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Step-by-Step Solution
Step 1: Recall the lens maker’s formula in a medium
For a thin lens with refractive index $ \mu_{\text{lens}} $ placed in a medium of refractive index $ \mu_{\text{medium}} $, the focal length $f_{\text{in medium}}$ is given by the modified lens maker’s formula:
$ \displaystyle \frac{1}{f_{\text{in medium}}} \;=\;\bigl(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1\bigr)\,\bigl(\frac{1}{R_1} - \frac{1}{R_2}\bigr)\,. $
For simplicity, take a symmetric thin convex lens with surface radii $R_1 = R$ and $R_2 = -R$, so
$ \displaystyle \frac{1}{f_{\text{in medium}}} \;=\;\bigl(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1\bigr)\,\left(\frac{1}{R} - \bigl(-\frac{1}{R}\bigr)\right)
= \bigl(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1\bigr)\,\left(\frac{2}{R}\right). $
Step 2: Focal length in air
When the lens is in air, $ \mu_{\text{medium}} = 1 $ and $ \mu_{\text{lens}} = \frac{3}{2} $. Its focal length is
$ \displaystyle \frac{1}{f} = \Bigl(\frac{\frac{3}{2}}{1} - 1\Bigr)\,\frac{2}{R}
= \Bigl(\frac{3}{2} - 1\Bigr)\,\frac{2}{R}
= \Bigl(\frac{1}{2}\Bigr)\,\frac{2}{R}
= \frac{1}{R}\,. $
Thus, $ f = R $ in air (from the above simplified derivation).
Step 3: Focal length in the first liquid (μL1 = 4/3)
For the first liquid, $ \mu_{\text{medium}} = \frac{4}{3} $ and $ \mu_{\text{lens}} = \frac{3}{2} $. Then,
$ \displaystyle \frac{1}{f_1}
= \Bigl(\frac{\frac{3}{2}}{\frac{4}{3}} - 1\Bigr)\,\frac{2}{R}
= \Bigl(\frac{3}{2} \times \frac{3}{4} - 1\Bigr)\,\frac{2}{R}
= \Bigl(\frac{9}{8} - 1\Bigr)\,\frac{2}{R}
= \Bigl(\frac{1}{8}\Bigr)\,\frac{2}{R}
= \frac{1}{4R}. $
Hence,
$ \displaystyle f_1 = 4R. $
Since $ R $ is the focal length in air (found to be $R$ from the simplified form above), $ f_1 = 4R $ is definitely larger than $ R $. Therefore, $ f_1 > f $.
Step 4: Focal length in the second liquid (μL2 = 5/3)
For the second liquid, $ \mu_{\text{medium}} = \frac{5}{3} $. Then,
$ \displaystyle \frac{1}{f_2}
= \Bigl(\frac{\frac{3}{2}}{\frac{5}{3}} - 1\Bigr)\,\frac{2}{R}
= \Bigl(\frac{3}{2} \times \frac{3}{5} - 1\Bigr)\,\frac{2}{R}
= \Bigl(\frac{9}{10} - 1\Bigr)\,\frac{2}{R}
= \Bigl(-\frac{1}{10}\Bigr)\,\frac{2}{R}
= -\frac{1}{5R}. $
So,
$ \displaystyle f_2 = -5R. $
A negative focal length indicates that the lens functions like a diverging lens in that medium. Hence $ f_2 $ is negative.
Step 5: Interpret the results
The results show:
$ f_1 = 4R > R = f $, so $ f_1 > f $.
$ f_2 = -5R $, which is negative.
This matches the condition “$ f_1 > f $ and $ f_2 $ becomes negative.” That is indeed the correct relation.
Answer
Option: $ f_1 > f $ and $ f_2 $ becomes negative
