A double convex lens made of glass \( (\mu = 1.5) \) has equal radii of curvature and a focal length of \( 24\ \mathrm{cm} \) in air. When the lens is immersed in a liquid of refractive index \( 1.25 \), its new focal length is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 60\ \mathrm{cm} \)
The lens maker’s formula uses the refractive index of the lens relative to the surrounding medium:
\( \displaystyle \frac{1}{f} = \left(\frac{\mu_{\mathrm{lens}}}{\mu_{\mathrm{medium}}} - 1\right)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right) \)
For a double convex lens with equal radii, \( R_{1} = +R \) and \( R_{2} = -R \), so the bracket is \( 2/R \).
In air:
\( \displaystyle \frac{1}{24\,\mathrm{cm}} = (1.5 - 1)\frac{2}{R} = \frac{1}{R} \ \Rightarrow\ R = 24\,\mathrm{cm} \)
In the liquid:
\( \displaystyle \frac{\mu_{\mathrm{lens}}}{\mu_{\mathrm{liquid}}} = \frac{1.5}{1.25} = 1.2 \)
\( \displaystyle \frac{1}{f'} = (1.2 - 1)\frac{2}{24\,\mathrm{cm}} = \frac{1}{60\,\mathrm{cm}} \)
\( f' = 60\,\mathrm{cm} \)
Shortcut: The curvature factor is the same in both media, so take the ratio directly:
\( \displaystyle \frac{f'}{f} = \frac{\mu - 1}{\mu/\mu_{l} - 1} = \frac{0.5}{0.2} = 2.5 \ \Rightarrow\ f' = 2.5\times24\,\mathrm{cm} = 60\,\mathrm{cm} \)