Question

Two identical equiconvex lenses, each of focal length ‘$f$’ are placed side by side in contact with each other with a layer of water in between them as shown in the figure. If refractive index of the material of the lenses is greater than that of water, how the combined focal length ‘$F$’ is related to ‘$f$’ ?

• A. $F > f $
• B. $\frac{f}{2} < F < f $
• C. $ F < \frac{f}{2} $
• D. $F = f $

Answer 1

Answer: (B)

The given combination of lenses

Here, $f_{1}=$ focal length of equiconvex lenses of glass.
$f_{2}=$ focal length of lens formed by water (concave).
The focal length of the combination.
$\therefore \frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{1}{f_{3}}$
$=\frac{1}{f_{1}}-\frac{1}{f_{2}}+\frac{1}{f_{1}}=\frac{2}{f_{1}}-\frac{1}{f_{2}}$
$\frac{1}{F}=\frac{2 f_{2}-f_{1}}{f_{1} f_{2}} $
$\Rightarrow F=\frac{f_{1} f_{2}}{2 f_{2}-f_{1}} \,\,\,\left[\because f_{3}=f_{1}\right]$
$F=\frac{f_{1}}{2-\frac{f_{1}}{f_{2}}} \ldots$(i)
Here, $\frac{1}{f_{1}}=\left(\mu_{g}-1\right) \frac{2}{R}$, for $L_{1}$ and $L_{3}$
and $\frac{1}{f_{2}}=\left(\mu_{w}-1\right)\left(-\frac{2}{R}\right)$, for $L_{2}$
Given, $\mu_{w} < \mu_{g}$
Thus, $\frac{f_{1}}{f_{2}} < 1$
So, $F > \frac{f_{1}}{2}\ldots$(ii)
From Eqs. (i) and (ii), we get
$\frac{f_{1}}{2} < F < f_{1} $ or $\frac{f}{2} < F < f $
$\left(\because f_{1}=f\right)$