JEE Advance - Physics (2020 - Paper 1 Offline - No. 13)
The filament of a light bulb has surface area 64 mm2
. 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−2K−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)
(Take Stefan-Boltzmann constant = 5.67 $$ \times $$ 10−8 Wm−2K−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)
power radiated by the filament is in the range 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
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 the range 2.75 $$ \times $$ 1011 to 2.85 $$ \times $$ 1011
Explanation
Here, A = 64 mm2, T = 2500K (where, A = surface area of filament, T = temperature of filament, d is distance of bulb from observer and Re = radius of pupil of eye)
Point source, d = 100 m, Re = 3 mm
(a) P = $$\sigma $$AeT$$-$$4
= 5.67 $$ \times $$ 10$$-$$8 $$ \times $$ 64 $$ \times $$ 10$$-$$6 $$ \times $$ 1 $$ \times $$ (2500)4
($$ \because $$ e = 1 for black body)
= 141.75 W
(b) Power reaching to the eye
= $${P \over {4\pi {d^2}}} \times (\pi R_e^2)$$
$$ = {{141.75} \over {4\pi \times {{(100)}^2}}} \times \pi \times {(3 \times {10^{ - 3}})^2}$$
$$ = 3.189375 \times {10^{ - 8}}$$ W
(c) $${\lambda _m}T = b$$
$${\lambda _m} \times 2500 = 2.9 \times {10^{ - 3}}$$
$$ \Rightarrow {\lambda _m} = 1.16 \times {10^{ - 6}}$$ = 1160 nm
(d) Power received by one eye of observer = $$\left( {{{hc} \over \lambda }} \right) \times {{\Delta N} \over {\Delta t}}$$
where, $${{\Delta N} \over {\Delta t}}$$ = number of photons.
entering into eye per second.
$$ = 3.189375 \times {10^{ - 8}}$$
$$ = {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1740 \times {{10}^{ - 9}}}} \times {{\Delta N} \over {\Delta t}}$$
$$ \Rightarrow {{\Delta N} \over {\Delta t}} = 2.79 \times {10^{11}}$$
Point source, d = 100 m, Re = 3 mm
(a) P = $$\sigma $$AeT$$-$$4
= 5.67 $$ \times $$ 10$$-$$8 $$ \times $$ 64 $$ \times $$ 10$$-$$6 $$ \times $$ 1 $$ \times $$ (2500)4
($$ \because $$ e = 1 for black body)
= 141.75 W
(b) Power reaching to the eye
= $${P \over {4\pi {d^2}}} \times (\pi R_e^2)$$
$$ = {{141.75} \over {4\pi \times {{(100)}^2}}} \times \pi \times {(3 \times {10^{ - 3}})^2}$$
$$ = 3.189375 \times {10^{ - 8}}$$ W
(c) $${\lambda _m}T = b$$
$${\lambda _m} \times 2500 = 2.9 \times {10^{ - 3}}$$
$$ \Rightarrow {\lambda _m} = 1.16 \times {10^{ - 6}}$$ = 1160 nm
(d) Power received by one eye of observer = $$\left( {{{hc} \over \lambda }} \right) \times {{\Delta N} \over {\Delta t}}$$
where, $${{\Delta N} \over {\Delta t}}$$ = number of photons.
entering into eye per second.
$$ = 3.189375 \times {10^{ - 8}}$$
$$ = {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1740 \times {{10}^{ - 9}}}} \times {{\Delta N} \over {\Delta t}}$$
$$ \Rightarrow {{\Delta N} \over {\Delta t}} = 2.79 \times {10^{11}}$$
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