An equiconvex lens \( \left(\mu = \displaystyle\frac{3}{2}\right) \) of focal length 20 cm in air is immersed in water of refractive index \( \left(\mu = \displaystyle\frac{4}{3}\right) \). What is the percentage change in focal length of the lens ?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 300%
By the lens maker's formula, the power of a lens depends on the refractive index of the lens relative to the surrounding medium; the radii of curvature do not change.
\( \displaystyle \frac{1}{f} = \left(\frac{\mu_{\mathrm{lens}}}{\mu_{\mathrm{medium}}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
Taking the ratio for water and air, the bracket with the radii cancels:
\( \displaystyle \frac{f_w}{f_a} = \frac{\mu - 1}{\mu/\mu_w - 1} = \frac{\frac{3}{2} - 1}{\frac{3/2}{4/3} - 1} = \frac{1/2}{9/8 - 1} = \frac{1/2}{1/8} = 4 \)
\( f_w = 4 \times 20\,\mathrm{cm} = 80\,\mathrm{cm} \)
\( \displaystyle \text{percentage change} = \frac{f_w - f_a}{f_a} \times 100 = \frac{80 - 20}{20} \times 100 = 300\% \)