In the problem, you're dealing with refraction at a convex surface. The light is coming from a point source located in air (with a refractive index, $$ \mu_1 = 1.0 $$) and entering a medium with a refractive index of $$ \mu_2 = 1.5 $$. The convex surface has a radius of curvature $$ R = 20 \, \text{cm} $$, and the source is placed $$ 100 \, \text{cm} $$ from the convex surface.

The formula used to find the image distance $$ v $$ due to refraction at a spherical surface is:

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

Substituting the given values:

$$ \frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20} $$

This equation allows us to solve for $$ v $$, the distance from the convex surface to the image. Upon solving, we found that $$ v = 100 \, \text{cm} $$, which means the image forms $$ 100 \, \text{cm} $$ on the other side of the convex surface, away from the point of refraction.

The distance from the object to the image isn't just $$ v $$, the distance from the surface to where the image forms. Since the object is $$ 100 \, \text{cm} $$ from the convex surface and the image also forms $$ 100 \, \text{cm} $$ from the convex surface but on the opposite side, the total distance between the object and the image is the sum of these distances:

• Distance from the object to the convex surface: $$ 100 \, \text{cm} $$
• Distance from the convex surface to the image: $$ 100 \, \text{cm} $$

Therefore, the total distance between the object and the image is $$ 100 \, \text{cm} + 100 \, \text{cm} = 200 \, \text{cm} $$.

This calculation accounts for the physical layout where the object and the image are on opposite sides of the convex surface, and to determine the distance between them, you sum the distances from each to the surface. This results in the image being formed $$ 200 \, \text{cm} $$ from the object, which means if you were to measure directly from the object to its image, the total distance covered would be $$ 200 \, \text{cm} $$.