JEE Advanced 2020 Paper 1

Section: Physics

Question : 8 of 54

Marks: +1, -0

The filament of a light bulb has surface area 64mm2. 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
(Take Stefan-Boltzmann constant =5.67×10−8Wm−2K−4, Wien's displacement constant =2.90×10−3 m−K, Planck's constant =6.63×10−34Js, speed of light in vacuum =3.00×108ms−1)

[JEE Adv 2020 P1]

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 3.15×10−8W to 3.25×10−8W
The wavelength corresponding to the maximum intensity of light is 1160nm
Taking the average wavelength of emitted radiation to be 1740nm, the total number of photons entering per second into one eye of the observer is in the range 2.75×1011 to 2.85×1011

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