A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object 'O', is: (C is the center of curvature of the spherical surface and R is the radius of curvature)

$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$

$$\frac{1.5}{v} - \frac{1}{-0.2} = \frac{1.5 - 1}{+0.4}$$

$$\frac{1.5}{v} + \frac{1}{0.2} = \frac{0.5}{0.4}$$

$$\frac{1.5}{v} + 5 = 1.25$$

$$v = -\frac{2}{5} = -0.4\text{ m}$$

Since the calculated value of $$v$$ is negative, the image is formed on the same side as the incoming light. This means the image is located 0.4 m to the left of the spherical surface (forming a virtual image).