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QJEEAdv20Ph3ebb: JEE Advanced 2020 Physics - Class 11 JEEAdv20Ph3ebb Solution: The filament of a light bulb has surface area 64 mm 2 . The…

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)

Multiple Choice Options

  • 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 ×\times 10−8 W to 3.25 ×\times 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 ×\times 1011 to 2.85 ×\times 1011

Step-by-Step Verified Solution

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 = σ\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

= P4πd2×(πRe2){P \over {4\pi {d^2}}} \times (\pi R_e^2)

=141.754π×(100)2×π×(3×10−3)2= {{141.75} \over {4\pi \times {{(100)}^2}}} \times \pi \times {(3 \times {10^{ - 3}})^2}

=3.189375×10−8= 3.189375 \times {10^{ - 8}} W

(c) λmT=b{\lambda _m}T = b

λm×2500=2.9×10−3{\lambda _m} \times 2500 = 2.9 \times {10^{ - 3}}

⇒λm=1.16×10−6\Rightarrow {\lambda _m} = 1.16 \times {10^{ - 6}} = 1160 nm

(d) Power received by one eye of observer = (hcλ)×ΔNΔt\left( {{{hc} \over \lambda }} \right) \times {{\Delta N} \over {\Delta t}}

where, ΔNΔt{{\Delta N} \over {\Delta t}} = number of photons.

entering into eye per second.

=3.189375×10−8= 3.189375 \times {10^{ - 8}}

=6.63×10−34×3×1081740×10−9×ΔNΔt= {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1740 \times {{10}^{ - 9}}}} \times {{\Delta N} \over {\Delta t}}

⇒ΔNΔt=2.79×1011\Rightarrow {{\Delta N} \over {\Delta t}} = 2.79 \times {10^{11}}

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