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 × 10 - 8 W m - 2 K - 4 , Wiens' disp

Step-by-step solution

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 ×