The radii of curvature of a biconvex lens are 20 cm and 30 cm respectively, the refractive index of the material of the lens is 1.5. Now the lens is immersed in a liquid of refractive index 1.8, the nature and the focal length of the lens will become
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Diverging, 72 cm
When a lens is placed in a liquid, the lens maker’s formula uses the refractive index of the lens relative to the surrounding medium. For the biconvex lens \( R_1 = +20 \) cm and \( R_2 = -30 \) cm.
\( \displaystyle \frac{1}{f} = \left(\frac{\mu_{\mathrm{lens}}}{\mu_{\mathrm{liquid}}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
\( \displaystyle \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{20} + \frac{1}{30} = \frac{1}{12}\,\mathrm{cm^{-1}} \)
\( \displaystyle \frac{1}{f} = \left(\frac{1.5}{1.8} - 1\right) \times \frac{1}{12} = -\frac{1}{6} \times \frac{1}{12} = -\frac{1}{72}\,\mathrm{cm^{-1}} \)
\( f = -72\,\mathrm{cm} \)
The negative focal length means the lens now behaves as a diverging lens. (In air it is converging, with f = 24 cm.)