Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

The focal length of a lens in a medium is governed by the Lens Maker's Formula:

$$\frac{1}{f} = (n_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$

Where $n_{rel} = \frac{n_{lens}}{n_{medium}}$.

Given Data:

• Refractive index of lens ($n_g$) = $1.52$
• Focal length in air ($f_a$) = $15.0\text{ cm}$
• Refractive index of air ($n_a$) = $1.0$
• Refractive index of liquid ($n_l$) = $1.65$

Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]

For air:

$$\frac{1}{f_a} = \left( \frac{n_g}{n_a} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \implies \frac{1}{15} = (1.52 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ $$\left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{15 \times 0.52} \quad \dots (i)$$

For liquid:

$$\frac{1}{f_l} = \left( \frac{n_g}{n_l} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$

Substituting $(i)$ into the equation for $f_l$:

$$\frac{1}{f_l} = \left( \frac{1.52}{1.65} - 1 \right) \times \frac{1}{15 \times 0.52}$$ $$\frac{1}{f_l} = \left( \frac{1.52 - 1.65}{1.65} \right) \times \frac{1}{7.8} = \left( \frac{-0.13}{1.65} \right) \times \frac{1}{7.8}$$ $$\frac{1}{f_l} = \frac{-0.13}{12.87} \approx -0.0101$$ $$f_l = -\frac{12.87}{0.13} = -99\text{ cm}$$

Step 3 (Final Concluded Answer): [0.5 Mark]

The focal length of the lens in the liquid is $-99\text{ cm}$. Since the focal length is negative, the nature of the lens changes from converging to diverging (concave lens behavior).

💡 EL10 Concept Insight: Think of a lens like a prism; it bends light because it's denser than the air around it. If you put the lens in a liquid that is even denser than the glass itself, the light bends the "wrong way," causing the lens to flip its behavior from a magnifier to a spreader of light.