The filament of a light bulb has surface area . 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 pupi

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 $= 5.67 \times 10^{- 8} W m^{- 2} K^{- 4}$ , 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 $= 3.00 \times 10^{8} {ms}^{- 1}$ ) [2016]

Smart Achievers · Accepted Answer

(2, 3, 4)

According to question,

Surface area of filament of light bulb, $A = 64 {mm}^{2}$

Temperature of filament, $T = 2500 K$

Distance of bulb (source) from observer, $d = 100 m$

Radius of the pupil of the observer's eye, $R_{e} = 3 mm = 3 \times 10^{- 3} m$

(1) Power radiated by the filament $P = σ A e T^{4}$

$= 5.67 \times 10^{- 8} \times 64 \times 10^{- 6} \times 1 \times {(2500)}^{4} = 141.75 W$ $(∵ e = 1 for black body)$

Hence, option (1) is incorrect.

(2) Radiated power entering into one eye of the observer,

$I = \frac{P}{4 π d^{2}} \times (π R_{e}^{2})$

$= \frac{141.75}{4 π {(100)}^{2}} \times π \times {(3 \times 10^{- 3})}^{2} = 3.189375 \times 10^{- 8} W$

Hence, option (2) is correct.

(3) From Wien's displacement law, $λ_{m} T = b$

or, $λ_{m} \times 2500 = 2.9 \times 10^{- 3}$

or, $λ_{m} = 1.16 \times 10^{- 6} = 1160 nm$

Hence, option (3) is correct.

(4) Total number of photons entering per second into one eye of the observer

$(\frac{h c}{λ}) \times N_{photons} = I$

$3.189375 \times 10^{- 8} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{1740 \times 10^{- 9}} \times N_{photons}$

$∴$ $N_{photons} = 2.79 \times 10^{11}$

Hence, option (4) is correct.

Solution

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

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

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

4 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 $2.85 \times 10^{11}$

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