Question

Two identical equi-convex 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 then 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

The correct option is B $\frac{f}{2}>F>f$

$\frac{1}{f_w}=({\mu }_w-1)(-\frac{2}{R})=-\frac{{\mu }_w-1}{{\mu }_l-1}\left(\frac{1}{f_l}\right)$
$\frac{1}{f_l}=({\mu }_l-1)\frac{2}{R}$
${\mu }_l={\mu }_w$
$\frac{1}{f_{eq}}=\frac{2}{f_l}-\frac{1}{f_l}\left(\frac{{\mu }_w-1}{{\mu }_l-1}\right)$
$=\frac{1}{f_l}(2-\frac{{\mu }_w-1}{{\mu }_l-1})$
${\mu }_l-1>{\mu }_w-1$
$\Rightarrow \frac{{\mu }_w-1}{{\mu }_l-1}<1$
$\frac{1}{f}<\frac{1}{f_{eq}}<\frac{2}{f}$