Two similar thin equi-convex lenses, of focal length f each, are kept coaxially in contact with each other such that the focal length of the combination is ${\mathrm{F}}_1$. When the space between the two lenses is filled with glycerin (which has the same refractive index $(\mathrm{\mu }=1.5)$ as that of glass) then the equivalent focal length is ${\mathrm{F}}_2$. The ratio ${\mathrm{F}}_1$:${\mathrm{F}}_2$ will be

$$\begin{gathered} \text{ According to lens maker's formula } \\ \frac{1}{\mathrm{f}}=(\mathrm{\mu }-1)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right) \\ \frac{1}{\mathrm{f}}=(\mathrm{\mu }-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{-\mathrm{R}}\right)=(1.5-1)\left(\frac{2}{\mathrm{R}}\right)=\frac{1}{\mathrm{R}} \end{gathered}$$

Two similar equi-convex lenses of focal length f each are held in contact with each other.

The focal length ${\mathrm{F}}_1$ of the combination is given by

$\frac{1}{{\mathrm{F}}_1}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{f}}=\frac{2}{\mathrm{f}}; {\mathrm{F}}_1=\frac{\mathrm{f}}{2}=\frac{\mathrm{R}}{2} .........\left(\mathrm{i}\right)$

For glycerin in between lenses, there are three lenses, one concave and two convex

Focal length of the curve lens is given 

$\frac{1}{{\mathrm{f}}^{\text{'}}}=(1.5-1)\left(\frac{-2}{\mathrm{R}}\right)=-\frac{1}{\mathrm{R}}$

Now, equivalent focal length of the combination is given by

$$\begin{gathered} \begin{array}{l}\frac{1}{{\mathrm{F}}_2}=\frac{1}{\mathrm{f}}+\frac{1}{{\mathrm{f}}^{\text{'}}}+\frac{1}{\mathrm{f}};\frac{1}{{\mathrm{F}}_2}=\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R}}\\ {\mathrm{F}}_2=\mathrm{R} ..........\left(\mathrm{ii}\right)\end{array} \\ \text{ Dividing equation (i) by (ii), we get }\frac{{\mathrm{F}}_1}{{\mathrm{F}}_2}=\frac{1}{2} \end{gathered}$$