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 W m⁻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 m s⁻¹):
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Question

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 W m⁻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 m s⁻¹):

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G1 Physics · Accepted Answer

Use Stefan-Boltzmann law P=eσA(T⁴−Tₛ⁴), Wien's law λₘT=b, or Newton's law of cooling dθ/dt=−k(θ−θₛ), as applicable. Evaluating the given data/statements gives B: radiated power entering into one eye of the observer is in the range 3.15 ×10⁻⁸ W to 3.25 ×10⁻⁸ W; C: the wavelength corresponding to the maximum intensity of light is 1160 nm; D: taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the rang…