The refractive index of a convex lens is \(1.5\). If it is immersed in a solution of refractive index \(1.6\), then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → the convex lens will behave as a diverging lens in the solution
By the lens maker’s formula, the power of a lens depends on its refractive index relative to the surrounding medium:
\( \displaystyle \frac{1}{f} = \left(\frac{n_{\mathrm{lens}}}{n_{\mathrm{medium}}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
In air:
\( \displaystyle \frac{1}{f_{\mathrm{air}}} = (1.5 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0.5\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
In the solution:
\( \displaystyle \frac{1}{f'} = \left(\frac{1.5}{1.6} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = -0.0625\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
\( \displaystyle f' = \frac{0.5}{-0.0625}\,f_{\mathrm{air}} = -8f_{\mathrm{air}} \)
The focal length becomes negative: the lens is diverging in the solution.
Why the other options are incorrect
Option (1) focal length of the given convex lens will increase — Incorrect. It misses that the lens becomes diverging.
Option (2) focal length of the given convex lens will decrease — Incorrect. It is “smaller” only in the signed sense.
Option (3) the convex lens will behave as a plane glass slab in solution — Incorrect. That needs equal refractive indices.
Note: The question does not say whether “focal length” means its signed value or its magnitude, so Option (1) or (2) could also be argued; Option (4) is the complete description.