To determine the image distance when a spherical surface separates two media with refractive indices of 1 and 1.5:

Use the lens maker's formula for a spherical surface:

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

Where:

$\mu_1$ = 1 (refractive index of the first medium)

$\mu_2$ = 1.5 (refractive index of the second medium)

$u$ = -0.2 m (object distance, negative as per convention)

$R$ = 0.4 m (radius of curvature)

Substituting the known values:

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

Simplify the equation:

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

Calculate:

$ \frac{1.5}{v} = -\frac{1.5}{0.4} $

Solve for $v$:

$ v = -0.4 \, \text{m} $

The negative sign indicates that the image is 0.4 meters to the left of the spherical surface.