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HW4ETS2221C

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  Homework ETS 2221C mdbedani@hotmail.com 1. A circle has a radius of 42 cm.Find angle θ  in rad(also deg) subtended at the centerof the circle by an arc length of (a)21 cm(b)5.5 m.Solution r  = 42 cm we have  l  =  θ × r  →  θ  =  lr a)  l  = 21 cm . →  θ  =  21 cm 42 cm  →  θ  = 0 . 5 rad  or θ  = 0 . 5 rad ×  180 ◦ πrad  →  θ  = 28 . 6 ◦ b)  l  = 5 . 5 m →  θ  =  550 cm 42 cm →  θ  = 13 . 095 rad  or θ  = 13 . 095 rad ×  180 ◦ πrad  →  θ  = 750 . 3 ◦ 2. Find the solid angle of a sphere of radius750 mm subtended by an area of (a)  15 m 2 (b) 25 cm 2 .Solution r  = 750 mm → r  = 0 . 75 m we have  Ω =  Arear 2 a)  Area  = 15 m 2 . →  Ω =  15 m 2 (0 . 75 m ) 2 →  Ω = 26 . 67 srad b)  Area  = 25 cm 2 . →  Ω =  25 cm 2 (0 . 75 m ) 2 →  Ω = 0 . 00444 srad 3. A G45(18 W) Halogen lamp has a radiusof 1.2 cm and its radiating energy uni-formly outward. The detector used tomeasure the power has an area of   5 mm 2 and is placed at a distance of 25 cm.Determine: (a)Radiant power (b)Radiantintensity (c) Radiant emmitance (d)Irradiance (e) Radiance.Solution r  = 1 . 2 cm  = 0 . 012 mA  = 5 mm 2 d  = 25 cm  = 0 . 25 m a)  φ e  :  Radiant Powerφ e  = 18 W  b)  I  e  :  Radiant Intensity →  I  e  =  ∆ φ e ∆Ω Ω = 4 πsr  (for a sphere) →  I  e  =  18 W  4 πsr →  I  e  = 1 . 43 W sr c)  M  e  :  Radiant Emmitance →  M  e  =  ∆ φ e Area ∆ φ e  = 18 W  ; Area  :  from sourceArea  = 4 πr 2 = 4 π (0 . 012 m ) 2 →  A  = 0 . 00181 m 2 →  M  e  =  18 W  0 . 00181 m 2 →  M  e  = 9 . 9448 KW m 2 1  d)  E  e  :  Irradiance →  E  e  =  ∆ φ e Area Area: total surface at a distance d Area  = 4 πd 2 →  E  e  =  18 W  4 π (0 . 25 m ) 2 →  E  e  = 22 . 93  W m 2 e)  L e  :  Radiance →  L e  =  ∆ φ e ∆Ω∆ A ∆Ω = 4 πsr ∆ A  = 4 πr 2 = 0 . 00181 m 2 →  L e  =  18 W  4 πsr ∗ 0 . 00181 m 2 →  L e  = 62 . 96  W m 2 sr 4. A light source emits 25 mW of opticalpower into a sphere. Find the radiant in-tensity of this light source.Solution φ e  : 25 mW I  e  =?  →  I  e  =  ∆ φ e ∆Ω ∆Ω = 4 πsr  (for a sphere) →  I  e  =  25 mW  4 πsr →  I  e  = 1 . 98 mW sr 5. Determine the maximum wavelength atwhich an osmium (Os) filament radiatesmost strongly when it has a temperatureof   4727 ◦ C  .Solution T   = 4727 ◦ C   = 4727 + 273 . 15 = 5000 . 15 K λ max  =? by Wien’s displacement Law: λ max  =  2898 µmT  ( K  ) ⇒ λ max  =  2898 µm 5000 . 15 ⇒  λ max  = 0 . 58 µm 6. An electric stove burner has a surfacetemperature of   734 ◦ F. Find the wave-length that radiates most strongly at thistemperature.Solution T   = 734 ◦ F   = 663 . 15 K  Obs  K   =  F  − 321 . 8  + 273 . 15 λ max  =? by Wien’s displacement Law: λ max  =  2898 µmT  ( K  )  ⇒ λ max  =  2898 µm 663 . 15 ⇒  λ max  = 4 . 37 µm 7. Calculate the spectral radiance (0.5 to2.5 µ m) of a black body at a temperatureof 5500 K. Plot your results.SolutionSpectral Radiance L λ  =  2 πhc 2 λ 5  .  1( e   hcλkT   − 1) (  W m 2 sr.µm ) for  λ  = 0 . 5 µm  ⇒ L λ  = 1 . 73 ∗ 10 − 4  W m 2 sr.µm for  λ  = 0 . 55 µm  ⇒ L λ  = 1 . 73 ∗ 10 − 4 for  λ  = 0 . 6 µm  ⇒ L λ  = 1 . 67 ∗ 10 − 4 for  λ  = 0 . 65 µm  ⇒ L λ  = 1 . 56 ∗ 10 − 4 for  λ  = 0 . 7 µm  ⇒ L λ  = 1 . 44 ∗ 10 − 4 for  λ  = 0 . 75 µm  ⇒ L λ  = 1 . 31 ∗ 10 − 4 for  λ  = 0 . 8 µm  ⇒ L λ  = 1 . 18 ∗ 10 − 4 2  for  λ  = 0 . 85 µm  ⇒ L λ  = 1 . 06 ∗ 10 − 4 for  λ  = 0 . 9 µm  ⇒ L λ  = 9 . 41 ∗ 10 − 5 for  λ  = 0 . 95 µm  ⇒ L λ  = 8 . 37 ∗ 10 − 5 for  λ  = 1 µm  ⇒ L λ  = 7 . 43 ∗ 10 − 5 for  λ  = 1 . 1 µm  ⇒ L λ  = 5 . 85 ∗ 10 − 5 for  λ  = 1 . 3 µm  ⇒ L λ  = 3 . 66 ∗ 10 − 5 for  λ  = 1 . 5 µm  ⇒ L λ  = 2 . 34 ∗ 10 − 5 for  λ  = 1 . 6 µm  ⇒ L λ  = 1 . 89 ∗ 10 − 5 for  λ  = 1 . 65 µm  ⇒ L λ  = 1 . 7 ∗ 10 − 5 for  λ  = 2 µm  ⇒ L λ  = 8 . 6 ∗ 10 − 6 for  λ  = 2 . 1 µm  ⇒ L λ  = 7 . 17 ∗ 10 − 6 for  λ  = 2 . 5 µm  ⇒ L λ  = 3 . 66 ∗ 10 − 6 3
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