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 = $$5.67 \times 10−8 Wm^{−2}K^{−4}$$, Wien’s displacement constant = $$2.90 \times 10^{−3} m^{-K}$$, Planck’s constant = $$6.63 \times 10^{−34} $$Js, speed of light in vacuum = $$3.00 \times 10^{8} ms^{−1}$$)