Sol. $A=64 \mathrm{~mm}^2, T=2500 \mathrm{~K}(A=$ surface area of filament, $T=$ temperature of filament, $d$ is distance of bulb from observer, $R_e=$ radius of pupil of eye) Point source $\mathrm{d}=100 \mathrm{~m}$ $$ \mathrm{R}_{\mathrm{e}}=3 \mathrm{~mm} $$ (A) $$ \begin{aligned} \mathrm{P}= & \sigma \mathrm{AeT}^4 \\ & =5.67 \times 10^{-8} \times 64 \times 10^{-6} \times 1 \times(2500)^4(\mathrm{e}=1 \text { black body }) \\ & =141.75 \mathrm{w} \end{aligned} $$ Option (A) is wrong (B) Power reaching to the eye $$ \begin{aligned} & =\frac{\mathrm{P}}{4 \pi \mathrm{~d}^2} \times\left(\pi \mathrm{R}_{\mathrm{e}}^2\right) \\ & =\frac{141.75}{4 \pi \times(100)^2} \times \pi \times\left(3 \times 10^{-3}\right)^2 \\ & =3.189375 \times 10^{-8} \mathrm{~W} \end{aligned} $$ Option (B) is correct (C) $$ \begin{aligned} \lambda_{\mathrm{m}} \mathrm{~T} & =\mathrm{b} \\ & \lambda_{\mathrm{m}} \\ \Rightarrow & \times 2500=2.9 \times 10^{-3} \\ \Rightarrow \lambda_{\mathrm{m}} & =1.16 \times 10^{-6} \\ & =1160 \mathrm{~nm} \end{aligned} $$ Option (C) is correct (D) Power received by one eye of observer $=\left(\frac{\mathrm{hc}}{\lambda}\right) \times \dot{\mathrm{N}}$ $\dot{\mathrm{N}}=$ Number of photons entering into eye per second $$ \begin{aligned} & \Rightarrow 3.189375 \times 10^{-8} \\ & \quad=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1740 \times 10^{-9}} \times \dot{\mathrm{N}} \end{aligned} $$ $$ \Rightarrow \dot{\mathrm{N}}=2.79 \times 10^{11} $$ Option (D) is correct