If some lightbulb has luminous flux $P$, and it is at the distance $r$, then the illuminance for this lightbulb is $E=frac{P}{4 pi r^2}$ For two lightbulbs at the distance $2r$, the total illuminance is a superposition of illuminance of each lightbulb

$$
begin{align*}
E^{‘}&=E_{1}+E_{2}\
E^{‘}&=frac{P}{4 pi (2r)^2}+frac{P}{4 pi (2r)^2}\
E^{‘}&= 2 frac{P}{4 pi 4r^2}\
E^{‘}&= frac{2}{4} frac{P}{4 pi r^2}\
E^{‘}&= frac{1}{2} frac{P}{4 pi r^2}\
E^{‘}&= frac{1}{2} E\
end{align*}
$$

So, we can see the illuminances are not the same. One lightbulb provides more illuminance than two identical lightbulbs at twice the distance.

If two lapms illuminate a screen equally, that means the illuminances of them are equal, $E_{A}=E_{B}$. The distance for lamp A is $r_{A}=5.0 hspace{0.5mm} mathrm{m}$, and the distance for lamp B is $r_{B}=3.0 hspace{0.5mm} mathrm{m}$. If we know that a lamp A is rated $75 hspace{0.5mm} mathrm{cd} Rightarrow P_{A}=75 cdot 4 hspace{0.5mm} pi mathrm{lm} Rightarrow P_{A}=942.5 hspace{0.5mm} mathrm{lm}$, then we can calculate
begin{align*}
E_{A}&=E_{B}\
frac{P_{A}}{4 hspace{0.5mm} pi r_{A}^{2}}&= frac{P_{B}}{4 hspace{0.5mm} pi r_{B}^{2}}\
P_{A} hspace{0.5mm} r_{B}^{2} &= P_{B} hspace{0.5mm} r_{A}^{2}\
P_{B}&= P_{A} frac{ r_{B}^{2}}{ r_{A}^{2}}\
P_{B}&= 942.5 hspace{0.5mm} mathrm{lm} frac{ (3.0 hspace{0.5mm} mathrm{m})^{2}}{ (5.0 hspace{0.5mm} mathrm{m})^{2}}\
P_{B}&= 942.5 hspace{0.5mm} mathrm{lm} frac{9.0}{ 25.0}\
P_{B}&= 339.3 hspace{0.5mm} mathrm{lm}\
end{align*}
The lamp B is rated $frac{339.3 hspace{0.5mm} mathrm{lm}}{4 pi}$
[ framebox[1.1width]{$ therefore 27.0 hspace{0.5mm} mathrm{cd} $}]

The lamp B is rated $27.0 hspace{0.5mm} mathrm{cd}$

First, we can write the equation for a case when a lightbulb provides only half the illuminance [E=frac{P}{4 pi r^2}] where $r=1.0mathrm{,,m}$ is the distance betweem a lightbulb and a desk. The luminous flux is the same for both cases, because it is the same source of light in both of the cases. To find double illuminance we can write
begin{align*}
E^{‘}&=2E\
frac{P}{4 pi r^{‘2}}&=2 frac{P}{4 pi r^2}\
r^2&=2r^{‘2}\
r^{‘}&= frac{1}{sqrt{2}}r\
r^{‘}&= frac{1}{sqrt{2}} cdot 1.0mathrm{,, m}\
r^{‘}&=0.7 mathrm{,,m}\
end{align*}
To provide the correct illuminance, the lightbulb should be at [ framebox[1.1width]{$ therefore r^{‘}=0.7mathrm{,,m} $}]

$$
r^{‘}=0.7 mathrm{,,m}
$$

the time it takes sound to travel 1 cm:

$t = dfrac{d}{c_{sound}} = dfrac{0.01}{343} = 2.915 times 10^{-5} s$

The distance light travels:

$$
d = c_{light} t = (3 times 10^8)*(2.915times 10^{-5}) = 8750 m
$$

$$
8750 m
$$

We know that the light travels from Earth to the Moon, and back for $t=2.562 hspace{0.5mm} mathrm{s}$. That means, it travels double distance from Earth to the Moon for that time, so we can write [2l=ct] where $l$ is the distance from Earth to the Moon, and $c=3 cdot 10^{8}hspace{0.5mm} mathrm{frac{m}{s}}$ is speed of light.
begin{align*}
2l&= c t / : 2\
l&= frac{ct}{2}\
l&= frac{3 cdot 10^{8}hspace{0.5mm} mathrm{frac{m}{s}} cdot 2.562 hspace{0.5mm} mathrm{s} }{2}\
l&= frac{7.686 cdot 10^{8}hspace{0.5mm} mathrm{m} }{2}\
l&=3.843 cdot 10^{8}hspace{0.5mm} mathrm{m}
end{align*}
The distance from Earth to the Moon is [ framebox[1.1width]{$ therefore l=3.843 cdot 10^{8}hspace{0.5mm} mathrm{m} $}]

$$
l=3.843 cdot 10^{8}hspace{0.5mm} mathrm{m}
$$

We know that the time takes for light to cross Earth’s orbit is $t=16.5 hspace{0.5mm} mathrm{min} Rightarrow t=990 hspace{0.5mm} mathrm{s}$, and the diameter of Earth’s orbit is $d=2.98 cdot 10^{11} hspace{0.5mm} mathrm{m}$. So, we can calculate the speed of light as

$$
begin{align*}
d&=ct / : t\
c&= frac{d}{t}\
c&= frac{ 2.98 cdot 10^{11} hspace{0.5mm} mathrm{m} }{990 hspace{0.5mm} mathrm{s}}\
c&= 3.01 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}
end{align*}
$$

The precise value of speed of light is $2.99 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}$, so the relative error is

$$
begin{align*}
RE&= frac{|c_{precise}-c|}{c_{precise}}\
RE&= frac{|2.99 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}} – 3.01 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}} |}{2.99 cdot 10^{8} hspace{0.5mm} frac{m}{s} }\
RE&= frac{0.02}{2.990} cdot 100 %\
RE&=0.007 cdot 100 %\
RE&=0.7 %
end{align*}
$$

We can see that the relative error is $0.7 %$, so it is negligible. We can say this method appears accurate.

This method appears accurate, because relative error is $0.7 %$.