Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bulb, in which most of the 50 watts goes to heat and only about 5-7 watts is emitted as visible light.) How far away would you have to put the light bulb for it to have the same apparent brightness as Alpha Centauri A in our sky? (Hint: Use 50 watts as L in the inverse square law for light, and use the apparent brightness given above for Alpha Centauri A. Then solve for the distance.)

Answer:0.049168726 light-yearsStep-by-step explanation:The apparent brightness of a star is [tex]\bf B=\displaystyle\frac{L}{4\pi d^2}[/tex] where L = luminosity of the star (related to the Sun) d = distance in ly (light-years) The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly. Hence the apparent brightness of  Alpha Centauri A is [tex]\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728[/tex] According to the inverse square law for light intensity [tex]\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}[/tex] where [tex]\bf I_1=[/tex] light intensity at distance [tex]\bf d_1[/tex]  [tex]\bf I_2=[/tex] light intensity at distance [tex]\bf d_2[/tex]   Let [tex]\bf d_2[/tex]  be the distance we would have to place the 50-watt bulb, then replacing in the formula [tex]\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}[/tex] Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.