Region I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of the image from the surface is:

$$\mu_1 = 1.25,\ \mu_2 = 1.4,\ u = -40\ \text{cm},\ R = -25\ \text{cm}$$

$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$

$$\frac{1.4}{v} - \frac{1.25}{-40} = \frac{1.4 - 1.25}{-25}$$ $$\implies \frac{1.4}{v} + \frac{1.25}{40} = \frac{0.15}{-25}$$

$$\implies \frac{1.4}{v} = -\frac{0.15}{25} - \frac{1.25}{40} = -\frac{3}{500} - \frac{1}{32}$$

$$\implies \frac{1.4}{v} = -\frac{24 + 125}{4000} = -\frac{149}{4000}$$ $$\implies v = -\frac{1.4 \times 4000}{149} = -\frac{5600}{149} \approx -37.58\ \text{cm}$$

$$|v| = 37.58\ \text{cm}$$