Two magnets A and B are identical and these are arranged as shown in the figure. The (short magnets are considered) ratio of distance $d_1$ and $d_2$ will be,

$$\begin{gathered} \mathrm{The} \mathrm{magnetic} \mathrm{needle} \mathrm{will} \mathrm{align} \mathrm{along} \mathrm{the} \mathrm{total} \mathrm{magnetic} \mathrm{field}. \\ \mathrm{In}\mathrm{Equilibrium}, \\ \mathrm{tan\theta }=\frac{{\mathrm{B}}_{\mathrm{axial}}}{{\mathrm{B}}_{\mathrm{equatorial}}}=\frac{{\mathrm{B}}_1}{{\mathrm{B}}_2} \end{gathered}$$

$$\begin{gathered} \Rightarrow {\mathrm{B}}_1={\mathrm{B}}_2\mathrm{tan\theta } \\ \Rightarrow \frac{{\mathrm{\mu }}_{\mathrm{o}}}{4\mathrm{\pi }}\frac{2\mathrm{M}}{{{\mathrm{d}}_1}^3}=\frac{{\mathrm{\mu }}_{\mathrm{o}}}{4\mathrm{\pi }}\frac{\mathrm{M}}{{{\mathrm{d}}_2}^3}\text{t}\mathrm{an\theta } \end{gathered}$$

$\Rightarrow \frac{{\mathrm{d}}_1}{{\mathrm{d}}_2}={\left[\frac{2}{\mathrm{tan\theta }}\right]}^{\frac{1}{3}}={\left[2\mathrm{cot\theta }\right]}^{\frac{1}{3}}$