Question

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

Answer 1

Let $R$ be the radius of curvature of each surface. Then

$\frac{1}{F}=(1.5-1)(\frac{1}{R}+\frac{1}{R})$

For the water lens,

$\frac{1}{f^{\prime }}=(\frac{4}{3}-1)(-\frac{1}{R}+\frac{1}{R})=\frac{1}{3}(-\frac{2}{f});\frac{1}{f^{\prime }}=-\frac{2}{3f}$

Now, using $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}$, we have

$\frac{1}{F}=\frac{1}{f}+\frac{1}{f}+\frac{1}{f^{\prime }}=\frac{2}{f}-\frac{2}{3f}=\frac{4}{3f}$

$\Rightarrow F=\frac{3f}{4}$.