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 pupil of the eyes of the observer to be circular with radius 3 mm. Then
(Take Stefan-Boltzmann constant , Wien's displacement constant , Planck's constant , speed of light in vacuum )-
(Take Stefan-Boltzmann constant , Wien's displacement constant , Planck's constant , speed of light in vacuum )-
- (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 to ✓ Correct
- (C)the wavelength corresponding to the maximum intensity of light is 1160 nm✓ Correct
- (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 to ✓ Correct
Answer: (B), (C), (D)
Solution
(A) Power radiated P = σAeT⁴ = 5.67×10⁻⁸ × 64×10⁻⁶ × 1 × (2500)⁴ = 141.75 W (not in the stated range). (B) Power reaching eye = P/(4πd²) × πR_e² = 141.75/(4π×100²) × π×(3×10⁻³)² = 3.19×10⁻⁸ W ✓. (C) Wien's law: λ_m T = b gives λ_m = 2.9×10⁻³/2500 = 1.16×10⁻⁶ m = 1160 nm ✓. (D) Power = (hc/λ)Ṅ gives Ṅ = 3.19×10⁻⁸/[(6.63×10⁻³⁴ × 3×10⁸)/(1740×10⁻⁹)] = 2.79×10¹¹ ✓.
JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · One or more correct options