The filament of a light bulb has 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 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 Wm⁻² K⁻⁴, Wien’s displacement constant= 2.90 × 10⁻³ m-K, Planck’s constant = 6.63 × 10⁻³⁴ Js, speed of light in vacuum= 3.00 × 10⁸ ms⁻¹)

A = 64 mm², T = 2500 K (A = surface area of filament, T = temperature of filament, d is distance of bulb from observer, R_e = radius of pupil of eye)
Point source d = 100 m

R_e = 3mm

$\begin{array}{l}\left(\mathrm{A}\right) \mathrm{P}=\sigma {\mathrm{AeT}}^4\\ =5.67\times {10}^{-8}\times 64\times {10}^{-6}\times 1\times \left(2500{)}^4\right(\mathrm{e}=1\text{ black body })\\ =141.75\mathrm{W}\end{array}$

Option (A) is wrong

$\left(B\right) Power reaching to the eye$

$\begin{array}{r}\begin{array}{cc} & =\frac{\mathrm{P}}{4\pi {\mathrm{d}}^2}\times \left(\pi {\mathrm{R}}_{\mathrm{e}}^2\right)\\ & =\frac{141.75}{4\pi \times (100{)}^2}\times \pi \times {\left(3\times {10}^{-3}\right)}^2\\ & =3.189375\times {10}^{-8}\mathrm{W}\\ & \text{ Option (B) is correct }\end{array}\\ \end{array}$

$\left(\mathrm{C}\right){\lambda }_{\mathrm{m}}\mathrm{T}=\mathrm{b}$

$\begin{array}{l}{\lambda }_{\mathrm{m}}\times 2500=2.9\times {10}^{-3}\\ \Rightarrow {\lambda }_{\mathrm{m}}=1.16\times {10}^{-6}\\ =1160\mathrm{nm}\end{array}$

Option (C) is correct

$\text{ (D) Power received by one eye of observer }=\left(\frac{\mathrm{hc}}{\lambda }\right)\times \dot{\mathrm{N}}$

$\dot{\mathrm{N}}=\text{ Number of photons entering into eye per second }$

$\begin{array}{l}\Rightarrow 3.189375\times {10}^{-8}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1740\times {10}^{-9}}\times \dot{\mathrm{N}}\\ \Rightarrow \dot{\mathrm{N}}=2.79\times {10}^{11}\end{array}$

Option (D) is correct