Question

Two identical glass $({\mu }_g=\frac{3}{2})$ equiconvex lenses of focal length f are kept in contact. The space between the two lenses is filled with water $({\mu }_w=\frac{4}{3})$ . The focal length of the combination is

• A. $f$
• B. $\frac{f}{2}$
• C. $\frac{4f}{3}$
• D. $\frac{3f}{4}$

Answer 1

The correct option is D $\frac{3f}{4}$

Let R be the radius of curvature of each surface. Then
$\frac{1}{f}=(1.5-1)(\frac{1}{R}+\frac{1}{R})$
$\therefore R=f$
For the water lens
$\frac{1}{f}=(\frac{4}{3}-1)(-\frac{1}{R}-\frac{1}{R})=\frac{1}{3}(-\frac{2}{R})$
or $\frac{1}{f}=-\frac{2}{3R}$
Now using $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}$ we have
$\frac{1}{F}=\frac{1}{f}+\frac{1}{f}+\frac{1}{f}$
$=\frac{2}{f}-\frac{2}{3f}=\frac{4}{3f}$ $\therefore F=\frac{3f}{4}$