The filament of a light bulb has surface area 64 mm2. 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 Wm−2 K−4, Wien’s displacement constant= 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)

A = 64 mm2, T = 2500 K (A = surface area of filament, T = temperature of filament, d is distance of bulb from observer, Re = radius of pupil of eye)Point source d = 100 mRe = 3mm(A) P=σAeT4=5.67×10−8×64×10−6×1×(2500)4(e=1 black body )=141.75WOption (A) is wrong(B) Power reaching to the eye =P4πd2×πRe2 =141.754π×(100)2×π×3×10−32 =3.189375×10−8W  Option (B) is correct  (C)λmT=bλm×2500=2.9×10−3⇒λm=1.16×10−6=1160nmOption (C) is correct (D) Power received by one eye of observer =hcλ×N˙N˙= Number of photons entering into eye per second ⇒3.189375×10−8=6.63×10−34×3×1081740×10−9×N˙⇒N˙=2.79×1011Option (D) is correct