Correct Option: (A) increases when the lens is dipped in water.

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

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

Where $\mu_{rel} = \frac{\mu_{lens}}{\mu_{medium}}$ is the refractive index of the lens material with respect to the surrounding medium.

Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]

• For an equiconvex lens, $R_1 = R$ and $R_2 = -R$. Thus, $\left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{2}{R}$.
• When the lens is in air ($\mu_a \approx 1$), $\mu_{rel} = \mu_g$.
• When the lens is dipped in water ($\mu_w \approx 1.33$), $\mu'_{rel} = \frac{\mu_g}{\mu_w}$.
• Since $\mu_w > 1$, it follows that $\frac{\mu_g}{\mu_w} < \mu_g$.
• Consequently, $(\mu'_{rel} - 1) < (\mu_{rel} - 1)$.

Step 3 (Final Concluded Answer): [0.25 Mark]

Since the term $(\mu_{rel} - 1)$ decreases when the lens is placed in water, the value of $\frac{1}{f}$ decreases. Therefore, the focal length $f$ must increase. Hence, option (A) is correct.

💡 EL10 Concept Insight:

Think of the lens as a "light bender"; it bends light because it's denser than the air around it. When you put it in water, the difference in density between the lens and the water is much smaller than the difference between the lens and air, so the lens loses some of its "bending power" and the focal length stretches out.