The filament of a light bulb has a surface area of $64 \ mm^2$. The filament can be considered as a black body at a temperature of $2500 \ K$,emitting radiation like a point source when viewed from afar. At night,the light bulb is observed from a distance of $100 \ m$. Assume the pupil of the observer's eye to be circular with a radius of $3 \ mm$. Then:
(Take Stefan-Boltzmann constant $= 5.67 \times 10^{-8} \ W \ m^{-2} \ K^{-4}$,Wien's displacement constant $= 2.90 \times 10^{-3} \ m \ K$,Planck's constant $= 6.63 \times 10^{-34} \ J \ s$,speed of light in vacuum $= 3.00 \times 10^8 \ m \ s^{-1}$)
$(A)$ The power radiated by the filament is in the range $642 \ W$ to $645 \ W$.
$(B)$ The radiated power entering into one eye of the observer is in the range $3.15 \times 10^{-8} \ W$ to $3.25 \times 10^{-8} \ W$.
$(C)$ 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 $2.75 \times 10^{11}$ to $2.85 \times 10^{11}$.

• A
  $A, B, C$
• B
  $A, B, D$
• C
  $A, C$
• D
  $B, C, D$