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PhysicsMultiple correct+4 / -2Question 8 of 54

JEE Advanced 2020 · Paper 1, Physics question 8

The filament of a light bulb has surface area 64 mm264\ \mathrm{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 =5.67×10−8 W m−2K−4= 5.67\times10^{-8}\ \mathrm{W\,m^{-2}K^{-4}}, Wien's displacement constant =2.90×10−3= 2.90\times10^{-3} m-K, Planck's constant =6.63×10−34= 6.63\times10^{-34} Js, speed of light in vacuum =3.00×108 m s−1= 3.00\times10^{8}\ \mathrm{m\,s^{-1}})

Choose one or more options
Answer and worked solution

Official answer: B, C, D

(A) Power.
P=σAT4=5.67×10−8×64×10−6×(2500)4=141.75 W.P=\sigma AT^4=5.67\times10^{-8}\times64\times10^{-6}\times(2500)^4=141.75\text{ W}.
This is not in 642-645 W, so (A) is false.

(B) Into one eye. The fraction of the isotropic output caught by the pupil is πr2/(4πd2)\pi r^2/(4\pi d^2):
141.75×(3×10−3)24×(100)2=3.19×10−8 W.141.75\times\frac{(3\times10^{-3})^2}{4\times(100)^2}=3.19\times10^{-8}\text{ W}.
This is in range, so (B) is true.

(C) Peak wavelength.
λm=2.90×10−32500=1.16×10−6 m=1160 nm.\lambda_m=\frac{2.90\times10^{-3}}{2500}=1.16\times10^{-6}\text{ m}=1160\text{ nm}.
(C) is true.

(D) Photons per second.
hcλ=6.63×10−34×3×1081740×10−9=1.143×10−19 J,N=3.19×10−81.143×10−19=2.79×1011.\frac{hc}\lambda=\frac{6.63\times10^{-34}\times3\times10^8}{1740\times10^{-9}}=1.143\times10^{-19}\text{ J},\qquad N=\frac{3.19\times10^{-8}}{1.143\times10^{-19}}=2.79\times10^{11}.
(D) is true.

(B), (C), (D)

Topics:Radiation and Stefan's lawWien's displacement lawPhotoelectric effect
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