The filament of a light bulb has surface area 64mm². 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 $= 5.67 \times 10^{- 8} W m^{- 2} K^{- 4},$ wien’s displacement constant $= 2.90 \times 10^{- 3} m - K,$ Planck’s constant $= 6.63 \times 10^{- 34} J s,$ speed of light in vacuum $= 3.00 \times 10^{8} m s^{- 1})$

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power radiated by the filament is in the range of 642 W to 645 W

Radiated power entering into one eye of the observer is in the range $3.15 \times 10^{- 8} W to 3.25 \times 10^{- 8} W$

The wavelength corresponding to the maximum intensity of light is 1160 nm

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

answer is B, C, D.

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Detailed Solution

$σ A e T^{4}$
$5.6 \times 10^{- 8} \times 64 \times 10^{- 6} \times 1 \times {(2500)}^{4}$
$14175 \times 10^{- 14} \times 10^{8} \times 10^{4}$
$(a) 141.75 W$
$(b) \frac{σ A e T^{4}}{4 π {(100)}^{2}} \times π {(3 \times 10^{- 3})}^{2} = 141.75 \times \frac{9 \times 10^{- 6}}{4 \times 10^{4}}$
$318.937 \times 10^{- 10}; 3.18937 \times 10^{- 8} ω$
$(c) λ T = b; λ = \frac{2.93 \times 10^{- 6}}{2500} = 1160 n m$
$(d) 3.18937 \times 10^{- 8} = (\frac{n}{\sec}) \frac{h c}{λ}$

$3.18937 \times \frac{10^{- 8} λ}{λ e} = n = 279.00 \times \frac{10^{- 8} \times 10^{- 9}}{10^{- 37} \times 10^{8}}$
$n = 279 \times 10^{- 17} \times 10^{34} \times 10^{- 8}$