A convex lens of glass (n = 1.5) has a focal length 18 cm in air. When it is immersed in water (n = 4/3), the focal length increases to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 72 cm
The lens maker’s formula for a lens of refractive index \( n_{\mathrm{l}} \) in a medium of refractive index \( n_{\mathrm{m}} \) is
\( \displaystyle \frac{1}{f} = \left(\frac{n_{\mathrm{l}}}{n_{\mathrm{m}}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
The lens and its radii are the same in air and in water, so the bracket with the radii cancels in the ratio of the two focal lengths.
\( \displaystyle \frac{f_{\mathrm{water}}}{f_{\mathrm{air}}} = \frac{1.5 - 1}{\frac{1.5}{4/3} - 1} \)
Here \( \frac{1.5}{4/3} = \frac{9}{8} \), so the denominator is \( \frac{9}{8} - 1 = \frac{1}{8} \).
\( \displaystyle \frac{f_{\mathrm{water}}}{f_{\mathrm{air}}} = \frac{0.5}{1/8} = 4 \)
\( f_{\mathrm{water}} = 4\times18\,\mathrm{cm} = 72\,\mathrm{cm} \)
The lens remains converging because glass is still optically denser than water.