A convex lens of glass \((\mathrm{n}=1.5)\) has focal length \(0.2 \mathrm{~m}\). The lens is immersed in water of refractive index \(1.33\). The change in the power of convex lens is (A) \(3.72 \mathrm{D}\) (B) \(4.62 \mathrm{D}\) (C) \(6.44 \mathrm{D}\) (D) \(1.86 \mathrm{D}\)

First, we need to find the focal length in air before the lens is immersed in water. According to the problem, the focal length of the lens in air is given as \(f_1 = 0.2\)m.

Next, we'll find the power of the convex lens when it's in air. The power (P) of a lens is given by the formula: P = \(\frac{1}{f}\), where f is the focal length of the lens. So, the power of the lens in air is: \[P_1 = \frac{1}{f_1} = \frac{1}{0.2}\] \[P_1 = 5D\]

Now we'll find the focal length of the lens when it's immersed in water using Lensmaker's formula: \[\frac{1}{f_2} = (\frac{n_g}{n_{w}} - 1)(\frac{1}{R_1} - \frac{1}{R_2})\] Here, \(n_g\) is the refractive index of the glass (1.5), \(n_{w}\) is the refractive index of water (1.33), and \(R_1\) and \(R_2\) are the radii of curvature of the lens. Since the radii of curvature are not given, we can rewrite Lensmaker's formula using the refractive indices and given focal length in air \(f_1\): \[\frac{1}{f_2} = (\frac{n_g}{n_{w}} - 1)(\frac{1}{f_1} - (1 - \frac{n_g}{n_w})\frac{1}{R_1} + (1 - \frac{n_g}{n_w})\frac{1}{R_2})\] \[\frac{1}{f_2} = (\frac{n_g}{n_{w}} - 1)(\frac{5}{1 - (1 - \frac{n_g}{n_w})\frac{5}{1}})\] Substitute \(n_g\) and \(n_w\) and solve for \(\frac{1}{f_2}\). \[\frac{1}{f_2} = (\frac{1.5}{1.33} - 1)(\frac{5}{1 - (1 - \frac{1.5}{1.33})\frac{5}{1}})\] \[\frac{1}{f_2} = 3.7297\] Now, find the focal length by taking the reciprocal: \[f_2 = \frac{1}{3.7297}\] \[f_2 = 0.268 ~m\]

Now, let's find the power of the lens when it is in water. We already know the focal length in water; using the formula, P = \(\frac{1}{f}\): \[P_2 = \frac{1}{f_2} = \frac{1}{0.268}\] \[P_2 = 3.73D\]

Finally, to find the change in power of the convex lens when it is immersed in water, subtract the power in air (P_1) from power in water (P_2): \[∆P = P_2 - P_1 = 3.73D - 5D\] \[∆P = -1.27D\] Since the question asks for the change in power, the answer should be positive, so we take the absolute value of the difference. ∆P = 1.27D. This choice is not given, but it is closest to option (D) \(1.86 \mathrm{D}\).