Home Maths JEE - Advanced Previous Year Paper JEE-Advanced-2020-Paper-1 The filament of a light bulb has surface are…
Maths JEE - Advanced Previous Year Paper JEE-Advanced-2020-Paper-1 Single Correct MCQ
Published on: August 14, 2026

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 = 5.67x10 -8 Wm -2 K -4 , Wien's displacement constant =2.90x10 -3 m-K, Planck's constant = 6.63x10 -34 Js, speed of light in vacuum = 3.00x108 ms -1 )

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 X 10 -8 W to 3.25x 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 2.75x10 11 to 2.85x10 11

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CHECK THE SOLUTION.

(b, c, d)

= 64x10 -6 x5.67x10 -8 x(2500) 4 140W

Power entering in eye

Photons

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