A thin convex lens of refractive index 1.5 has focal length \( f_1 \). When immersed in a liquid of refractive index \( \displaystyle\frac{5}{3} \), its focal length is found to be \( f_2 \). The relationship between \( f_1 \) and \( f_2 \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( f_2 = -5f_1 \)
By the lens maker's formula, the power of a lens depends on the refractive index of the lens relative to the surrounding medium. Write \( k = 1/R_1 - 1/R_2 \); it does not change when the lens is immersed.
\( \displaystyle \frac{1}{f} = \left(\frac{n_{\mathrm{lens}}}{n_{\mathrm{medium}}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
In air:
\( \displaystyle \frac{1}{f_1} = (1.5 - 1)\,k = 0.5\,k \)
In the liquid:
\( \displaystyle \frac{1}{f_2} = \left(\frac{1.5}{5/3} - 1\right)k = (0.9 - 1)\,k = -0.1\,k \)
\( \displaystyle \frac{f_2}{f_1} = \frac{0.5\,k}{-0.1\,k} = -5 \;\Rightarrow\; f_2 = -5f_1 \)
The minus sign shows that the convex lens becomes a diverging lens in a liquid denser than its glass.