Q: 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 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

(Stefan’s Boltzmann constant = 5.67 x 10^{-8} W m^{-2} K^{-4} , Wien’s displacement constant = 2.90 x 10^{-3} m-K , PlancK ‘ s constant = 6.63 x 10^{-34} J-s , Speed of light in vacuum = 3 x 10^{8} m/s )

(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 x 10^{-8} W to 3.25 x 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 x 10^{11} to 2.85 x 10^{11} .

Ans: (b , c , d)

Solution: Let A = surface area of filament , T = temperature of filament , d = distance of bulb from observer and R_{e} = pupil of eye

A = 64 mm^{2} , T = 2500 K , d = 100 m , R_{e} = 3 mm

(a) $P = \sigma A e T^4 $

$P = 5.67 \times 10^{-8} \times 64 \times 10^{-6} \times 1 \times (2500)^4$ (Since e = 1 foe black body)

P = 141.7 W

(b) Power reaching to the eye

$= \frac{P}{4 \pi d^2} \times (\pi R_e^2 ) $

$\frac{141.7}{4 \pi (100)^2}\times \pi \times (3 \times 10^{-3})^2$

= 3.189375 W

(c) $\lambda_m T = b $

$\lambda_m \times 2500 = 2.9 \times 10^{-3}$

$\lambda_m = 1.16 \times 10^{-6}$

= 1160 nm

(d) Power received by one eye of observer

$= \frac{h c}{\lambda} \times \frac{\Delta N}{\Delta t} $

Where , $\frac{\Delta N}{\Delta t}$ = no. of photons entering eye per second

$3.189375 \times 10^{-8} = \frac{6.63 \times 10^{-34}\times 3 \times 10^8}{1740 \times 10^{-9}} \times \frac{\Delta N}{\Delta t}$

$\frac{\Delta N}{\Delta t} = 2.79 \times 10^{11}$