwise the left-hand and right-hand sides of these
expressions, it can be readily shown that e3 =
5.15. 01 = arctan n = 53°.
5.16. n1/n2 = 1/1/-112 — 1 =1.25.
5.17. x= [1 --V (1— sin2 0)/(n2 — sin2 0)] d sin 0 = 3.1 cm.
5.18. h' = (hn2 cos3 0)/(n2 — sin2 0)3/2.
5.21. 0 = 83°.
5.22. From 37 to 58°.
5.23. a = 8.7°.
2 sin (8/2)
5.24. Aa —
An = 0.44°.
- 171— n2 sin2(0/2)
5.27. (a) f = 11341 — p2) = 10 cm; (b) f = 111132132 -
=
= 2.5 cm.
5.28. I' = P 1012 (f — sr = 2.0.103 cd.
5.29. Suppose S is a point source of light and S' its image
(Fig. 38). According to Fermat's principle the optical paths of all
rays originating at S and converging at S' are equal. Let us draw
335