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 = 5.67x10 -8 Wm -2 K -4 , Wien's displacement constant =2.90x10 -3 m-K, Planck's constant = 6.63x10 -34 Js, speed of light in vacuum = 3.00x108 ms -1 )
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(b, c, d)


= 64x10 -6 x5.67x10 -8 x(2500) 4
140W
Power entering in eye


Photons

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