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Two identical glass ($\mu_g = 3/2$) equiconvex lenses of focal length f each 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
$\frac{4f}{3}$
B
$\frac{3f}{4}$
C
$\frac{f}{3}$
D
f
Explanation
The water forms a concave lens between two convex lenses.
Detailed Solution
$f_1 = f_3 = \frac{R}{2\left(\frac{3}{2} - 1\right)} = R = f$ (given)
Water lens: $f_2 = \frac{-R}{2\left(\frac{4}{3} - 1\right)} = -\frac{3}{2}R = -\frac{3}{2}f$
$\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} = \frac{1}{f} - \frac{2}{3f} + \frac{1}{f} = \frac{4}{3f}$
$f_{eq} = \frac{3f}{4}$