A convex lens has a focal length of 0.3 m an | vedclass

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(B) According to the Lens Maker's formula: $\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$,where $\mu$ is the relative refracti

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$1.2$

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(B) According to the Lens Maker's formula: $\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} ight]$,where $\mu$ is the relative refractive index $\mu = \frac{\mu_L}{\mu_M}$.For air: $\frac{1}{f_a} = \left( \frac{3/2}{1} - 1 ight) K = \frac{1}{2} K$,where $K = \left[ \frac{1}{R_1} - \frac{1}{R_2} ight]$.Given $f_a = 0.3 \ m$,so $\frac{1}{0.3} = \frac{1}{2} K \implies K = \frac{2}{0.3} = \frac{20}{3}$.For water: $\frac{1}{f_w} = \left( \frac{3/2}{4/3} - 1 ight) K = \left( \frac{9}{8} - 1 ight) K = \frac{1}{8} K$.Substituting $K$: $\frac{1}{f_w} = \frac{1}{8} 	imes \frac{20}{3} = \frac{20}{24} = \frac{5}{6}$.Therefore,$f_w = \frac{6}{5} = 1.2 \ m$.

$A$ convex lens has a focal length of $0.3 \ m$ and a refractive index of $3/2$. Find the focal length of the lens in $m$ when it is immersed in water,which has a refractive index of $4/3$.

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