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(B) Given: Refractive index of glass $\mu_g = 3/2$,refractive index of water $\mu_w = 4/3$.The focal length $f$ of an equiconvex glass lens is given b

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$\frac{3f}{4}$

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(B) Given: Refractive index of glass $\mu_g = 3/2$,refractive index of water $\mu_w = 4/3$.The focal length $f$ of an equiconvex glass lens is given by the lens maker's formula:$\frac{1}{f} = (\mu_g - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\frac{3}{2} - 1) \frac{2}{R} = \frac{1}{2} \cdot \frac{2}{R} = \frac{1}{R}$.Thus,$f = R$.The water lens formed between the two convex lenses is a biconcave lens with radii of curvature $R$. Its focal length $f_w$ is given by:$\frac{1}{f_w} = (\mu_w - 1) \left( -\frac{1}{R} - \frac{1}{R} \right) = (\frac{4}{3} - 1) \left( -\frac{2}{R} \right) = \frac{1}{3} \cdot (-\frac{2}{R}) = -\frac{2}{3R}$.Since $R = f$,we have $\frac{1}{f_w} = -\frac{2}{3f}$.The combination consists of two convex lenses and one concave water lens in contact. The equivalent focal length $f_{eq}$ is:$\frac{1}{f_{eq}} = \frac{1}{f} + \frac{1}{f_w} + \frac{1}{f} = \frac{1}{f} - \frac{2}{3f} + \frac{1}{f} = \frac{3 - 2 + 3}{3f} = \frac{4}{3f}$.Therefore,$f_{eq} = \frac{3f}{4}$.

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

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