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: Understanding the Question:
We are given the focal length of a glass lens in air and need to find its new focal length when it is completely immersed in a specific liquid.
Step 2: Key Formula or Approach:
The Lens Maker's Formula relates the focal length ($f$) to the relative refractive index of the lens material with respect to the surrounding medium:
$$\frac{1}{f} = \left( \frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$
Let $K = \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. We will find $K$ using the air parameters, then use it for the liquid medium.
Step 3: Detailed Explanation:
Case 1: Lens in air
$f_a = 15\ \text{cm}$
$\mu_{\text{lens}} = \frac{3}{2}$
$\mu_{\text{medium}} = \mu_{\text{air}} = 1$
$$\frac{1}{15} = \left( \frac{3/2}{1} - 1 \right) K$$
$$\frac{1}{15} = \left( \frac{3}{2} - 1 \right) K = \frac{1}{2} K$$
$$K = \frac{2}{15}$$
Case 2: Lens in liquid
$\mu_{\text{lens}} = \frac{3}{2}$
$\mu_{\text{medium}} = \frac{9}{5}$
$$\frac{1}{f_l} = \left( \frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1 \right) K$$
$$\frac{1}{f_l} = \left( \frac{3/2}{9/5} - 1 \right) K$$
$$\frac{1}{f_l} = \left( \frac{3 \times 5}{2 \times 9} - 1 \right) K = \left( \frac{15}{18} - 1 \right) K$$
Simplify the fraction ($\frac{15}{18} = \frac{5}{6}$):
$$\frac{1}{f_l} = \left( \frac{5}{6} - 1 \right) K = -\frac{1}{6} K$$
Substitute the value of $K = \frac{2}{15}$:
$$\frac{1}{f_l} = -\frac{1}{6} \times \frac{2}{15} = -\frac{2}{90} = -\frac{1}{45}$$
$$f_l = -45\ \text{cm}$$
The negative sign indicates that the converging (convex) lens behaves as a diverging (concave) lens in this liquid.
Step 4: Final Answer:
The focal length in the liquid is $-45\ \text{cm}$, matching option (D).
