Question

The focal length of a convex lens is $f$ in air. When it is completely immersed in water of refractive index $\frac{4}{3}$, its focal length becomes (take refractive index of glass = 1.5):

• A. \( f \)
• B. \( 2f \)
• C. \( 4f \)
• D. \( \frac{f}{2} \)

Hint:

Keep this highly recurring benchmark result memorized for speed! Whenever a standard glass lens (\(\mu = 1.5\)) is completely immersed in water (\(\mu = 1.33\)), its focal length always increases exactly by a factor of 4 (\(f_{\text{water}} = 4f_{\text{air}}\)). Remembering this constant numeric multiplier will save you over a minute of algebraic scratchwork during a time-pressured exam.

Answer 1

The Correct Option is C

Concept: The focal length (\(f\)) of a thin spherical lens depends on the radii of curvature of its two surfaces (\(R_1\) and \(R_2\)) as well as the relative refractive index of the lens material with respect to its surrounding medium. This relationship is defined by the Lens Maker's Formula: \[ \frac{1}{f} = \left(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] When a lens is immersed in a liquid medium like water instead of air, the relative refractive index decreases, which reduces the bending power of the lens interfaces and subsequently increases its focal length.

Step 1: Writing the Lens Maker's Formula for the lens in air.
For air, the refractive index of the surrounding medium is \(\mu_{\text{air}} = 1\). The refractive index of the glass lens is given as \(\mu_{\text{g}} = 1.5 = \frac{3}{2}\). Substituting these parameters into the Lens Maker's formula: \[ \frac{1}{f} = \left(\frac{\mu_{\text{g}}}{\mu_{\text{air}}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] \[ \frac{1}{f} = (1.5 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0.5 \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{2}\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad \cdots (1) \]

Step 2: Writing the Lens Maker's Formula for the lens immersed in water.
Let the new focal length of the lens in water be \(f_{\text{w}}\). The refractive index of water is \(\mu_{\text{w}} = \frac{4}{3}\). Substituting these values: \[ \frac{1}{f_{\text{w}}} = \left(\frac{\mu_{\text{g}}}{\mu_{\text{w}}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] \[ \frac{1}{f_{\text{w}}} = \left(\frac{1.5}{\frac{4}{3}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(\frac{\frac{3}{2}}{\frac{4}{3}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] \[ \frac{1}{f_{\text{w}}} = \left(\frac{9}{8} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{8}\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad \cdots (2) \]

Step 3: Comparing the two conditions to find $f_{\text{w}}$.
Dividing equation (1) by equation (2) eliminates the structural geometric constant group \(\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\): \[ \frac{\left(\frac{1}{f}\right)}{\left(\frac{1}{f_{\text{w}}}\right)} = \frac{\frac{1}{2}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)}{\frac{1}{8}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)} \] \[ \frac{f_{\text{w}}}{f} = \frac{\frac{1}{2}}{\frac{1}{8}} = \frac{8}{2} = 4 \] Rearranging to solve for the final immersed focal length \(f_{\text{w}}\): \[ f_{\text{w}} = 4f \]