Question 1[Maximum number: 2]A star of luminosity 3.8×1031 W3.8 \times 10^{31} \mathrm{~W}3.8×1031 W is a distance of 1.8×1024 m1.8 \times 10^{24} \mathrm{~m}1.8×1024 m from the Earth.Calculate the radiant flux intensity at the Earth of the radiation emitted by the star.radiant flux intensity = Wm−2\mathrm{Wm}^{-2}Wm−2Mark as masteredShow AnswerF=L/(4πd2)F=L /\left(4 \pi d^{2}\right)F=L/(4πd2)=(3.8×1031)/[4π×(1.8×1024)2]=9.3×10−19 W m−2\begin{aligned} =\left(3.8 \times 10^{31}\right) /\left[4 \pi \times\left(1.8 \times 10^{24}\right)^{2}\right] =9.3 \times 10^{-19} \mathrm{~W} \mathrm{~m}^{-2} \end{aligned}=(3.8×1031)/[4π×(1.8×1024)2]=9.3×10−19 W m−2Add to Test