Q122 / 180 · NEET 2016 (Phase 2) · Physics

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

Solution

$\frac{1}{f} = (μ_{g} - 1) \frac{2}{R} = \frac{1}{R}, (μ_{g} = \frac{3}{2}) \Rightarrow R = f$

$\frac{1}{f_{1}} = - (μ_{w} - 1) \frac{2}{R} = \frac{- 2}{3 R} = \frac{- 2}{3 f}$ .

$\frac{1}{f_{eq}} = \frac{1}{f} + \frac{1}{f} + \frac{1}{f_{1}} = \frac{1}{R} + \frac{1}{R} - \frac{2}{3 R}$

$\frac{1}{f_{eq}} = \frac{1}{f} + \frac{1}{f} - \frac{2}{3 f}$

$\frac{1}{f_{eq}} = \frac{2}{f} - \frac{2}{3 f}$

$\frac{1}{f_{eq}} = \frac{4}{3 f}$

$f_{eq} = \frac{3 f}{4}$ .

Hence, the correct answer is option: 'D'.