N lamps each of resistance r , are fed by a machine of resistance R...

As each are containing $n$ lamps, hence, Resistance of each arc $=n r$ Number of arcs $=N / n$ Total resistance $S$ is given by $$\begin{aligned} \frac{1}{S} & =\Sigma \frac{1}{n r}=\frac{N}{n}\left(\frac{1}{n r}\right) \\ S & =\frac{n^2 r}{N}\end{aligned}$$ $\therefore$ Total resistance $=R+S=R+\left(n^2 r\right) N$ If $E$ is the emf of the machine, current entering the arcs is $E /(R+S)$ and in each arc is $n E /(R+S) N$. Hence, current passing through each lamp $I=\frac{n E}{N\left(R+n^2 r / N\right)}=\frac{E}{N}$\left[\frac{R}{n}+\frac{n r}{N}\right]^{-1}$Now, heat produced per second in the lamps is$H=N_r I^2$Since, light emitted is proportional to$H^2$therefore light produced is maximum when$H^2$and hence$H$is maximum or \left[\frac{R}{n}+\frac{n r}{N}\right]$ is minimum. Hence, we can write, $\frac{R}{n}+\frac{n r}{N}=\left[\left(\frac{R}{n}\right)^{1 / 2}-\left(\frac{n r}{N}\right)^{1 / 2}\right]^2+2\left(\frac{R r}{N}\right)^{1 / 2}$ This is minimum when $\sqrt{R / n}-\sqrt{n r / N}=0$ or very small or $n$ is closely equal to $(N R / r)^{1 / 2}$.