A convex lens \( \left(n_{g} = 1.5\right) \) of focal length 10 cm in air when immersed in a liquid behaves as a diverging lens of focal length 45 cm. The refractive index of the liquid:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1.68
By the lens maker's formula, the focal length depends on the refractive index of the glass relative to the surrounding medium. Let \( K = \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \), which does not change.
In air:
\( \displaystyle \frac{1}{f_a} = (n_g - 1)\,K \;\Rightarrow\; \frac{1}{10} = 0.5\,K \;\Rightarrow\; K = 0.2\,\mathrm{cm^{-1}} \)
In the liquid of refractive index \( n_l \) the lens is diverging, so \( f_l = -45\,\mathrm{cm} \):
\( \displaystyle \frac{1}{f_l} = \left(\frac{n_g}{n_l} - 1\right) K \;\Rightarrow\; -\frac{1}{45} = \left(\frac{1.5}{n_l} - 1\right)(0.2) \)
\( \displaystyle \frac{1.5}{n_l} - 1 = -\frac{1}{9} \;\Rightarrow\; \frac{1.5}{n_l} = \frac{8}{9} \)
\( \displaystyle n_l = \frac{1.5 \times 9}{8} = 1.6875 \approx 1.68 \)
A convex lens becomes diverging only when the liquid is optically denser than the glass (\( n_l \gt 1.5 \)), which also rules out 1.25 and 1.33 at once.