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The filament of a light bulb has surface area $64 m m^{2}$ . 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
$(σ = 5.67 \times 10^{- 8} w m^{- 2} k^{- 4}; b = 2.9 \times 10^{- 3} m k; h = 6.63 \times 10^{- 34} J S; C = 3 \times 10^{8} m / s)$

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power radiated by the filament is in the range 642w to 645w

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 1160nm

Taking the average wave length 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 is B, C, D.

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Detailed Solution

$P = \frac{Q}{t} = σ e A T^{4} P = 64 \times 10^{- 6} \times 5.67 \times 10^{- 8} \times {(2500)}^{4} = 140 w$

Power entering in eye $= \frac{140}{4 π {(100)}^{2}} π {(3 \times 10^{- 3})}^{2} = \frac{140 \times 9 \times 10^{- 10}}{4} = 3.15 \times 10^{- 8} w$

$λ Τ = b \Rightarrow λ = \frac{2.9 \times 10^{- 3}}{2500} = 1160 n m$

No.of photos $n \times \frac{1240}{1740} \times 1.6 \times 10^{- 19} = 3.15 \times 10^{- 8} = 2.76 \times 10^{11}$

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