To solve the problem, we need to understand how the focal length of a convex lens changes when placed in water.
1. Focal Length and Refractive Index:
The focal length of a lens is related to the refractive index of the lens material and the medium in which the lens is placed. The formula for the focal length of a lens in a medium is:
$ \frac{1}{f} = (n_{\text{lens}} - n_{\text{medium}}) \times \frac{1}{R} $
Where $n_{\text{lens}}$ is the refractive index of the lens material, $n_{\text{medium}}$ is the refractive index of the medium (water in this case), and $R$ is the radius of curvature of the lens.
2. Effect of Placing the Lens in Water:
When a convex lens is placed in water, the refractive index of the medium changes. Water has a refractive index of approximately 1.33, which is lower than the refractive index of glass (the common material of a convex lens).
Thus, the difference between the refractive index of the lens and the water decreases, resulting in an increase in the focal length.
Final Answer:
If a convex lens is placed in water, its focal length $ \mathbf{increases} $.
| List-I | List-II | ||
| P | If \(n = 2\) and \(\alpha = 180°\), then all the possible values of \(\theta_0\) will be | I | \(30\degree\) or \(0\degree\) |
| Q | If \(n = √3\) and \(\alpha= 180°\), then all the possible values of \(\theta_0\) will be | II | \(60\degree\) or \(0\degree\) |
| R | If \(n = √3\) and \(\alpha= 180°\), then all the possible values of \(\phi_0\) will be | III | \(45\degree\) or \( 0\degree\) |
| S | If \(n = \sqrt2\) and \(\theta_0 = 45°\), then all the possible values of \(\alpha\) will be | IV | \(150\degree\) |
| \[0\degree\] | |||