Answer:
0.049168726 light-years
Step-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.
