The filament of a light bulb has surface area 64 mm 2 . The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the l

Question

The filament of a light bulb has surface area $64 {mm}^{2} .$ The filament can be considered as a black body at temperature $2500 K$ emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of $100 m$ . Assume the pupil of the eyes of the observer to be circular with radius $3 mm$ . Then:

(Take Stefan-Boltzmann constant $= 5.67 \times 10^{- 8} W m^{- 2} K^{- 4},$ Wiens' displacement constant $= 2 .90 \times 10^{- 3} m K$ , Planck's constant $= 6 .60 \times 10^{- 34} J s$ , speed of light in vacuum $= 3 .00 \times 10^{8} m s^{- 1}$ )

TestHub · Accepted Answer

P = σ A e T 4 P = 5 . 6 × 10 - 8 × 64 × 10 - 6 × 1 × ( 2500 ) 4 P = 14175 × 10 - 14 × 10 8 × 10 4 (a) P = 141 . 75 W (b) σ A e T 4 4 π ( 100 ) 2 × π 3 × 10 - 3 2 = 141 . 75 × 9 × 10 - 6 4 × 10 4 318 . 937 × 10 - 10 3 . 18937 × 10 8 W (c) λ T = b λ = 2 . 93 × 10 - 6 2500 = 1160 nm (d) 3 . 18937 × 10 - 8 = n sec h c λ 3 . 18937 × 10 - 8 λ λ e = n = 279 . 00 × 10 - 8 × 10 - 9 10 - 34 × 10 8 n = 279 × 10 - 17 × 10 34 × 10 - 8 n = 2 . 79 × 10 11

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