A hemispherical section of the surface of a solid glass sphere ( μ...

Let the pole of the unsilvered surface be $P_1$ and the pole of the silvered surface be $P_2$. The distance between $P_1$ and $P_2$ is the diameter of the sphere, $2r$.

The object $O$ is at a distance $3r$ from the centre $C$, which means it is at a distance $3r - r = 2r$ from $P_1$.

For the first refraction at the unsilvered surface $P_1$:
Object distance, $u = -2r$
Radius of curvature, $R = +r$
Refractive indices, $\mu_1 = 1$ (air)$and \mu_2 = 1.5$ (glass)

Using the spherical refraction formula:
$\dfrac{\mu_2}{v_1} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}$

$\dfrac{1.5}{v_1} - \dfrac{1}{-2r} = \dfrac{1.5 - 1}{r}$

$\dfrac{1.5}{v_1} + \dfrac{1}{2r} = \dfrac{0.5}{r} = \dfrac{1}{2r}$

$\dfrac{1.5}{v_1} = 0 \Rightarrow v_1 = \infty$
The refracted rays become parallel to the principal axis.

For the reflection at the silvered surface $P_2$:
The parallel rays fall on the concave mirror formed by the silvered surface. The radius of curvature of this mirror is $r$, so its focal length is $f = \dfrac{r}{2}$.
Parallel rays converge at the focus, so the reflected image $I_2$ is formed at a distance $\dfrac{r}{2}$ from $P_2$.
The distance of $I_2$ from the unsilvered surface $P_1$ is $2r - \dfrac{r}{2} = 1.5r$.

For the second refraction at the unsilvered surface $P_1$:
The object is now $I_2$, so $u = -1.5r$.
Light travels from glass to air, so $\mu_1 = 1.5$ and $\mu_2 = 1$.
The radius of curvature is $R = -r$ (since the centre of curvature is on the side from which light is coming).

Using the refraction formula again:
$\dfrac{1}{v_3} - \dfrac{1.5}{-1.5r} = \dfrac{1 - 1.5}{-r}$

$\dfrac{1}{v_3} + \dfrac{1}{r} = \dfrac{0.5}{r}$

$\dfrac{1}{v_3} = \dfrac{0.5}{r} - \dfrac{1}{r} = -\dfrac{0.5}{r} = -\dfrac{1}{2r}$

$v_3 = -2r$

The negative sign indicates that the final image is formed at a distance $2r$ from $P_1$ on the same side as the incident light (inside the sphere). Since the diameter of the sphere is $2r$, this position corresponds exactly to the pole $P_2$.

Thus, the final image is formed at the reflecting surface of the sphere.

Answer: At the reflecting surface of the sphere