Step 1 : Understanding the Question:
This problem asks us to calculate the change in the focal length of a convex lens when it is transferred from air into water. The focal length of any lens depends on the refractive index of the lens material relative to the surrounding medium, as well as the radii of curvature of the lens surfaces. Immersing the lens in a denser medium like water reduces the relative refractive index, which decreases the light-bending capability of the lens and consequently increases the focal length.
Step 2 : Key Formulas and Approach:
The quantitative relationship between the focal length of a lens, its refractive index, the refractive index of the medium, and its curvature is given by the Lens Maker's Formula:
\[ \frac{1}{f} = \left(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
We will first apply this formula for the lens in air (where the medium's refractive index is 1).
Next, we will apply the formula for the lens immersed in water (where the medium's refractive index is $4/3$).
By taking the ratio of these two equations, we can eliminate the geometric factor containing the radii of curvature and find the new focal length in terms of the original focal length.
Step 3 : Detailed Explanation:
We begin by analyzing the initial state of the glass lens in air. Here, the refractive index of the surrounding medium is $\mu_{\text{air}} = 1$, and the refractive index of the glass is $\mu_{\text{g}} = 1.5 = \frac{3}{2}$.
Substituting these values into the Lens Maker's Formula gives: $\frac{1}{f} = \left(\frac{3/2}{1} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{2} \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. We can label this as Equation 1.
We then analyze the second state where the lens is completely immersed in water. The refractive index of the water is $\mu_{\text{w}} = \frac{4}{3}$, and the refractive index of the lens remains $\mu_{\text{g}} = \frac{3}{2}$. Let the new focal length in water be denoted as $f_{\text{w}}$.
Substituting these values into the formula yields: $\frac{1}{f_{\text{w}}} = \left(\frac{3/2}{4/3} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
Simplifying the relative refractive index term gives: $\frac{\mu_{\text{g}}}{\mu_{\text{w}}} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8}$.
Substituting this back into the equation for $f_{\text{w}}$ gives: $\frac{1}{f_{\text{w}}} = \left(\frac{9}{8} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{8} \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. We can label this as Equation 2.
To find the relationship between $f$ and $f_{\text{w}}$, we divide Equation 1 by Equation 2. This mathematical step cancels out the geometric term $\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ completely.
Performing the division gives: $\frac{1/f}{1/f_{\text{w}}} = \frac{f_{\text{w}}}{f} = \frac{1/2}{1/8} = \frac{8}{2} = 4$.
Rearranging this equation to solve for $f_{\text{w}}$ gives: $f_{\text{w}} = 4f$.
Step 4 : Final Answer:
The focal length of the convex lens when completely immersed in water is $4f$, which corresponds to Option (C).