Irodov – Problems in General Physics

4.222. (P)I (S) = (e 2 1m) 2 416a.

4.223. (P)/(S) =

(e ((:)/ m)20;)2.^4

4.224. R = 3P/163-tcypMc x 0.6 p.m.

5.1. (a) 3 and 9 mW; (b) cp = 1 /2 (V1 + V 2 ) clue = 1.6 lm,

where A = 1.6 mW/lm, V 1 and V2 are the values of relative spectral

response of an eye for the given wavelengths.

5.2. a = V l-LoicoAcD/23-cr^2 Vx, hence Em = 1.1 V/m, H m^

= 3.0 mA/m. Here A = 1.6 mW/lm, V x is the relative spectral

response of an eye for the given wavelength.

0.—(R/02 I

5.3. (a) (E)= 1 / 2 E 0 ; (b) (E):= (^) 1—R11 R 2 — 50 lx.
5.4. M = 2 / 3 o-cL o.
5.5. (a) cD = nLAS sin^2 0; (b) M =
5.6. h R, E = LS/4R^2 = 40 lx.
5.7. I = / 0 /cos 3 0, = nI 0 R 2 1h 2 = 3.10 2 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, ) = 7.10 2 lm/m 2.
5.12. E 0 = aLR 2 /h,^2 = 25 lx.
5.13. e' = e — 2 (en) n.
5.14. Suppose n 1 , n 2 , n 3 are the unit vectors of the normals to
the planes of the given mirrors, and e^0 , e^1 , e^2 , e^3 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) n^1 , e^2 =e — 2 (et^1 n^2 ) n^2 , e^3 = e^2 — 2 (e^2 n^3 ) ns.
Summing termwise the left-hand and right-hand sides of these
expressions, it can be readily shown that e^3 =
5.15. 0 1 = arctan n = 53°.
5.16. n 1 /n^2 = 1/1/-^112 — 1 =1.25.
5.17. x= [1 --V (1— sin 2 0)/(n 2 — sin 2 0)] d sin 0 = 3.1 cm.
5.18. h' = (hn 2 cos 3 0)/(n 2 — sin 2 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 —n^2 sin^2 (0/2)
  5.27. (a) f = 11341 — p2) = 10 cm; (b) f = 11113 213 2 - =
  = 2.5 cm.
  5.28. I' = P 1012 (f — sr = 2.0.10 3 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

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