Two identical glass (μg = 3/2) equiconvex lenses of focal length f each are kept in contact. The space between that two lenses in filled with water (μw = 4/3). The focal length of the combination is
3f / 4
f / 3
f
4f / 3
Explanation:
$\dfrac{1}{f} = \left(\dfrac{3}{2}–1\right)\dfrac{2}{R} = \dfrac{1}{R}$
$\dfrac{1}{f} = \left(\dfrac{4}{3}–1\right)\left\{–\dfrac{2}{R}\right\} = –\dfrac{2}{3R}$
$\dfrac{1}{f_{eq}} = \dfrac{1}{f}–\dfrac{2}{3f} + \dfrac{1}{f} = \dfrac{3 – 2 + 3}{3f} = \dfrac{4}{3f}$
$f_{eq} = \dfrac{3f}{4}$