To solve this question, we need to determine the new focal length of the thin convex lens when it is immersed in a liquid with a different refractive index. We will use the lens maker's formula for this purpose.
The lens maker's formula is given by:
\[\frac{1}{f} = \left( \frac{n_{\text{lens}}}{n_{\text{medium}}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]Where:
Initially, in air (refractive index 1), the focal length is given as \(+20 \, \text{cm}\). Using the lens maker's formula for air:
\[\frac{1}{20} = \left( \frac{1.5}{1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]From this, we can determine that:
\[\frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{20 \times 0.5} = \frac{1}{10}\]Now, when the lens is immersed in a liquid with refractive index 1.6, we apply the lens maker's formula again:
\[\frac{1}{f_{\text{new}}} = \left( \frac{1.5}{1.6} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]Substituting the known values:
\[\frac{1}{f_{\text{new}}} = \left( \frac{1.5}{1.6} - 1 \right) \times \frac{1}{10}\]Simplify to find the new focal length:
\[\frac{1.5}{1.6} = 0.9375\]\[\frac{1}{f_{\text{new}}} = (0.9375 - 1) \times \frac{1}{10}\]
\[\frac{1}{f_{\text{new}}} = -0.0625 \times \frac{1}{10}\]
\[\frac{1}{f_{\text{new}}} = -0.00625\]
Thus, the new focal length \(f_{\text{new}}\) is:
\[f_{\text{new}} = -\frac{1}{0.00625} = -160 \, \text{cm}\]Therefore, the correct answer is \(-160 \, \text{cm}\), indicating that the focal length of the lens becomes negative, meaning it behaves as a diverging lens in the liquid.