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(D) Given: Focal length in air $f_{a} = 0.15 \,m$, refractive index of glass $\mu_{g} = \frac{3}{2}$, refractive index of water $\mu_{w} = \frac{4}{3}

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$0.6$

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(D) Given: Focal length in air $f_{a} = 0.15 \,m$, refractive index of glass $\mu_{g} = \frac{3}{2}$, refractive index of water $\mu_{w} = \frac{4}{3}$.According to the Lens Maker's formula: $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight)$, where $\mu_{rel} = \frac{\mu_{lens}}{\mu_{medium}}$.For air: $\frac{1}{f_{a}} = \left( \frac{\mu_{g}}{\mu_{a}} - 1 ight) C$, where $C = \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight)$.$\frac{1}{0.15} = \left( \frac{3/2}{1} - 1 ight) C = \frac{1}{2} C \implies C = \frac{2}{0.15} = \frac{40}{3}$.For water: $\frac{1}{f_{w}} = \left( \frac{\mu_{g}}{\mu_{w}} - 1 ight) C$.$\frac{1}{f_{w}} = \left( \frac{3/2}{4/3} - 1 ight) C = \left( \frac{9}{8} - 1 ight) C = \frac{1}{8} C$.Substituting $C = \frac{40}{3}$: $\frac{1}{f_{w}} = \frac{1}{8} 	imes \frac{40}{3} = \frac{5}{3}$.Therefore, $f_{w} = \frac{3}{5} = 0.6 \,m$.

$A$ convex lens made of glass has a focal length of $0.15 \,m$ in air. If the refractive index of glass is $\frac{3}{2}$ and that of water is $\frac{4}{3}$, what is the focal length of the lens when immersed in water (in $\,m$)?

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