Inside a vessel filled with liquid a converging lens is placed as shown in figure. The lens has focal length $15\ \text{cm}$ when in air and has refractive index $\frac{3}{2}$. If the liquid has refractive index $\frac{9}{5}$, the focal length of lens in liquid is
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When a lens is placed in a medium with a refractive index higher than its own material ($\mu_{\text{medium}} > \mu_{\text{lens}}$), it changes its nature! A convex lens becomes concave (diverging, negative focal length) and vice versa.
Step 1: Understand the question.
A convex lens has focal length $15\ \text{cm}$ in air and refractive index $\tfrac{3}{2}$. It is dipped in a liquid of refractive index $\tfrac{9}{5}$. We must find the lens's focal length inside the liquid.
Step 2: Recall the lens maker's formula.
The focal length depends on how much the lens bends light compared to the medium around it:
\[ \frac{1}{f} = \left(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1\right) K \]
Here $K = \left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$ depends only on the shape of the lens, so it stays the same in air and in liquid.
Step 3: Find $K$ using the air case.
In air, $\mu_{\text{medium}} = 1$ and $f = 15\ \text{cm}$:
\[ \frac{1}{15} = \left(\frac{3/2}{1} - 1\right) K = \frac{1}{2} K \]
So $K = \dfrac{2}{15}$.
Step 4: Set up the liquid case.
In the liquid, $\mu_{\text{medium}} = \tfrac{9}{5}$:
\[ \frac{1}{f_l} = \left(\frac{3/2}{9/5} - 1\right) K \]
Step 5: Simplify the bracket.
First, $\dfrac{3/2}{9/5} = \dfrac{3\times 5}{2\times 9} = \dfrac{15}{18} = \dfrac{5}{6}$. So:
\[ \frac{1}{f_l} = \left(\frac{5}{6} - 1\right) K = -\frac{1}{6} K \]
Step 6: Put in $K$ and solve.
\[ \frac{1}{f_l} = -\frac{1}{6}\times \frac{2}{15} = -\frac{2}{90} = -\frac{1}{45} \]
So $f_l = -45\ \text{cm}$. The minus sign means the convex lens now acts like a diverging lens because the liquid bends light more than the glass. This matches option (4).
\[ \boxed{f_l = -45\ \text{cm}} \]
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