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 :

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

Answer 1

Answer: (D)

$\frac{3f}{4}$

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

$\frac{1}{f}=(1.5-1)(\frac{1}{r}+\frac{1}{r})$

$⟹R=f$

Now, for the water lens

$\frac{1}{f^{\prime }}=(\frac{4}{3}-1)(-\frac{1}{R}-\frac{1}{R})=\frac{1}{3}\times \left(\frac{-2}{f}\right)$

$⟹\frac{1}{f^{\prime }}=\frac{-2}{3f}$

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

$\frac{1}{F}=\frac{1}{f}+\frac{1}{f}+\frac{1}{f^{\prime }}$

$⟹F=\frac{3f}{4}$