Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

The filament of a light bulb had 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 100m. assume the pupil of the eyes of the observer to be circular with the radius 3 mm Then....

solution : given, surface area, A = 64 mm² = 64 × 10^-6 m²

temperature, T = 2500 K

distance of bulb from the observer , d = 100 m

radius of pupil, Rp = 3 mm = 0.003 m

for black body, emissivity , e = 1

now power radiated by the filament, P = σAeT⁴

= 5.67 × 10^-8 × 64 × 10^-6 × 1 × (2500)⁴

= 141.75 watt.

option (A) is wrong.

power reaching to the eyes = intensity of radiated energy × cross sectional area of eyes

= P/4πd² × πRp²

= 141.75/(4 × 100²) × (0.003)²

= 3.19 × 10^-8 Watt

option (B) is correct. [ use approximation]

from Wien's displacement law, λ_m × T = b

⇒λ_m × 2500 = 2.9 × 10^-3

⇒λ_m = 2.9 × 10^-3/2500 = 1160 nm

so option (C) is also correct.

power received by one eye of the observer ⇒3.19 × 10^-8 = energy of one photon × number of photons entering into eyes per second

⇒3.19 × 10^-8 = (hc/λ) × n

let's take λ = 1740 nm

⇒3.19 × 10^-8 = 6.63 × 10^-34 × 3 × 10^8/(1740 × 10^-9 × n)

⇒n = (3.19 × 10^-8 × 1740 × 10^-9)/(6.63 × 10^-34 × 3 × 10^8)

= 2.79 × 10¹¹

so option (D) is also correct.

Therefore options (B), (C) and (D) are correct choices.