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 $=5.67 \times 10^{-8}\, Wm ^{-2} K ^{-4}$, 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 $=3.00 \times 108\, ms ^{-1}$ )

• 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: (B), (C), (D)

$ P = A \sigma T ^{4}$
$=64 \times 10^{-6} \times 5.67 \times 10^{-8} \times(2500)^{4} \approx 140\, W$
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
$n \frac{12420}{1740}=\frac{3.15 \times 10^{-8}}{1.6 \times 10^{-19}}$
$=2.75 \times 10^{11}$