Two identical glass ( μ _ g = (3)/(2)) equiconvex lenses of focal...

Two identical glass

\left({\mu}_{g} = \dfrac{3}{2}\right)

equiconvex lenses of focal length,

f

are kept in contact. The space between the two lenses is filled with water \left({\mu}_{w} = \dfrac{4}{3}\right). The focal length of the combination is

Options

A $\dfrac{f}{3}$
B $f$
C $\dfrac{4f}{3}$
D $\dfrac{3f}{4}$

Correct answer

D. $\dfrac{3f}{4}$

Step-by-step solution

$\dfrac{1}{f} = \left({\mu}_{g} - 1\right)\dfrac{2}{R} = \dfrac{1}{R}, \left({\mu}_{g} = \dfrac{3}{2}\right) ⇒ R = f$

$\dfrac{1}{{f}_{1}} = - \left({\mu}_{w} - 1\right)\dfrac{2}{R} = \dfrac{- 2}{3R} = \dfrac{- 2}{3f}$ .

$\dfrac{1}{{f}_{eq}} = \dfrac{1}{f} + \dfrac{1}{f} + \dfrac{1}{{f}_{1}} = \dfrac{1}{R} + \dfrac{1}{R} - \dfrac{2}{3R}$

$\dfrac{1}{{f}_{eq}} = \dfrac{1}{f} + \dfrac{1}{f} - \dfrac{2}{3f}$

$\dfrac{1}{{f}_{eq}} = \dfrac{2}{f} - \dfrac{2}{3f}$

$\dfrac{1}{{f}_{eq}} = \dfrac{4}{3f}$

${f}_{eq} = \dfrac{3f}{4}$ .

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