Question

The filament of a light bulb has surface area $64m{m}^2$. The filament can be considered as a black body at temperature $2500K$ emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of $100m$. Assume the pupil of the eyes of the observer to be circular with radius $3mm$. Then
(Take Stefan-Boltzmann constant $=5.67\times 10^{-8}W{m}^{-2}{K}^{-4}$, 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 $=3.00\times 108m{s}^{-1}$ )

• A. power radiated by the filament is in the range $642W$ to $645W$
• 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 $1160nm$
• D. taking the average wavelength of emitted radiation to be $1740nm$, 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: (B), (C), (D)

$P=A\sigma T^4$
$=64\times 10^{-6}\times 5.67\times 10^{-8}\times (2500{)}^4\approx 140W$
Power entering in eye
$=\frac{140}{4\pi (100{)}^2}\pi {\left(3\times 10^{-3}\right)}^2$
$=\frac{140\times 9\times 10^{-10}}{4}$
$=3.15\times 10^{-8}W$
$\lambda T=b\Rightarrow \lambda =\frac{2.93\times 10^{-6}}{2500}$
$\Rightarrow \lambda =1160$
Photons
$n\frac{12420}{1740}=\frac{3.15\times 10^{-8}}{1.6\times 10^{-19}}$
$=2.75\times 10^{11}$