The given lens is equivalent to the combination of two biconcave lenses of different focal lengths. When a beam parallel to principal axis falls on this lens, after refraction, they appear to diverge from their foci. Hence the difference of their focal lengths will be equal to the distance between the diverging points.

$\frac{1}{{\mathrm{f}}_1}=\left({\mathrm{\mu }}_1-1\right)\left(\frac{1}{-\mathrm{R}}-\frac{1}{+\mathrm{R}}\right)$

$\Rightarrow {\mathrm{f}}_1=\frac{-\mathrm{R}}{2\left({\mathrm{\mu }}_1-1\right)}$

$\text{ similarly, }{\mathrm{f}}_2=\frac{-\mathrm{R}}{2\left({\mathrm{\mu }}_2-1\right)}$

Required distance

$\mathrm{d}=\left|{\mathrm{f}}_1\right|-\left|{\mathrm{f}}_2\right|=\frac{\mathrm{R}}{2}\left[\frac{1}{{\mathrm{\mu }}_1-1}-\frac{1}{{\mathrm{\mu }}_2-1}\right]$$=\frac{\mathrm{R}}{2}\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{\left({\mathrm{\mu }}_1-1\right)\left({\mathrm{\mu }}_2-1\right)}$