Question

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 =$567 \times 10^{-8}$, $Wm ^{-2} K ^{-4}$, Wien's displacement constant =$290 \times 10^{-3}$,$m - K$, Planck's constant =$663 \times 10^{-34} \,Js$, speed of light in vacuum =$300 \times 108\, ms ^{-1} )$

• A. power radiated by the filament is in the range $642\, W$to $645\, W$
• B. radiated power entering into one eye of the observer is in the range $3.15 \times 10^{-8} \, W$ to $3.25 \times 10^{-8}\, W$
• 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 range $2.75 \times 10^{11}$ to $2.85 \times 10^{11}$

Answer 1

Answer: (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 range $2.75 \times 10^{11}$ to $2.85 \times 10^{11}$

Answer 2

The Correct Option is (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 range $2.75 \times 10^{11}$ to $2.85 \times 10^{11}$