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(A) The Lens Maker's formula is given by: $\frac{1}{f_m} = \left( \frac{\mu_g}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a

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$f_1 > f$ and $f_2$ becomes negative

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(A) The Lens Maker's formula is given by: $\frac{1}{f_m} = \left( \frac{\mu_g}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a convex lens in air,$\frac{1}{f} = (\mu_g - 1) \left( \frac{2}{R} \right) = (\frac{3}{2} - 1) \frac{2}{R} = \frac{1}{R}$. Thus,$f = R$.In liquid $1$ with $\mu_{L1} = \frac{4}{3}$: $\frac{1}{f_1} = \left( \frac{3/2}{4/3} - 1 \right) \frac{2}{R} = \left( \frac{9}{8} - 1 \right) \frac{2}{R} = \frac{1}{8} \cdot \frac{2}{R} = \frac{1}{4R}$. So,$f_1 = 4R = 4f$. Since $f_1 > f$,the focal length increases.In liquid $2$ with $\mu_{L2} = \frac{5}{3}$: $\frac{1}{f_2} = \left( \frac{3/2}{5/3} - 1 \right) \frac{2}{R} = \left( \frac{9}{10} - 1 \right) \frac{2}{R} = -\frac{1}{10} \cdot \frac{2}{R} = -\frac{1}{5R}$. So,$f_2 = -5R = -5f$. Since $f_2$ is negative,the lens behaves as a concave lens.

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