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 = $$5.67 \times 10−8 Wm^{−2}K^{−4}$$, Wien’s displacement constant = $$2.90 \times 10^{−3} m^{-K}$$, Planck’s constant = $$6.63 \times 10^{−34} $$Js, speed of light in vacuum = $$3.00 \times 10^{8} ms^{−1}$$)

The filament behaves like an ideal black‐body radiator.
Given data: surface area $$A = 64\ \text{mm}^2 = 64\times 10^{-6}\,\text{m}^2 = 6.4\times 10^{-5}\,\text{m}^2$$, temperature $$T = 2500\ \text{K}$$, distance to observer $$R = 100\ \text{m}$$, radius of pupil $$r = 3\ \text{mm} = 3\times 10^{-3}\,\text{m}$$.

1. Power radiated by the filament
Stefan-Boltzmann law: $$P = \sigma A T^{4}$$ where $$\sigma = 5.67\times 10^{-8}\,\text{W\,m}^{-2}\text{K}^{-4}$$.
$$T^{4} = (2500)^4 = 3.90625\times 10^{13}$$
$$P = (5.67\times 10^{-8})(6.4\times 10^{-5})(3.90625\times 10^{13})$$
$$P = 141.8\ \text{W}\;(\text{approximately})$$
The stated range in Option A is $$642-645\ \text{W}$$, which does not agree with our value. Hence Option A is incorrect.

2. Power that enters one eye
The bulb appears as a point source; power is spread uniformly over the sphere of radius $$R$$.
Surface area of that sphere: $$4\pi R^{2} = 4\pi(100)^2 = 1.2566\times 10^{5}\,\text{m}^2$$.
Intensity at the observer: $$I = \dfrac{P}{4\pi R^{2}} = \dfrac{141.8}{1.2566\times 10^{5}} = 1.13\times 10^{-3}\,\text{W\,m}^{-2}$$.
Area of one pupil: $$A_{\text{pupil}} = \pi r^{2} = \pi(3\times 10^{-3})^{2} = 2.8274\times 10^{-5}\,\text{m}^2$$.
Power entering one eye: $$P_{\text{eye}} = I A_{\text{pupil}} = (1.13\times 10^{-3})(2.8274\times 10^{-5})$$ $$P_{\text{eye}} \approx 3.19\times 10^{-8}\,\text{W}$$.
This lies inside the range $$3.15\times 10^{-8}\,\text{W} - 3.25\times 10^{-8}\,\text{W}$$ quoted in Option B, so Option B is correct.

3. Wavelength of maximum intensity
Wien’s law: $$\lambda_{\text{max}}T = b,\qquad b = 2.90\times 10^{-3}\,\text{m·K}$$.
$$\lambda_{\text{max}} = \dfrac{b}{T} = \dfrac{2.90\times 10^{-3}}{2500} = 1.16\times 10^{-6}\,\text{m} = 1160\ \text{nm}$$.
Option C states 1160 nm, therefore Option C is correct.

4. Number of photons entering one eye per second
Assume average wavelength $$\lambda_{\text{avg}} = 1740\ \text{nm} = 1.74\times 10^{-6}\,\text{m}$$.
Energy of one photon: $$E = \dfrac{hc}{\lambda_{\text{avg}}} = \dfrac{(6.63\times 10^{-34})(3.00\times 10^{8})}{1.74\times 10^{-6}} = 1.14\times 10^{-19}\,\text{J}$$.
Number of photons per second: $$N = \dfrac{P_{\text{eye}}}{E} = \dfrac{3.19\times 10^{-8}}{1.14\times 10^{-19}} \approx 2.8\times 10^{11}\ \text{s}^{-1}$$.
Option D gives the range $$2.75\times 10^{11} - 2.85\times 10^{11}$$, matching our result; hence Option D is correct.

5. Final conclusion
The correct statements are:
Option B (power entering one eye), Option C (wavelength of maximum intensity), and Option D (number of photons).

Answer: Option B, Option C, Option D.